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It contains solutions of the solutions of Applied Mathematics II All the previous years papers from CSVTU since 2006 onwards
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CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Unit I – Complex Numbers
(May-Jun-2006)
1. Prove that
nninn
ii n
2sin
2cos
cossin1cossin1
.
Sol: We have
1 sin cos1 sin cos
nii
2
2
1 cos sin2 2
1 cos sin2 2
2cos 2sin cos4 2 4 2 4 2
2cos 2sin cos4 2 4 2 4 2
2cos co4 2
n
n
i
i
i
i
s sin4 2 4 2
2cos cos sin4 2 4 2 4 2
cos sin4 2 4 2
cos sin4 2 4 2
cos4 2
n
n
i
i
i
i
sin cos sin4 2 4 2 4 2
cos sin cos sin4 2 4 2 4 2 4 2
cos sin4 2 4 2
n n
i i
n n n n n n n ni i
n n n ni
2
cos sin2 2
n nn i n
2. If , prove that . )sin()tan( ivuiyx v
uy
xtanhtan
2sinh2sin
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Ans: We have
sin 2 sinh 2tancos 2 cosh 2 cos 2 cosh 2
sin sin cosh cos sin
x i yx iyx y x y
u iv u v i u hv
Equating real and imaginary part, we get
sin 2 sin coshcos 2 cosh 2
sinh 2 cos sincos 2 cosh 2
x u vx y
y u hvx y
Dividing both the equation we get sin 2 sin cosh
sinh 2 cos sinx u vy u hv
sin 2 cot sinh 2 cothsin 2 tan
sinh 2 tan
x u y vx uy hv
Find sum of the series Ans: Let
2 3
1 1cos cos 2 cos32 31 1sin sin 2 sin 32 3
1 1 1 1cos cos 2 cos3 sin sin 2 sin 32 3 2 3
1 12 3
log 1
log 1 cos
i i i
i
c
s
c is
c is e e e
c is e
c is
2 2 1
sin1 sinlog 1 cos sin tan2 1 cos
i
c is i
...................3cos312cos
21cos
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 2 1
2
1
2 1
2sin cos1 2 2log 1 cos 2cos sin tan2 2cos
2
sin1 2log 2 1 2cos tan2 cos
21 log 2.2cos tan tan2 2 2
log cos2 2
log cos2
2
c is i
c is i
c is i
c is i
c
s
Equating real and imaginary parts we get
log cos2
c and
2s
(Nov-Dec-2006)
3. If ........iii A iB
prove that 2 2tan ,2
BA B and A B eA
.
Ans: ........iii A iB
A iBi A iB
log log( )A iB i A iB
1 2 2 1log(1) tan log tan BA iB i A B iA
2 2 1log tan2
BA iB i A B iA
2 2 1log tan2 2A B Bi A B i
A
1 2 2tan , log2 2A B B A B
A
2 21tan , log2 2 2A B B A B
A
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 2tan , log2A B B A B
A
2 2tan ,2
BA B A B eA
(Proved)
6. Show that
2 2 3 4 1 21 1 1sin sin 2 sin sin 3 sin sin 4 sin ....... tan (sin / (1 cos sin ))2 3 4
Ans: Let 2 2 3 41 1 1sin sin 2 sin sin 3 sin sin 4 sin .......2 3 4
S
2 3 41 1 1cos sin cos 2 sin cos3 sin cos 4 sin .......2 3 4
C
2 3sin sinsin (cos sin ) (cos 2 sin 2 ) (cos3 sin 3 ) ....
2 3C iS i i i
2 3
2 3sin sinsin ....2 3
i i iC iS e e e
2 3(sin ) (sin )sin ....2 3
i ii e eC iS e
log(1 sin )iC iS e
2log(1 sin cos sin )C iS i
log (cos sin )C iS r i -------------(1)
Where 2 4(1 sin cos ) sinr and 2
1 sintan1 sin cos
So, equation (1) becomes log iC iS re
logC iS r i
2
2 4 1 sinlog (1 sin cos ) sin tan1 sin cos
C iS i
2
1 sintan1 sin cos
S
2
2 2 3 4 11 1 1 sinsin sin 2 sin sin 3 sin sin 4 sin ....... tan2 3 4 1 sin cos
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
7. If cos( ) ii re prove that: 1 sin( )log2 sin( )
.
Ans: cos( ) ii re
cos cosh sin sinh cos sini r ir
cos cosh cos , sin sinh sinr r
cos sincosh , sinhcos sin
r r
coscosh cos
sinsinhsin
r
r
cosh cos sinsinh sin cos
cos sinsin cos
e ee e
cos sin sin coscos sin sin cos
e e e ee e e e
2 sin( )2 sin( )
ee
2 sin( )
sin( )e
sin( )2 logsin( )
1 sin( )log2 sin( )
(Proved)
(May-Jun 2007)
8. If 12cos xx
and 12cos yy
, show that one of the value of 1m nm nx y
x y is
2 cos( )m n .
Ans: 12cos xx
22 cos 1x x 2 2 cos 1 0x x
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
22cos 4cos 4
2x
2cos 2 sin cos sin
2ix i
----------------(1)
12cos yy
22 cos 1y y
2 2 cos 1 0y y
22cos 4cos 4
2y
2cos 2 sin cos sin
2iy i
----------------(2)
Taking +ve sign we get
cos sin , cos sinm mx m i m x m i m and
cos sin , cos sinn ny n i n y n i n
cos sin cos sinm nx y m i m n i n
cos sinm nx y m n i m n and
1 cos sinm nm nx y m n i m n
x y
Now, 1m n m n m n
m nx y x y x yx y
1 cos sin cos sinm nm nx y m n i m n m n i m n
x y
1 2cosm nm nx y m n
x y (Proved)
9. If tan( ) ii e prove that:
i. 12 2
n
.
ii. 1 log tan2 4 2
Ans: tan( ) ii e
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
tan( ) cos sini i ------------(1)
So, tan( ) cos sini i ----------------(2)
(i) Now, tan 2 tan i i
tan tantan 2
1 tan tani i
i i
cos sin cos sintan 2
1 cos sin cos sini i
i i
2 costan 2 tan
1 1 2
22
n
1
2 2n
(ii) tan 2 tani i i
tan tantan 2
1 tan tani i
ii i
cos sin cos sintan 2
1 cos sin cos sini ii
i i
tanh 2 sini i tanh 2 sin
2 2
2 2
sin1
e ee e
2 2 2 2
2 2 2 2
1 sin1 sin
e e e ee e e e
24
2
cos( / 2) sin( / 2)cos( / 2) sin( / 2)
e
2 cos( / 2) sin( / 2)cos( / 2) sin( / 2)
e
2 1 tan( / 2)
1 tan( / 2)e
22
n 1
2 2n
(Proved)
10. Find the sum of the series2 3
sin sin 2 sin 3 .......2 3x xx .
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Ans: Let 2 3
sin sin 2 sin 3 .......2 3x xS x
2 3
cos cos 2 cos3 .......2 3x xC x
2 3
(cos sin ) (cos 2 sin 2 ) (cos3 sin 3 ) .......2 3x xC iS x i i i
2 3
3 .......2 3
i i ix xC iS xe e e
2 3( ) ( ) .......
2 3
i ii xe xeC iS xe
log(1 )iC iS xe log(1 cos sin )C iS x ix
log (cos sin )C iS r i -------------(1)
Where 2 2 2(1 cos ) sinr x x 1 sintan1 cos
xx
21 2 cosr x x
So, equation (1) becomes log iC iS re logC iS r i
2 2 2 1 sin(1 cos ) sin tan1 cos
xC iS x x ix
1 sintan1 cos
xSx
2 3
1 sinsin sin 2 sin 3 ....... tan2 3 1 cosx x xx
x
(Ans)
(Nov-Dec 2007)
11. State De’Moivre’s Theorem. Ans: De’ Moivre’s Theorem: - For any real value of n, (cos sin ) cos sinni n i n .
12. Find sum of the series
Ans: Let 2 2 3 41 1 1sin sin 2 sin sin 3 sin sin 4 sin .......2 3 4
S
2 3 41 1 1cos sin cos 2 sin cos3 sin cos 4 sin .......2 3 4
C
........sin4sin41sin3sin
31sin2sin
21sin 4322
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 3sin sinsin (cos sin ) (cos 2 sin 2 ) (cos3 sin 3 ) ....
2 3C iS i i i
2 3
2 3sin sinsin ....2 3
i i iC iS e e e
2 3(sin ) (sin )sin ....2 3
i ii e eC iS e
log(1 sin )iC iS e
2log(1 sin cos sin )C iS i
log (cos sin )C iS r i -------------(1)
Where 2 4(1 sin cos ) sinr and 2
1 sintan1 sin cos
So, equation (1) becomes log iC iS re
logC iS r i
2
2 4 1 sinlog (1 sin cos ) sin tan1 sin cos
C iS i
2
1 sintan1 sin cos
S
2
2 2 3 4 11 1 1 sinsin sin 2 sin sin 3 sin sin 4 sin ....... tan2 3 4 1 sin cos
13. If 0coscoscossinsinsin , prove that
2 2 2 2 2 2cos cos cos sin sin sin . and cos2 cos2 cos2 sin2 sin2 sin2 .
Ans: - Let cos sincos sincos sin
x iy iz i
From the given information we will have 0sincossincossincos0 iiizyx …………1
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
0)sinsin(sin)coscos(cos i
0sinsinsin,0coscoscos
Now, sincos
1sincos
1sincos
1111iiizyx
sincossincossincos111 iiizyx
00)sinsin(sin)coscos(cos111 iizyx
0111
zyx ………………………………………………2
22 2 2 2( )x y z x y z xy yz zx
20 2( )xy yz zx ………….from …1
1 1 12 0xyzx y z
(from 2)………………………….3
2 2 22 2 2 cos sin cos sin cos sinx y z i i i
By applying the De’ Moivre’s Theorem we will have
2 2 22 2 2 cos sin cos sin cos sinx y z i i i
= (cos 2 sin 2 ) cos 2 sin 2 cos 2 sin 2i i i
From 3 we will have
(cos 2 sin 2 ) cos 2 sin 2 cos 2 sin 2i i i =0
Therefore cos 2 cos 2 cos 2 sin 2 sin 2 sin 2 =0
cos 2 cos 2 cos 2 = 2 2 2 2 2 2cos sin cos sin cos sin =0
Therefore 2 2 2 2 2 2cos cos cos sin sin sin =0….Hence Proved
14. If , show that and .
Ans: tan( ) ii e
tan( ) cos sini i ------------(1)
So, tan( ) cos sini i ----------------(2)
(i) Now, tan 2 tan i i
iei )tan(22
1
n
24tanlog
21
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
tan tantan 2
1 tan tani i
i i
cos sin cos sintan 2
1 cos sin cos sini i
i i
2 costan 2 tan
1 1 2
22
n
1
2 2n
(ii) tan 2 tani i i
tan tantan 2
1 tan tani i
ii i
cos sin cos sintan 2
1 cos sin cos sini ii
i i
tanh 2 sini i tanh 2 sin
2 2
2 2
sin1
e ee e
2 2 2 2
2 2 2 2
1 sin1 sin
e e e ee e e e
24
2
cos( / 2) sin( / 2)cos( / 2) sin( / 2)
e
2 cos( / 2) sin( / 2)cos( / 2) sin( / 2)
e
2 1 tan( / 2)
1 tan( / 2)e
22
n 1
2 2n
(Proved)
(May-Jun 2008)
15. State De’Moivre’s Theorem. Ans: De’ Moivre’s Theorem: - For any real value of n, (cos sin ) cos sinni n i n .
16. Find sum of the series ........sin4sin41sin3sin
31sin2sin
21sin 4322
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Ans: Let 2 2 3 41 1 1sin sin 2 sin sin 3 sin sin 4 sin .......2 3 4
S
2 3 41 1 1cos sin cos 2 sin cos3 sin cos 4 sin .......2 3 4
C
2 3sin sinsin (cos sin ) (cos 2 sin 2 ) (cos3 sin 3 ) ....
2 3C iS i i i
2 3
2 3sin sinsin ....2 3
i i iC iS e e e
2 3(sin ) (sin )sin ....2 3
i ii e eC iS e
log(1 sin )iC iS e
2log(1 sin cos sin )C iS i
log (cos sin )C iS r i -------------(1)
Where 2 4(1 sin cos ) sinr and 2
1 sintan1 sin cos
So, equation (1) becomes log iC iS re
logC iS r i
2
2 4 1 sinlog (1 sin cos ) sin tan1 sin cos
C iS i
2
1 sintan1 sin cos
S
2
2 2 3 4 11 1 1 sinsin sin 2 sin sin 3 sin sin 4 sin ....... tan2 3 4 1 sin cos
17. If 0coscoscossinsinsin , prove that
2 2 2 2 2 2cos cos cos sin sin sin . and cos2 cos2 cos2 sin2 sin2 sin2 .
Ans: - Let
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
cos sincos sincos sin
x iy iz i
From the given information we will have 0sincossincossincos0 iiizyx …………1
0)sinsin(sin)coscos(cos i
0sinsinsin,0coscoscos
Now, sincos
1sincos
1sincos
1111iiizyx
sincossincossincos111 iiizyx
00)sinsin(sin)coscos(cos111 iizyx
0111
zyx ………………………………………………2
22 2 2 2( )x y z x y z xy yz zx
20 2( )xy yz zx ………….from …1
1 1 12 0xyzx y z
(from 2)………………………….3
2 2 22 2 2 cos sin cos sin cos sinx y z i i i
By applying the De’ Moivre’s Theorem we will have
2 2 22 2 2 cos sin cos sin cos sinx y z i i i
= (cos 2 sin 2 ) cos 2 sin 2 cos 2 sin 2i i i
From 3 we will have
(cos 2 sin 2 ) cos 2 sin 2 cos 2 sin 2i i i =0
Therefore cos 2 cos 2 cos 2 sin 2 sin 2 sin 2 =0
cos 2 cos 2 cos 2 = 2 2 2 2 2 2cos sin cos sin cos sin =0
Therefore 2 2 2 2 2 2cos cos cos sin sin sin =0….Hence Proved
18. Find the sum of the series . 2 3
sin sin 2 sin 3 .......2 3x xx
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Ans: Let 2 3
2 3
2 3 2 3
2 2 3 3
1 1cos cos 2 cos32 31 1sin sin 2 sin 32 3
1 1 1 1cos cos 2 cos3 sin sin 2 sin 32 3 2 3
1 12 3
i i i
c x x x
s x x x
c is x x x x x
c is xe x e x e
c
2 2 2 1
2 1
log 1
log 1 cos sin1 sinlog 1 cos sin tan2 1 cos1 sinlog 1 2 cos tan2 1 cos
iis xe
c is x ixxc is x x i
xxc is x x i
x
Equating real and imaginary parts we get
2
1
1 log 1 2 cos2
sintan1 cos
c x x
xsx
(Nov - Dec 2008)
19. If 12cos xx
, find the value of 푥 .
Ans:
2
2
12cos
2 cos 1 0
2cos 4 4
cos sin
xx
x x
csox
x i
By De-moivers theorem
cos sin
cos sin
pp
p
x i
x p i p
20. If 2r rx Cis
, show that 1 2lim . ...... 1.
nn
x x x .
Ans: - Given that 2r rx Cis
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Now, 1 2 1 2lim . ...... lim . ........2 2 2n nn n
x x x Cis Cis Cis
1 2 1 2lim . ...... lim .......2 2 2n nn n
x x x Cis
1 2 1 2 3lim . ...... .......2 2 2nn
x x x Cis
1 21 1lim . ...... ( ) cos sin 1 012 1
2
nnx x x Cis Cis i i
1 2lim . ...... 1nnx x x
(Ans).
21. Separate 푡푎푛 (푥 + 푖푦) into real and imaginary parts.
Ans: Let 1
1
tan ( ) (1)tan ( ) (2)
i x iyi x iy
Adding (1) and (2) , we get
1 1
1
12 2
12 2
12 2
2 tan ( ) tan ( )( ) ( )2 tan1 ( )( )
22 tan1
22 tanh1
1 2tan2 1
x iy x iyx iy x iy
x iy x iyx
x y
xx y
xx y
Subtracting equation (2) from equation (1), we get 1 1
1
12 2
12 2
12 2
2 tan ( ) tan ( )( ) ( )2 tan
1 ( )( )22 tan
122 tanh
11 2tanh2 1
i x iy x iyx iy x iyi
x iy x iyyi i
x yyi i
x yy
x y
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Thus real part of 1tan ( )x iy = 12 2
1 2tan2 1
xx y
and imaginary part of 1tan ( )x iy =
12 2
1 2tanh2 1
yx y
.
22. Find sum of the series
Ans: - Let 퐶 = 1 − + ..푐표푠2휃 − . .
. .푐표푠3휃 + … … … … … … . .∞
Then 푆 = 0− + .
.푠푖푛2휃 − . .
. .푠푖푛3휃 + … … … … … … . .∞
Now 퐶 + 푖푠 = 1− (푐표푠휃 + 푖푠푖푛휃) + .
.(푐표푠2휃 + 푖푠푖푛2휃) − . .
. .(푐표푠3휃 + 푖푠푖푛3휃)+. .∞
⇒ 퐶 + 푖푠 = 1 − 푒 + ..푒 − . .
. .푒 +. .∞
⇒ 퐶 + 푖푠 = 1 + 푒 +−1
2 . − 12− 1
1.2푒 −
−12 . −1
2 − 1 . −12 − 2
1.2.3푒 +. .∞
⇒ 퐶 + 푖푠 = 1 + 푒 ⇒ 퐶 + 푖푠 = (1 + 푐표푠휃 + 푖푠푖푛휃)
⇒ 퐶 + 푖푠 = 2푐표푠 + 푖2푠푖푛 푐표푠 = 2푐표푠 푐표푠 + 푖푠푖푛
⇒ 퐶 + 푖푠 = 2푐표푠 푐표푠 − 푖푠푖푛 = 푐표푠 2푐표푠 − 푖푠푖푛 2푐표푠
So, ⇒ 퐶 = 푐표푠 2푐표푠
⇒ 1 − + ..푐표푠2휃 − . .
. .푐표푠3휃 + … … . .∞ = 푐표푠 2푐표푠 (Ans).
(May-Jun 2009)
23. State De’Moivre’s Theorem. Ans: De’ Moivre’s Theorem: - For any real value of n, (cos sin ) cos sinni n i n .
24. Find all the roots of the equation: (i) cos 2z . Ans: cos 2z
22 4 4 1 02
iz iziz iz iz ize e e e e e
4 16 4 4 2 3 2 3
2 2ize
2 3 2 log 2 3iz Log n i
2 log 2 3z n i
...................6cos6.4.25.3.14cos
4.23.12cos
211
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 log 2 3z n i (Ans)
(ii) tanh 2z . Ans: tanh 2z
2 2 2z z
z z z zz z
e e e e e ee e
3 0z ze e 2 3ze 2 ( 3) (3) ( 1) 2 log 3z Log Log Log n i i 2 log 3 (2 1)z n i
1 1log32 2
z n i
(Ans)
25. Separate 1sin (cos sin )i into real and imaginary parts, where is a positive acute angle.
Ans: Let 1sin (cos sin )i x iy
(cos sin ) sin( )i x iy (cos sin ) sin cosh cos sinhi x y i x y
cos sin cosh ..........(1) sin cos sinh ..........(2)x y x y By squaring and adding (1) and (2) we get 2 2 2 2sin cosh cos sinh 1x y x y
2 2 2 2sin (1 sinh ) cos sinh 1x y x y
2 2 2 2 2sin sinh cos sinh 1 sinx y x y x
2 2 2 2sinh (sin cos ) cosy x x x
2 2cos sinhx y Now, from (2) we get
2 2 2 2 2 4sin cos sinh cos cos cosx y x x x
2cos sinx as is positive acute angle.
cos sinx
1cos sinx ………….(3)
By using (2) and (3) we get
sin sin sinh y
sinh sin sin2
y ye ey
2 2 sin 1y ye e
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 sin 4sin 4 2 sin 2 1 sin sin 1 sin
2 2ye
log sin 1 siny ………….(4)
So, 1 1sin (cos sin ) cos sin log sin 1 sini i (Ans)
26. Find the sum of the series ( 1) ( 1)( 2)sin sin 2 sin 3 .......1.2 1.2.3
n n n n nn .
Ans: Let ( 1) ( 1)( 2)sin sin 2 sin 3 .......1.2 1.2.3
n n n n nS n and
( 1) ( 1)( 2)cos cos 2 cos3 .......1.2 1.2.3
n n n n nC n
Then,
( 1) ( 1)( 2)(cos sin ) (cos 2 sin 2 ) (cos3 sin 3 ) ...1.2 1.2.3
n n n n nC iS n i i i
2 3( 1) ( 1)( 2) ...1.2 1.2.3
i in n n n nC iS ne e e
2 3( 1) ( 1)( 2)1 ... 11.2 1.2.3
i in n n n nC iS ne e e
1 1niC iS e
1 11
niC iS
e
1 11 cos sin nC iS
i
2
2 2 2
1 12sin 2sin cos
nC iSi
2 2 2
1 12 sin sin cos nn n
C iSi
2 2 2 2 2
1 12 sin cos sin
nn nC iS
i
2 2 2 2 2
1 1 12 sin cos sinn nC iS
n i n
2 2 2 2
2
cos sin1
2 sinn n
n i nC iS
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 2 2 2
2 2
cos sin1
2 sin 2 sinn n n n
n nC iS i
2 2
2
sin2 sinn n
nS
2 2
2
sin( 1) ( 1)( 2)sin sin 2 sin 3 .......1.2 1.2.3 2 sinn n
nn n n n nn
(Ans)
(Nov-Dec 2009)
27. If 2r rx Cis
. Find the value of 1 2lim . ...... nn
x x x
.
Ans: - Given that 2r rx Cis
Now, 1 2 1 2lim . ...... lim . ........2 2 2n nn n
x x x Cis Cis Cis
1 2 1 2lim . ...... lim .......2 2 2n nn n
x x x Cis
1 2 1 2 3lim . ...... .......2 2 2nn
x x x Cis
1 21 1lim . ...... ( ) cos sin 1 012 1
2
nnx x x Cis Cis i i
1 2lim . ...... 1nnx x x
(Ans).
28. If ........iii A iB
prove that 2 2tan ,2
BA B and A B eA
.
Ans: -........iii A iB
A iBi A iB log log( )A iB i A iB
1 2 2 1log(1) tan log tan BA iB i A B iA
2 2 1log tan2
BA iB i A B iA
2 2 1log tan2 2A B Bi A B i
A
1 2 2tan , log2 2A B B A B
A
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 21tan , log2 2 2A B B A B
A
2 2tan , log2A B B A B
A
2 2tan ,2
BA B A B eA
(Proved)
OR
29. Find the sum to infinite of the series: 1 1.3 1.3.51 cos cos 2 cos3 .......2 2.4 2.4.6
Ans: - Let 퐶 = 1 − + ..푐표푠2휃 − . .
. .푐표푠3휃 + … … … … … … . .∞
Then 푆 = 0− + .
.푠푖푛2휃 − . .
. .푠푖푛3휃 + … … … … … … . .∞
Now 퐶 + 푖푠 = 1− (푐표푠휃 + 푖푠푖푛휃) + .
.(푐표푠2휃 + 푖푠푖푛2휃) − . .
. .(푐표푠3휃 + 푖푠푖푛3휃)+. .∞
⇒ 퐶 + 푖푠 = 1 − 푒 + ..푒 − . .
. .푒 +. .∞
⇒ 퐶 + 푖푠 = 1 + 푒 +−1
2 . − 12− 1
1.2푒 −
−12 . −1
2 − 1 . −12 − 2
1.2.3푒 +. .∞
⇒ 퐶 + 푖푠 = 1 + 푒 ⇒ 퐶 + 푖푠 = (1 + 푐표푠휃 + 푖푠푖푛휃)
⇒ 퐶 + 푖푠 = 2푐표푠 + 푖2푠푖푛 푐표푠 = 2푐표푠 푐표푠 + 푖푠푖푛
⇒ 퐶 + 푖푠 = 2푐표푠 푐표푠 − 푖푠푖푛 = 푐표푠 2푐표푠 − 푖푠푖푛 2푐표푠
So, ⇒ 퐶 = 푐표푠 2푐표푠
⇒ 1 − + ..푐표푠2휃 − . .
. .푐표푠3휃 + … … . .∞ = 푐표푠 2푐표푠 (Ans).
30. If log tan4 2
u
, proved that:
i. tanh tan2 2u
.
Ans: - 푢 = 푙표푔 푡푎푛 + ⇒ 푢 = 푙표푔 푡푎푛 + ⇒ 푒 = 푡푎푛 +
⇒/
/ = ⇒/ /
/ / = ⇒ 푡푎푛ℎ = 푡푎푛 (Proved)
ii. log tan4 2
iui
Ans: - From Ans (i)
⇒ 푡푎푛ℎ = 푡푎푛 ⇒ 푡푎푛 = 푡푎푛ℎ ⇒ = 푡푎푛ℎ 푡푎푛
CSVTU II SEMESTER SOLUTION ALL UNITS
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⇒ = 푙표푔 ⇒ 휃 = 푙표푔.
⇒ 휃 = −푖푙표푔 푡푎푛 + 푖 (Ans)
May –June (2010)
31. What is the imaginary part of )34( iLog .
Ans: - 2 2 1( ) log 2 tan yLog x iy x y i nx
Here, Imaginary part of
xyniyxLog 1tan2)(
So, Imaginary part of 1 3(4 3) 2 tan4
Log i n
32. If ...........)1( 2210 xPxPPx n show that,
(i)
4cos2........... 2/
420nPPP n .
Ans: - ...........)1( 2210 xPxPPx n
...........)1( 2210 xPxPPx n
...........2)1()1( 44
220 xPxPPxx nn
Putting ix , we get ...........2)1()1( 420 PPPii nn
...........24
sin4
cos24
sin4
cos2 4202/12/1
PPPii
nn
...........24
sin4
cos24
sin4
cos2 4202/2/
PPPninnin nn
...........24
cos22 4202/ PPPnn
4
cos2........... 2/420
nPPP n (Proved).
(ii)
4sin2........... 2/
531nPPP n
Ans: - ...........)1( 2210 xPxPPx n
...........)1( 2210 xPxPPx n
...........2)1()1( 55
331 xPxPxPxx nn
Putting ix , we get
CSVTU II SEMESTER SOLUTION ALL UNITS
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...........2)1()1( 531 iPiPiPii nn
...........24
sin4
cos24
sin4
cos2 5312/12/1
PPPiii
nn
...........24
sin4
cos24
sin4
cos2 5312/2/
PPPininnin nn
...........24
sin22 5312/ PPPinin
4
sin2........... 2/531
nPPP n (Proved).
33. If iivu )(sin 1 , prove that 2sin and 2cosh are the roots of the equation 0)1( 2222 uvuxx
Ans: - To prove 2sin and 2cosh are the roots of the equation
0)1( 2222 uvuxx
is same as to prove 2 2 2 2 2 2 2sin cosh (1 ), sin .coshu v u
1sin ( ) sin( )u iv i u iv i sin .cosh cos .sinhu iv i -------------------(1) From(1) modulus square is
2 2 2 2 2 2sin .cosh cos .sinhu v
2 2 2 2 2 2sin .cosh (1 sin ).(cosh 1)u v
2 2 2 2 2 2 2 2sin .cosh cosh 1 sin .cosh sinu v
2 2 2 2cosh 1 sinu v
2 2 2 2sin cosh (1 )u v (Proved -1) Form (1) sin .cosh , cos .sinhu v
2 2 2sin .coshu (proved – 2)
Hence 2sin and 2cosh are the roots of the equation 0)1( 2222 uvuxx .
34. Find sum of the series: ...............5sin5
3sin3
sin53
ccc .
Ans: - Let 3 5
sin sin 3 sin 5 ...............3 5c cS c
So, 3 5
cos cos3 cos5 ...............3 5c cC c
3 5
cos sin cos3 sin 3 cos5 cos5 ...............3 5c cC is c
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3 5
3 5 .......3 3
i i ic cC iS ce e e
3 5
.......3 3
i ii ce ce
C iS ce
As
xxxxxx
11log
21tanh.................
531
53
1 1 1log log 1 log 12 1 2
ii i
i
ceC iS ce cece
2 2 2 1
2 2 2 1
1 sin(1 cos ) sin tan2 1 cos
1 sin(1 cos ) sin tan2 1 cos
cC iS c c ic
cc c ic
2 2 2 2 2 2
1 1
1 (1 cos ) sin (1 cos ) sin21 sin sintan tan2 1 cos 1 cos
C iS c c c c
c cic c
1 11 sin sintan tan2 1 cos 1 cos
c cSc c
1 11 sin sintan tan2 1 cos 1 cos
c cSc c
1 12
sin sin1 1 2 sin1 cos 1 costan tansin sin2 2 11 .
1 cos 1 cos
c ccc cS c c c
c c
3 5
12
1 2 sinsin sin 3 sin 5 ............... tan3 5 2 1c c cc
c
(Ans)
Nov Dec (2010)
35. Find :
4
5
(cos sin )(sin cos )
ii
Sol 4
5
(cos sin )(sin cos )
ii
We know from De’ Moivre’s Theorem: - For any real value of n,(cos sin ) cos sinni n i n .
CSVTU II SEMESTER SOLUTION ALL UNITS
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4
55 5 5
(cos sin ) (cos 4 sin 4 ) (cos 4 sin 4 )(sin cos ) (sin cos ) cos sin
i i ii i i i
4
55 5
(cos sin ) (cos 4 sin 4 ) (cos 4 sin 4 )(sin cos ) (sin cos ) cos sin
i i ii i i i
54
5 5
(cos 4 sin 4 ) cos sin(cos sin ) (cos 4 sin 4 )(sin cos ) (sin cos )
i ii ii i i
54
5 5
(cos 4 sin 4 ) cos5 sin 5(cos sin ) (cos 4 sin 4 )(sin cos ) (sin cos )
i ii ii i i
4
5 5
(cos 4 5 sin 4 5(cos sin ) (cos 4 sin 4 )(sin cos ) (sin cos )
ii ii i i
4
5 5
(cos sin ) (cos 4 sin 4 ) (cos 4 5 sin 4 5(sin cos ) (sin cos )
i i i ii i
4
5 5
(cos sin ) (cos 4 sin 4 ) sin 4 5 (cos 4 5(sin cos ) (sin cos )
i i i ii i
36. If x = cos sini , y= cos sini , z= cos sini and x+ y +z =0, then prove that
1 1 1 0.x y z
Ans: - Let cos sincos sincos sin
x iy iz i
From the given information we will have 0sincossincossincos0 iiizyx …………1
0)sinsin(sin)coscos(cos i
0sinsinsin,0coscoscos
Now, sincos
1sincos
1sincos
1111iiizyx
sincossincossincos111 iiizyx
00)sinsin(sin)coscos(cos111 iizyx
0111
zyx ………………………………………………2 Hence Proved
CSVTU II SEMESTER SOLUTION ALL UNITS
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37. If ( ) p x iya ib m then prove that one of the value of y/x is 2 2
2 tan ( / )log ( )e
b aa b
.
Ans:
2 2 1
log( ) loglog( ) log
1 log tan log log2
p x iy
p x iy
a ib m
a ib mp a ib x iy m
bp a b i x m iy ma
Equating real and imaginary parts
2 2
1
1log log ...........................(1)2
log tan ............................(2)
x m p a b
by m pa
Dividing (2 )by (1)
1 1
2 22 2
tan 2 tan
1 loglog2
b bpy a ax a bp a b
38. Sum the series 2 3
sin sin( ) sin( 2 ) sin( 3 ).......2 3
x xx
Ans: let
2
2
2
cos cos cos 2 ........................2!
sin sin sin 2 ........................2!
cos sin cos sin cos 2 sin 2 .........................2!
xc x
xs x
xc is i x i i
i xic is e xe
2
2
cos sin
sincos
cos
2........................................
2!21 ........................
2!
cos sin sin sin
x i
i xx
x
ie
xi iic is e xe e
ic is e e
c is e ec is e x i x
equating real and imaginary parts
CSVTU II SEMESTER SOLUTION ALL UNITS
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cos
cos
cos sin
sin sin
x
x
c e x
s e x
May June (2011)
39. Find real and imagiant parts of exp(z2).
Ans: We have
22
2 2 2
2 2 2
2 2 2
2 2 2 2 2
exp exp
exp exp 2
exp exp .exp 2
exp exp . cos 2 sin 2
exp cos 2 exp sin 2 exp
z x iy
z x y ixy
z x y ixy
z x y xy i xy
z xy x y i xy x y
Hence real part is 2 2cos 2 expxy x y and imaginary part is 2 2sin 2 expxy x y .
40. If (1 + 푥) = 푃 + 푃 푥 + 푃 푥 + 푃 푥 + ⋯… … ..
Show that
(i) p0 - p2 + p4 - ---------------------- = 2n/2 cos n/4 , (iii) p1 – p3 + p5 – ---------------------- = 2n/2 sin n/4 .
(i)
4cos2........... 2/
420nPPP n .
Ans: - ...........)1( 2210 xPxPPx n
...........)1( 2210 xPxPPx n
...........2)1()1( 44
220 xPxPPxx nn
Putting ix , we get ...........2)1()1( 420 PPPii nn
...........24
sin4
cos24
sin4
cos2 4202/12/1
PPPii
nn
...........24
sin4
cos24
sin4
cos2 4202/2/
PPPninnin nn
CSVTU II SEMESTER SOLUTION ALL UNITS
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...........24
cos22 4202/ PPPnn
4
cos2........... 2/420
nPPP n (Proved).
(ii)
4sin2........... 2/
531nPPP n
Ans: - ...........)1( 2210 xPxPPx n
...........)1( 2210 xPxPPx n
...........2)1()1( 55
331 xPxPxPxx nn
Putting ix , we get ...........2)1()1( 531 iPiPiPii nn
...........24
sin4
cos24
sin4
cos2 5312/12/1
PPPiii
nn
...........24
sin4
cos24
sin4
cos2 5312/2/
PPPininnin nn
...........24
sin22 5312/ PPPinin
4
sin2........... 2/531
nPPP n (Proved).
41. If ........iii A iB
prove that 2 2tan ,2
BA B and A B eA
.
Ans: -........iii A iB
A iBi A iB log log( )A iB i A iB
1 2 2 1log(1) tan log tan BA iB i A B iA
2 2 1log tan2
BA iB i A B iA
2 2 1log tan2 2A B Bi A B i
A
1 2 2tan , log2 2A B B A B
A
2 21tan , log2 2 2A B B A B
A
2 2tan , log2A B B A B
A
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 2tan ,2
BA B A B eA
(Proved)
42. Find the sum to infinite of the series:
Cos + Cos ( ) + Cos ( 2 ) +….. to n terms.
Cos + Cos ( ) + Cos ( 2 ) +….. to n terms.
\
cos cos cos 2 ........................
sin sin sin 2 ........................
cos sin cos sin cos 2 sin 2 ............................
c nterms
s nterms
c is i i i nterms
iic is e e
2
........................................2
. . ................................
1
1
ie nterms
i ii i ic is e e e e e nterms
inie ec is ie
We know that
12
2 2 2 21 2 sin2
2 2 2 21 2 sin2
sin cos2 2
in
i n
i i i iie e e e e i
n n nin i i i ne e e e e i
nc is e ec
Equating real and imaginary parts we get
1cos sin cos
2 2 2
1sin sin cos
2 2 2
n nc ec
n ns ec
Nov Dec (2011)
43. State the De Moiver’s theorem.
CSVTU II SEMESTER SOLUTION ALL UNITS
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Ans: De’ Moivre’s Theorem: - For any real value of n, (cos sin ) cos sinni n i n
44. Prove that 1(1 cos sin ) (1 cos sin ) 2 cos cos2 2
n n n n ni i .
Ans:
2 2
(1 cos sin ) (1 cos sin )
2cos .2sin cos 2cos .2sin cos2 2 2 2 2 2
2 cos cos sin 2 cos cos sin2 2 2 2 2 2
2 cos cos sin 2 cos cos sin2 2 2 2 2 2
n n
n n
n nn n n n
n n n n
i i
i i
i i
n n n ni i
1
2 cos 2cos2 2
2 cos cos2 2
n n
n n
n
n
45. If , prove that .
Ans: We have
sin 2 sinh 2tancos 2 cosh 2 cos 2 cosh 2
sin sin cosh cos sin
x i yx iyx y x y
u iv u v i u hv
Equating real and imaginary part, we get
sin 2 sin coshcos 2 cosh 2
sinh 2 cos sincos 2 cosh 2
x u vx y
y u hvx y
Dividing both the equation we get sin 2 sin cosh
sinh 2 cos sinx u vy u hv
sin 2 cot sinh 2 cothsin 2 tan
sinh 2 tan
x u y vx uy hv
46. Find the sum to infinite of the series: 1 1.3 1.3.51 cos cos 2 cos3 .......2 2.4 2.4.6
Ans: - Let 퐶 = 1 − + ..푐표푠2휃 − . .
. .푐표푠3휃 + … … … … … … . .∞
)sin()tan( ivuiyx v
uy
xtanhtan
2sinh2sin
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Then 푆 = 0− + ..푠푖푛2휃 − . .
. .푠푖푛3휃 + … … … … … … . .∞
Now 퐶 + 푖푠 = 1− (푐표푠휃 + 푖푠푖푛휃) + .
.(푐표푠2휃 + 푖푠푖푛2휃) − . .
. .(푐표푠3휃 + 푖푠푖푛3휃)+. .∞
⇒ 퐶 + 푖푠 = 1 − 푒 + ..푒 − . .
. .푒 +. .∞
⇒ 퐶 + 푖푠 = 1 + 푒 +−1
2 . − 12− 1
1.2푒 −
−12 . −1
2 − 1 . −12 − 2
1.2.3푒 +. .∞
⇒ 퐶 + 푖푠 = 1 + 푒 ⇒ 퐶 + 푖푠 = (1 + 푐표푠휃 + 푖푠푖푛휃)
⇒ 퐶 + 푖푠 = 2푐표푠 + 푖2푠푖푛 푐표푠 = 2푐표푠 푐표푠 + 푖푠푖푛
⇒ 퐶 + 푖푠 = 2푐표푠 푐표푠 − 푖푠푖푛 = 푐표푠 2푐표푠 − 푖푠푖푛 2푐표푠
So, ⇒ 퐶 = 푐표푠 2푐표푠
⇒ 1 − + ..푐표푠2휃 − . .
. .푐표푠3휃 + … … . .∞ = 푐표푠 2푐표푠 (Ans).
May –June (2012)
47. Prove that : 4cos sin cos8 sin8
sin cosi ii
Sol. L.H.S.
=
4cos sinsin cos
ii
4
4
4
cos sin
s sin2 2
cos sin
s 4 sin 42 2
i
co i
i
co i
4
4
4
cos sins 2 4 sin 2 4
cos sin
s sin
ico i
ico i
CSVTU II SEMESTER SOLUTION ALL UNITS
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4 4
8
cos sin cos sin
cos sin
cos8 sin8
i i
i
i
48. If 12cos xx
and 12cos yy
show that : 1 2 cos( )m nm nx y m n
x y
Ans: 12cos xx
22 cos 1x x 2 2 cos 1 0x x
22cos 4cos 4
2x
2cos 2 sin cos sin
2ix i
----------------(1)
12cos yy
22 cos 1y y
2 2 cos 1 0y y
22cos 4cos 4
2y
2cos 2 sin cos sin
2iy i
----------------(2)
Taking +ve sign we get
cos sin , cos sinm mx m i m x m i m and
cos sin , cos sinn ny n i n y n i n
cos sin cos sinm nx y m i m n i n
cos sinm nx y m n i m n and
1 cos sinm nm nx y m n i m n
x y
Now, 1m n m n m n
m nx y x y x yx y
1 cos sin cos sinm nm nx y m n i m n m n i m n
x y
CSVTU II SEMESTER SOLUTION ALL UNITS
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1 2cosm nm nx y m n
x y (Proved)
49. If tan( ) ii e , show that : 12 2
n
and 1 log tan2 4 2
Ans: tan( ) ii e
tan( ) cos sini i ------------(1)
So, tan( ) cos sini i ----------------(2)
(i) Now, tan 2 tan i i
tan tantan 2
1 tan tani i
i i
cos sin cos sintan 2
1 cos sin cos sini i
i i
2 costan 2 tan
1 1 2
22
n
1
2 2n
(ii) tan 2 tani i i
tan tantan 2
1 tan tani i
ii i
cos sin cos sintan 2
1 cos sin cos sini ii
i i
tanh 2 sini i tanh 2 sin
2 2
2 2
sin1
e ee e
2 2 2 2
2 2 2 2
1 sin1 sin
e e e ee e e e
24
2
cos( / 2) sin( / 2)cos( / 2) sin( / 2)
e
2 cos( / 2) sin( / 2)cos( / 2) sin( / 2)
e
2 1 tan( / 2)
1 tan( / 2)e
CSVTU II SEMESTER SOLUTION ALL UNITS
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22
n 1
2 2n
(Proved)
50. Sum of the following series: 1 1cos cos 2 cos3 ..........2 3
Ans: Let
2 3
1 1cos cos 2 cos32 31 1sin sin 2 sin 32 3
1 1 1 1cos cos 2 cos3 sin sin 2 sin 32 3 2 3
1 12 3
log 1
log 1 cos
i i i
i
c
s
c is
c is e e e
c is e
c is
2 2 1
sin1 sinlog 1 cos sin tan2 1 cos
i
c is i
2 2 1
2
1
2 1
2sin cos1 2 2log 1 cos 2cos sin tan2 2cos
2
sin1 2log 2 1 2cos tan2 cos
21 log 2.2cos tan tan2 2 2
log cos2 2
log cos2
2
c is i
c is i
c is i
c is i
c
s
Dec –Jan (2012)
51. Define De Moiver’s theorem Ans: De’ Moivre’s Theorem: - For any real value of n, (cos sin ) cos sinni n i n
CSVTU II SEMESTER SOLUTION ALL UNITS
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52. If and are the roots of the equation 2 2sin sin 1 0.z z then prove that i. 2cos cosn n nn ec
ii. 2cosn n nec Ans: It is given those roots of the equation 2 2sin sin 1 0.z z Are and therefor by applying the property of relations between roos and
coeffecfint of the equation we will have
cosb eca
And 2cos ec ………………………..1
As 2cos ec therefore 2cosn n n nec ……………………2
cosb eca
; 2cos ec
2cos ec
From above 2cos cosec ec
2 2cos cos 0ec ec 2 2cos cos 4cos
2ec ec ec
1 3cos2iec
1 3 1 3cos & cos2 2i iec ec
cos cos sin & cos cos sinec r i ec r i
221
31 3 21 & tan 12 2 3
2
where r
1 3 1 3cos & cos2 2
cos cos sin & cos cos sinn nn n n n
i iec ec
ec i ec i
2cos cosn n nec
In the given problem instead of 2cos cosn n nn ec it should be
2cos cosn n nec
53. f ( ) p x iya ib m then prove that one of the value of y/x is 2 2
2 tan ( / )log ( )e
b aa b
Ans:
CSVTU II SEMESTER SOLUTION ALL UNITS
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log( ) log
p x iy
p x iy
a ib m
a ib m
2 2 1
log( ) log
1 log tan log log2
p a ib x iy m
bp a b i x m iy ma
Equating real and imaginary parts
2 2
1
1log log ...........................(1)2
log tan ............................(2)
x m p a b
by m pa
Dividing (2 )by (1)
1 1
2 22 2
tan 2 tan
1 loglog2
b bpy a ax a bp a b
54. Find sum of the series
Ans: Let 2 2 3 41 1 1sin sin 2 sin sin 3 sin sin 4 sin .......2 3 4
S
2 3 41 1 1cos sin cos 2 sin cos3 sin cos 4 sin .......2 3 4
C
2 3sin sinsin (cos sin ) (cos 2 sin 2 ) (cos3 sin 3 ) ....
2 3C iS i i i
2 3
2 3sin sinsin ....2 3
i i iC iS e e e
2 3(sin ) (sin )sin ....2 3
i ii e eC iS e
log(1 sin )iC iS e
2log(1 sin cos sin )C iS i
log (cos sin )C iS r i -------------(1)
Where 2 4(1 sin cos ) sinr and 2
1 sintan1 sin cos
So, equation (1) becomes log iC iS re
........sin4sin41sin3sin
31sin2sin
21sin 4322
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logC iS r i
2
2 4 1 sinlog (1 sin cos ) sin tan1 sin cos
C iS i
2
1 sintan1 sin cos
S
2
2 2 3 4 11 1 1 sinsin sin 2 sin sin 3 sin sin 4 sin ....... tan2 3 4 1 sin cos
Unit II – Differential Equations of higher order
April -May, 2006
1. Solve: 2
2 32 2 cos 2x xd y y x e e x
dx
Ans: 2 2 32 cos 2x xD y x e e x -----------(1)
Here Auxiliary equation is 2 2 0 2D D i
So, cos 2 sin 2hy A x B x
2 32
1. . cos 22
x xpP I y x e e x
D
3 22 2
1 1 cos 2( 3) 2 ( 1) 2
x xpy e x e x
D D
3
222
1 1 cos 211 2 36 1
11
xx
pey x e x
D DD D
3
22 2
1 1 cos 2611 4 2 31
11
xx
pey x e x
D D D
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 32 23 2
22
6 66 2 131 ... cos 211 11 121 1331 4 169
xx
p
D D D De D D Dy x e xD
3
2 22
6 1 36 2 131 ... cos 211 11 11 121 4 169
xx
pe D Dy D x e x
D
3
2 22
6 47 2 131 ... cos 211 11 121 4 4 169
xx
pe D Dy D x e x
3
2 6 47 2 132 2 0 0 ... cos 211 11 121 233
xx
pe Dy x x e x
3
2 12 94 4sin 2 13cos 211 11 121 233
x x
pe x ey x x x
So, Solution h py y y
3
2 12 94cos 2 sin 2 4sin 2 13cos 211 11 121 233
x xe x ey A x B x x x x
2. Solve: 3 2
3 23 2
12 2 10( )d y d yx x y xdx dx x
Ans: 3 2
3 23 2
12 2 10d y d yx x y xdx dx x
-------------(1)
Let logzx e z x
So, 2 3
2 32 3', '( ' 1) , '( ' 1)( ' 2)dy d y d yx D x D D y x D D D y
dx dx dx where
' dDdz
Putting all these values in (1) we get
'( ' 1)( ' 2) 2 '( ' 1) 2 10( )z zD D D D D y e e
3 2' ' 2 10( )z zD D y e e -------(2)
Its auxiliary equation is 3 2 2 0m m
2( 1)( 2 2) 0m m m
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 41,
2m
1, 1m i
So, . . cos sin )z zhC F y Ae e B z C z
. . cos(log ) sin(log )hAC F y x B x C xx
-------------(3)
Now, 3 2
1. . 10( )' ' 2
z zpP I y e e
D D
3 2 3 2
1 110 10' ' 2 ' ' 2
z zpy e e
D D D D
3 2 3 2
1 110 10 (1)1 1 2 ( ' 1) ( ' 1) 2
z zpy e e
D D
2
15 10 (1)'( ' 4 ' 5)
z zpy e e
D D D
1
21 4 15 10 1 ' ' (1)5 ' 5 5
z zpy e e D D
D
1 45 2 1 ' ..... (1)
' 5z z
py e e DD
15 2 (1)
'z z
py e eD
5 2z zpy e ze
2log5p
xy xx
So, Solution h py y y
2 logcos(log ) sin(log ) 5A xy x B x C x xx x
(Ans)
3. Solve the following simultaneous equation: 2 , 2 , 2dx dy dzy z xdt dt dt
(Ans.) The given equation is:
2dx ydt
………..(1), 2 ..........(2)dy zdt
,
2 ..........(3)dz xdt
Differentiating (1)w . r. “t.”, we get 2
2 2 2(2 )d x dy zdt dt
using(2)
CSVTU II SEMESTER SOLUTION ALL UNITS
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Differentiating again w. r. t. “t”,, we get 3
3 4 4(2 )d x dz xdt dt
3
3 2
21 2 3
21 2 3 2 3
21 2 3 3
21 2
( 8) 0,
8 0 ( 2)() 2 4) 0
2, 1 3
cos( 3 )12
1 2 cos( 3 ) 3 sin( 3 )2
2 2cos cos( 3 ) sin sin( 3 )3 3
cos 3
t t
t t t
t t
t t
dD x whereDdt
D or D D D
D i
x c e c e t cdxydt
c e c e t c c e t c
c e c e t c t c
c e c e t
323
c
From(2)….. 21 2 3 2 3
21 2 3
12
1 2 22 cos( 3 ) 3 sin( 3 )2 3 3
4cos( 3 )3
t t t
t t
dyzdt
c e c e t c c e t c
c e c e t c
Nov-Dec 2006
4. Solve the following : sin , cosdx dxy t x tdt dt
, Given that , 2, 0, , 0x y when t
Ans: Given simultaneous differential equation are
sindx y tdt
-------------(1)
cosdy x tdt
------------(2)
From equation (2) we get cosdyx tdt
---------(3)
Differentiating (3) with respect to t we get
2
2 sindx d y tdt dt
---------(4)
From (1) and (4) we get
2
2sin sind yt y tdt
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2
2 2sind y y tdt
2( 1) 2sinD y t -----------(5)
Its Auxiliary equation is 2 1 0 1,1m m
So, . . t thC F y Ae Be
And 2
1. . ( 2sin )1pP I y t
D
2
1( 2sin ) sin1 1py t t
So, h py y y
sint ty Ae Be t -------(6)
Putting the value of y in (2) we get
cos cost tAe Be t x t
t tx Ae Be ---------(7)
Given that 2, 0, 0x y when t
So, equation (6) and (7) becomes 0 A B and 2 A B
1, 1A B
So, solution t tx e e and sint ty e e t (Ans)
5. Solve the differential equation: 3 2
3 23 2
12 2 10( )d y d yx x y xdx dx x
Ans: 3 2
3 23 2
12 2 10d y d yx x y xdx dx x
-------------(1)
Let logzx e z x
So, 2 3
2 32 3', '( ' 1) , '( ' 1)( ' 2)dy d y d yx D x D D y x D D D y
dx dx dx where ' dD
dz
Putting all these values in (1) we get
'( ' 1)( ' 2) 2 '( ' 1) 2 10( )z zD D D D D y e e
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3 2' ' 2 10( )z zD D y e e -------(2)
Its auxiliary equation is 3 2 2 0m m
2( 1)( 2 2) 0m m m
2 41,
2m
1, 1m i
So, . . cos sin )z zhC F y Ae e B z C z
. . cos(log ) sin(log )hAC F y x B x C xx
-------------(3)
Now, 3 2
1. . 10( )' ' 2
z zpP I y e e
D D
3 2 3 2
1 110 10' ' 2 ' ' 2
z zpy e e
D D D D
3 2 3 2
1 110 10 (1)1 1 2 ( ' 1) ( ' 1) 2
z zpy e e
D D
2
15 10 (1)'( ' 4 ' 5)
z zpy e e
D D D
1
21 4 15 10 1 ' ' (1)5 ' 5 5
z zpy e e D D
D
1 45 2 1 ' ..... (1)
' 5z z
py e e DD
15 2 (1)
'z z
py e eD
5 2z zpy e ze
2log5p
xy xx
So, Solution h py y y
2 logcos(log ) sin(log ) 5A xy x B x C x xx x
(Ans)
6. Using method of variation of parameters: 2
2 4 tan 2d y y xdx
Ans: Homogeneous equation is 4 0y y
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Its characteristics equation is 2 4 0 2i Hence, Homogeneous solution is cos 2 sin 2hy A x B x
1 2cos 2 , sin 2y x y x
1 1 2 2
2 2
cos 2 sin 22cos 2 2sin 2 2
2sin 2 2cos 2y y x x
W x xx xy y
Wrdxyy
Wrdxyyy p
.. 12
21
sin 2 .tan 2 cos 2 tan 2cos 2 sin 2
2 2px xdx x xdxy x x
cos 2 sin 2(sec 2 cos 2 ) sin 2 .
2 2px xy x x dx x dx
cos 2 log(sec 2 tan 2 ) sin 2 sin 2 cos 2
2 2 2 2 2px x x x x xy
cos 2 .log(sec 2 tan 2 )
4px x xy
So, Solution cos 2 .log(sec 2 tan 2 )cos 2 sin 2
4h px x xy y y A x B x
(Ans)
April -May, 2007
7. Solve the differential equation2
2 2 sinxd y dy y xe xdx dx
.
Ans: 2
2 2 sinxd y dy y xe xdx dx
Its auxiliary equation is 2 22 1 0 ( 1) 0 1,1m m m m
So, . . x xhC F y Ae Bxe
Now, 2
1. . sin( 1)
xpP I y xe x
D
2 2
1 1sin sin( 1 1)
x xpy e x x e x x
D D
1 sinx
py e x xdxD
1 cos sinxpy e x x x
D
cos sinxpy e x x x dx
sin cos cosxpy e x x x x
sin 2cosxpy e x x x
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So, general solution h py y y
sin 2cosx x xy Ae Bxe e x x x (Ans)
8. Solve by method of variation of parameters of : - 2
2 sind y y x xdx
.
Ans: Homogeneous equation is 0y y
Its characteristics equation is 2 1 0 2i Hence, Homogeneous solution is cos sinhy A x B x
1 2cos 2 , sin 2y x y x
1 1 2 2
2 2
cos sincos sin 1
sin cosy y x x
W x xx xy y
Wrdxyy
Wrdxyyy p
.. 12
21
sin . sin cos . sincos sin
1 1px x xdx x x xdxy x x
2cos sin . sin sin cospy x x x dx x x x xdx
cos sin(1 cos 2 ) sin 2 .
2 2px xy x x dx x x dx
2cos sin 2 cos 2 sin cos 2 sin 2
2 2 2 4 2 2 4px x x x x x x x xy
2 cos cos sin 2 cos cos 2 sin cos 2 sin sin 2
4 4 8 4 8px x x x x x x x x x x xy
2 cos sin 2 sin cos 2cos cos cos 2 sin sin 2
4 4 8p
x x x x xx x x x x xy
2 cos sin cos
4 4 8px x x x xy
So, Solution 2 cos sin coscos sin
4 4 8h px x x x xy y y A x B x (Ans)
9. Solve the differential equation2
2 22 3 4 (1 )d y dyx x y x
dx dx .
Ans: 2
2 22 3 4 (1 )d y dyx x y x
dx dx ---------(1)
Let logzx e z x and 2
22' , '( ' 1)dy d yx D y x D D y
dx dx where ' dD
dz
CSVTU II SEMESTER SOLUTION ALL UNITS
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Putting all the values in (1) we get
2'( ' 1) 3 ' 4 (1 )zD D D y e 2 2' 4 ' 4 1 2 z zD D y e e -----------(2)
Its auxiliary equation is 2 24 4 0 ( 2) 0 2, 2m m m m
So, 2 2 2 2. . logz zhC F y Ae BZe Ax Bx x
Now, 22
1. . (1 2 )( ' 2)
z zpP I y e e
D
0 22 2 2
1 1 12( ' 2) ( ' 2) ( ' 2)
z z zpy e e e
D D D
22 2 2
1 1 1.1 2 1(0 2) (1 2) ( ' 2 2)
z zpy e e
D
22
1 12 14 '
z zpy e e
D 21 2 1
4z z
py e e dzdz
2 2 2 21 1 (log )2 2
4 2 4 2
zz
pz e x xy e x
So, general solution is 2 2
2 2 1 (log )log 24 2h p
x xy y y Ax Bx x x (Ans).
Nov-Dec 2007
10. Solve: 2
2 3 4 0d y dy ydx dx
Sol. It’s symbolic form is: 2
2
3 4 0( 3 4) 0D y Dy yD D y
It’s auxiliary equation is :
2 3 4 0D D 4, 1D
Hence C.F.= 4
1 2x xy c e c e (Ans.)
11. Solve: 2
2 5 6 sin 3d y dy y xdx dx
Sol. It’ s symbolic form is 2 5 6 tan 2D D y x
It’s auxiliary equation is: 2 5 6 0D D
( 2)( 3) 02,3
D DD
Hence C.F.= y=2 3
1 2x xc e c e
And P.I.=
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 2
2 2
1 1sin 3 sin 3( 5 6) 3 5 6
1 1 5 3sin 3 [ ][ ]sin 35 3 5 3 5 35 3 5 3sin 3 sin 3
25 9 25( 3 ) 91 15cos3 3sin 3(5 3)sin 3
234 2345cos3 sin 3
78
x xD D D
Dx xD D D
D Dx xD
x xD x
x x
Hence complete solution is:y=C.F.+P.I.
2 31 2
1 (5cos3 sin 3 )78
x xy c e c e x x
12. Solve by method variation of parameters:
2
2 4 tan 2d y y xdx
Ans: Homogeneous equation is 4 0y y
Its characteristics equation is 2 4 0 2i Hence, Homogeneous solution is cos 2 sin 2hy A x B x
1 2cos 2 , sin 2y x y x
1 1 2 2
2 2
cos 2 sin 22cos 2 2sin 2 2
2sin 2 2cos 2y y x x
W x xx xy y
Wrdxyy
Wrdxyyy p
.. 12
21
sin 2 .tan 2 cos 2 tan 2cos 2 sin 2
2 2px xdx x xdxy x x
cos 2 sin 2(sec 2 cos 2 ) sin 2 .
2 2px xy x x dx x dx
cos 2 log(sec 2 tan 2 ) sin 2 sin 2 cos 2
2 2 2 2 2px x x x x xy
cos 2 .log(sec 2 tan 2 )
4px x xy
So, Solution cos 2 .log(sec 2 tan 2 )cos 2 sin 2
4h px x xy y y A x B x
(Ans)
13. Solve the simultaneously equation:
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2 5 , 4 3tdx dxx y e x y tdt dt
Sol. The given equation can be expressed as:
( 5) 3 ............(1)2 ( 5) ......(2)t
D x y tx D y e
To eliminate y , operating equation (1) by (D+5) and equation (2) by 3 then subtracting , we get
2
( 5)( 4) 6 ( 5) 3( 9 14) 1 5 3
t
t
D D x x D t eD D x t e
…………………..(3)
The root of auxiliary equation of the equation corresponding homogeneous equation 2( 9 14) 0D D x
of the equation (3) is given by 2( 9 14) 0, 2, 7
D Dor D
Hence the complementary function of equation (3) is: 2 7
1 2. . t tC F c e c e
The particular integral of equation (3) is
2
2 2
2
1. . (1 5 3 )( 9 14)
1 1(1 5 ) 3( 9 14) ( 9 14)
1 9 31 ............. (1 5 )14 14 14 1 9 14
1 91 5 .514 14 81 315
14 14 8
t
t
t
t
t
P I t eD D
t eD D D D
D eD t
et
et
Hence the general solution of equation (3) is:
2 71 2
2 71 2
. . . .1 315
14 14 85, 2 7
14 8
tt t
tt t
x C F P Iex c e c e t
dx eNow c e c edt
…………………….(4)
Substituting the above values of x and dx/dt in equation (1), we get;
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2 7 2 71 2 1 2
2 71 2
5 20 1243 2 7 4 414 8 14 196 2
1 3 5 272 3 .......................(5)3 7 8 98
t tt t t t
t t t
e ey c e c e c e c e t t
y c e c e t e
Since the degree of D in the determinant 4 3
, 22 5
Dis
D
it follows that the number of
independent constant in general solution must be two. Hence (4) and (5) together constitute the general solution of the given system.
April -May, 2008
14. State Cauchy’s Linear equation.
Ans: - A Differential equation of the form
Xyadx
ydxadx
ydxadx
ydx nn
nn
n
nn
n
nn
........2
22
21
11
1 is known as Cauchy’s
Linear differential equation.
15. Solve2
22 4 sin2xd y dy
y e xdx dx
.
(ANS). It’s symbolic form is : 2 2( 4 1) sin 2xD D y e x
It’s auxiliary equation is: 2( 4 1) 0D D
4 16 4( ) 2 2 32
D
21 2. cosh 2 3 sinh 2 3xC F e C C
And P.I.=
2 2
2 2
1 1sin 2 sin 2( 4 1) 1 4 1 1
x xe x e xD D D D
2 2
2 2sin 2 sin 21 2 4 4 1 2 2
x xe ex xD D D D D
2 2
2 sin 2 sin 24 2 2 16 2 2
x xe ex xD D
2 2
sin 2 sin 214 2 14 2
x xe ex xD D
CSVTU II SEMESTER SOLUTION ALL UNITS
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22
2
2 14sin 2 sin 2
14 2 4 196
xx e De x xD D
2
2
2 14sin 2
4 2 196
xe Dx
2 2 14sin 2
212
xe Dx
2 4cos 2 14sin 2212
xe x x
Hence complete solution is:y=C.F.+P.I.
22
1 2
4cos 2 14sin 2cosh 2 3 sinh 2 3
212
xx e x x
y e C C
16. Solve by method of variation of parameters of : -2
2 4 4tan2 .d y
y xdx
Ans: Homogeneous equation is 4 0y y
Its characteristics equation is 2 4 0 2i Hence, Homogeneous solution is cos 2 sin 2hy A x B x
1 2cos 2 , sin 2y x y x
1 1 2 2
2 2
cos 2 sin 22cos 2 2sin 2 2
2sin 2 2cos 2y y x x
W x xx xy y
Wrdxyy
Wrdxyyy p
.. 12
21
sin 2 .tan 2 cos 2 tan 2cos 2 sin 2
2 2px xdx x xdxy x x
cos 2 sin 2(sec 2 cos 2 ) sin 2 .
2 2px xy x x dx x dx
cos 2 log(sec 2 tan 2 ) sin 2 sin 2 cos 2
2 2 2 2 2px x x x x xy
cos 2 .log(sec 2 tan 2 )
4px x xy
So, Solution cos 2 .log(sec 2 tan 2 )cos 2 sin 2
4h px x xy y y A x B x
(Ans)
17. Solve 2 0dydx
x ydt dt
and 5 3 0dydx
x ydt dt
CSVTU II SEMESTER SOLUTION ALL UNITS
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(Ans). The given equation are:
2 0.....................(1)dydx
x ydt dt
5 3 0.....................(2)dydx
x ydt dt
Subtracting equation (2) from equation (1), we get
3 2 0...............(3)dx x ydt
Writing equation (2)and (3) symbolically (i.e;D=d/dt),we have 5 ( 3) 0..........(4)( 3) 2 0..........(5)
x D yD x y
Operating equation (5)by D+3 ,we get 2( 9) 2( 3) 0.........(6)D x D y
Now, multiplying equation(4) by 2 and adding in equation (6),we get 2( 1) 0D x ………………………….(7)
This is the linear equation in x with constant coefficient s for which the auxiliary equation is 2( 1) 0D
D i
Hence the general solution of linear equation is: 0
1 2
1 2
( cos sin )( cos sin )..........................(8)
tx e c t c tx c t c t
From equation (8)
1 2
1 2
sin cos
sin cos
dx c t c tdtDx c t c t
Substituting these values in equation(5) we get
1 2 1 2
2 1 1 2
1 1( 3 ) ( sin cos ) 3( cos sin )2 21 1( 3 )cos ( 3 )sin2 2
y Dx x c t c t c t c t
y c c t c c t
Nov-Dec 2008 18. Define the linear differential equation. Ans. A differential equation in which the dependent variable & its derivatives occur only in the first
degree & are not multiplied together is known as Linear Differential equation.
General form of Linear Differential Equation of nth order is:
XyPdx
ydPdx
ydPdx
ydnn
n
n
n
n
n
........2
2
21
1
1 , where XPPP n ,,........,, 21 are functions of x.
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19. Solve the differential equation:
2
2 3 2xd y dy ey e
dx dx
Ans: - Its symbolic form is 2 3 2xeD D y e .
Its Auxiliary equation is 2 3 2 0D D 1, 2D
Hence, Complementary function is 2x xcy Ae Be
Now, 2
1 1 1 13 2 ( 2)( 1) ( 2) ( 1)
x x xe e ePI e e eD D D D D D
2 2 21 12 2
x x x x x x x xx x x xe e e ePI e e e dx e e e e e e dx e e e dxD D
2 x xePI e e
So, Solution 2 2x x xc p
xey y y Ae Be e e (Ans)
20. Solve by method variation of parameters:
2 3
2 26 9xd y dy ey
dx dx x
Ans: - 2
3
2
2
96xey
dxdy
dxyd x
Its Homogeneous equation is 0962
2
ydxdy
dxyd
Its symbolic form is 0)96( 2 yDD
Its characteristics equation is 0)96( 2 DD 3,3 D
Hence, Homogeneous solution is 213)( ByAyeBxAy x
h
xx xeyey 32
31 ,
xxxxxxx
xx
exexeexeee
xeeyyyy
W 6666333
33
21
21 3333''
Wrdxyy
Wrdxyyy p
.. 12
21
dxxe
eexedx
xe
exeey
x
x
xx
x
x
xx
p 2
3
6
33
2
3
6
33 ..
dxx
xedxx
ey xxp 2
33 11
xxexey x
ex
p1log 33 x
ex
p exey 33 log
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So, Solution xe
xxxph exeBxeAeyyy 3333 log (Ans).
21. Solve the simultaneously equation:
5 2 , 2 0dx dyx y t x ydt dt
being x=y=0 when t=0.
Ans: - + 5푥 − 2푦 = 푡, + 2푥 + 푦 = 0 Its symbolic form is (퐷 + 5)푥 − 2푦 = 푡------------------(1) 2푥 + (퐷 + 1)푦 = 0-------------------(2) Now, eq(1) X (D+1) + eq(2) X 2 we get ⇒ [(퐷 + 1)(퐷 + 5) + 4]푥 = (퐷 + 1)푡 + 0 ⇒ (퐷 + 6퐷 + 9)푥 = 1 + 푡, which is linear differential equation. Its auxiliary equation is (퐷 + 6퐷 + 9) = 0 ⇒ 퐷 = −3,−3 So, 퐶.퐹. = (퐶 + 퐶 푡)푒
Now, 푃. 퐼. = ( ) (1 + 푡) = 1 + (1 + 푡) = 1− 2 + 3 + … … … … . (1 + 푡)
⇒ 푃. 퐼. = 1 + 푡 − = 푡 + = +
So, 푥 = (퐶 + 퐶 푡)푒 + + ---------------------------------(3) Now, from (1) 2푦 = (퐷 + 5)푥 − 푡 = 퐷푥 + 5푥 − 푡 ⇒ 2푦 = 5(퐶 + 퐶 푡)푒 + + − 3(퐶 + 퐶 푡)푒 + 퐶 푒 + − 푡
⇒ 푦 = (2퐶 + 퐶 ) + 푡퐶 푒 − + ---------------(4) Given that at 푡 = 0, 푥 = 푦 = 0, so equation (3) and (4) becomes 퐶 + = 0 ⇒ 퐶 = − and
(2퐶 + 퐶 ) + = 0 ⇒ 퐶 = − − 2퐶 == − + = − = − So, Solution is
푥 = (− − 푡)푒 + + and 푦 = − + + 푡푒 − +
⇒ 푥 = − {(1 + 6푡)푒 − (3푡 + 1)} and 푦 = − {(4 + 6푡)푒 + (6푡 − 4)}
April -May, 2009
22. Explain briefly the method of variation of parameter.
Ans: Let us solve 2
2
d y dyP Qy Xdx dx
by variation of parameter method.
Let its complementary function is 1 1 2 2cy c y c y
Then find out wronskian 1 21 2
1 2
( , )' '
y yW y y
y y
Now, particular integral 2 11 2
1 2 1 2
. .( , ) ( , )py X y Xy y dx y dx
W y y W y y
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Then general solution is c py y y (Ans).
23. Solve the differential equation 2 2 32 cos 2x xD y x e e x .
Ans: 2 2 32 cos 2x xD y x e e x -----------(1)
Here Auxiliary equation is 2 2 0 2D D i
So, cos 2 sin 2hy A x B x
2 32
1. . cos 22
x xpP I y x e e x
D
3 22 2
1 1 cos 2( 3) 2 ( 1) 2
x xpy e x e x
D D
3
22 2
1 1 cos 2611 2 31
11
xx
pey x e x
D D D D
3
22 2
1 1 cos 2611 4 2 31
11
xx
pey x e x
D D D
2 32 23 2
22
6 66 2 131 ... cos 211 11 121 1331 4 169
xx
p
D D D De D D Dy x e xD
3
2 22
6 1 36 2 131 ... cos 211 11 11 121 4 169
xx
pe D Dy D x e x
D
3
2 22
6 47 2 131 ... cos 211 11 121 4 4 169
xx
pe D Dy D x e x
3
2 6 47 2 132 2 0 0 ... cos 211 11 121 233
xx
pe Dy x x e x
3
2 12 94 4sin 2 13cos 211 11 121 233
x x
pe x ey x x x
So, Solution h py y y
3
2 12 94cos 2 sin 2 4sin 2 13cos 211 11 121 233
x xe x ey A x B x x x x
24. Solve the equation : - 2
22(1 ) (1 ) sin 2 log(1 )d y dyx x y x
dx dx .
Ans: 2
22(1 ) (1 ) sin 2 log(1 )d y dyx x y x
dx dx ……(1) is a Legendre’s equation.
Now let us put 1 tx e we get
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2
22(1 ) ( 1) , (1 )d y dyx D D y x Dy
dx dx where
dDdt
So, equation (1) becomes ( 1) sin 2D D y Dy y t
2( 1) sin 2D y t which is linear differential equation with constant coefficients.
Its auxiliary equation is 2( 1) 0D D i
So, 1 2. . cos sinC F c t c t ……………..(2)
Now, 2 2
1 sin 2 sin 2. . sin 21 2 1 3
t tP I tD
So, solution 1 2sin 2. . . . cos sin
3ty C F P I c t c t (Ans)
25. Solve the simultaneous equation: -
2 , 2t tdx dyy e x edt dt
.
Ans: 2 , 2t tdx dyy e x edt dt
2 ...............(1) 2 ..............(2)t tDx y e Dy x e where dDdt
(1) 2 (2) D we get
22 4 2 2 t tDx y D y Dx e e
2( 4) 2 t tD y e e …………..(3)
Here Its auxiliary equation is 2( 4) 0 2D D i
So, 1 2. . cos 2 sin 2C F c t c t ……………..(4)
Now, 2 2 2
1 1 1. . 2 21 1 1
t t t tP I e e e eD D D
2 2
1 1. . 21 1 ( 1) 1 2
tt t t eP I e e e
So, solution 1 2. . . . cos 2 sin 22
tt ey C F P I c t c t e
……………..(5)
Now putting the value of y in (2) we get 2 tDy x e
1 22 sin 2 2 cos 2 22
tt tec t c t e x e
1 22 2 sin 2 2 cos 22
tt ex c t c t e
1 2sin 2 cos 22 4
t te ex c t c t
………………(6)
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So, solution 1 2 1 2sin 2 cos 2 , cos 2 sin 22 4 2
t t tte e ex c t c t y c t c t e
(Ans)
Nov-Dec 2009
26. Solve the equation: 2( 2) 0D D y . Ans: - 2( 2) 0D D y
Its auxiliary equation is 2( 2) 0 1, 2D D D
So, solution is 2x xy Ae Be (Ans). 27. Solve the following differential equation:
2
22 logd y dyx x y x
dx dx .
Ans: - 푥 − 푥 + 푦 = 푙표푔푥
--------------------(1)
Given equation is Cauchy’s Homogeneous Linear differential equation.
Put 푥 = 푒 ⇒ 푡 = 푙표푔푥, if 퐷 = , 푥 = 퐷푦, 푥 = 퐷(퐷 − 1)푦
So, equation (1) is 퐷(퐷 − 1)푦 − 퐷푦 + 푦 = 푡 ⇒ (퐷 − 2퐷 + 1)푦 = 푡 Which is linear differential equation with constant coefficients. So, its Auxiliary equation is 퐷 − 2퐷 + 1 = 0 ⇒ 퐷 = 1,1 So, 퐶.퐹. = 퐶 푒 + 퐶 푡푒 ------------------------(2) Now, 푃. 퐼. =
( )푡 = (1− 퐷) 푡 = (1 + 2퐷 + 3퐷 + … … … … . )푡
⇒ 푃. 퐼. = 푡 + 2 So, solution is 푦 = 퐶.퐹. +푃. 퐼. = 퐶 푒 + 퐶 푡푒 + 푡 + 2 ⇒ 푦 = 퐶 푥 + 퐶 푥푙표푔푥 + 푙표푔푥 + 2 (Ans).
OR 28. Solve the simultaneous differential equation:
5 2 , 2 0dx dyx y t x ydt dt
, given that 0x y when 0t .
Ans: - + 5푥 − 2푦 = 푡, + 2푥 + 푦 = 0 Its symbolic form is (퐷 + 5)푥 − 2푦 = 푡------------------(1) 2푥 + (퐷 + 1)푦 = 0-------------------(2) Now, eq(1) X (D+1) + eq(2) X 2 we get ⇒ [(퐷 + 1)(퐷 + 5) + 4]푥 = (퐷 + 1)푡 + 0 ⇒ (퐷 + 6퐷 + 9)푥 = 1 + 푡, which is linear differential equation. Its auxiliary equation is (퐷 + 6퐷 + 9) = 0 ⇒ 퐷 = −3,−3 So, 퐶.퐹. = (퐶 + 퐶 푡)푒
Now, 푃. 퐼. = ( ) (1 + 푡) = 1 + (1 + 푡) = 1− 2 + 3 + … … … … . (1 + 푡)
⇒ 푃. 퐼. = 1 + 푡 − = 푡 + = +
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So, 푥 = (퐶 + 퐶 푡)푒 + + ---------------------------------(3) Now, from (1) 2푦 = (퐷 + 5)푥 − 푡 = 퐷푥 + 5푥 − 푡 ⇒ 2푦 = 5(퐶 + 퐶 푡)푒 + + − 3(퐶 + 퐶 푡)푒 + 퐶 푒 + − 푡
⇒ 푦 = (2퐶 + 퐶 ) + 푡퐶 푒 − + ---------------(4) Given that at 푡 = 0, 푥 = 푦 = 0, so equation (3) and (4) becomes 퐶 + = 0 ⇒ 퐶 = − and
(2퐶 + 퐶 ) + = 0 ⇒ 퐶 = − − 2퐶 == − + = − = − So, Solution is
푥 = (− − 푡)푒 + + and 푦 = − + + 푡푒 − +
⇒ 푥 = − {(1 + 6푡)푒 − (3푡 + 1)} and 푦 = − {(4 + 6푡)푒 + (6푡 − 4)}
29. Solve the following differential equation:2
2 4 sinhd y y x xdx
.
Ans: - − 4푦 =
⇒ − 4푦 =
Its Symbolic form is ⇒ (퐷 − 4)푦 =
Its Auxiliary equation is ⇒ (퐷 − 4) = 0 ⇒ 퐷 = ±2
So, 퐶.퐹. = 퐶 푒 + 퐶 푒
Now, 푃. 퐼. = = 푥푒 − 푥푒
⇒ 푃. 퐼. = 푒( )
푥 − 푒( )
푥 = 푒 푥 − 푒 푥
⇒ 푃. 퐼. = 푥 + 푥
⇒ 푃. 퐼. = 1 + + … … 푥 + 1 + + … … 푥
⇒ 푃. 퐼. = 푥 + + 푥 − = ( ) − ( )
⇒ 푃. 퐼. = [−3푥푠푖푛ℎ푥 − 2푐표푠ℎ푥]
So, solution is 푦 = 퐶.퐹. +푃. 퐼. = 퐶 푒 + 퐶 푒 + [−3푥푠푖푛ℎ푥 − 2푐표푠ℎ푥] (Ans).
April -May, 2010 30. Define Cauchy and Legendre Linear differential equation.
CSVTU II SEMESTER SOLUTION ALL UNITS
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Ans: - A Differential equation of the form
Xyadx
ydxadx
ydxadx
ydx nn
nn
n
nn
n
nn
........2
22
21
11
1 is known as Cauchy’s
Linear differential equation. A Differential equation of the form
Xyadx
ydbaxadx
ydbaxadx
ydbax nn
nn
n
nn
n
nn
........)()()( 2
22
21
11
1 is
known as Legendre’s Linear differential equation.
31. Solve 3
3 3 2xd y dy ey e
dx dx .
Ans: - Its symbolic form is 2 3 2xeD D y e .
Its Auxiliary equation is 2 3 2 0D D 1, 2D
Hence, Complementary function is 2x xcy Ae Be
Now, 2
1 1 1 13 2 ( 2)( 1) ( 2) ( 1)
x x xe e ePI e e eD D D D D D
2 2 21 12 2
x x x xx x e x e x x x e x x ePI e e e dx e e e e e e dx e e e dxD D
2 xx ePI e e
So, Solution 2 2 xx x x ec py y y Ae Be e e (Ans)
32. Solve, by the method of variation of parameters: xeDD x log)12( 2 . Ans: - Its Auxiliary equation is 2 2 1 0D D 1, 1D
Hence, Complementary function is xcy A Bx e
1 2,x xy e y xe
1 2 2 2 2 2
1 2
x xx x x x
x x x
y y e xeW e xe xe e
e e xey y
WXdxyy
WXdxyyy p
.. 12
21
2 2
. log . logx x x xx x
p x x
xe e xdx e e xdxy e xee e
log logx xpy e x xdx xe xdx
2 2
log log2 4
x xp
x xy e x xe x x x
2 2 23log 2 log 3
2 4 4x x x
px x xy e x e e x
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So, Solution 2
( ) 2log 34
x xc p
xy y y A Bx e e x (Ans)
33. Solve the simultaneous equation: -
0 ydtdxt , 0 x
dtdyt , given that 0)1(,1)1( yx .
Ans: - 0 ydtdxt -------------------(1), 0 x
dtdyt -------------(2)
Differentiating (1) w.e.t. t we get 2
2 0dx d x dytdt dt dt
, by multiplying t both side we get
2
22 0dx d x dyt t t
dt dt dt --------------------(3)
By putting (2) in (3) we get2 2
2 22 2 0d x dx d xt t t x
dt dt dt -----(4) which is Cauchy
Linear Differential Equation.
Put logzt e z t and 2
22 ( 1) ,d x dxt D D x t Dx
dt dt where
dDdz
So, eqn (4) becomes 2( 1) 0 ( 1) 0D D x Dx x D x
Its Auxiliary equation is 2( 1) 0 1,1D D
So, z z Bx Ae Be Att
-------(5)
Putting in (1) we get 2
dx B By t t A Atdt t t
-----------(6)
So, Solution is Bx Att
and By Att
(Ans).
Given x(1) = 1 and y(-1) = 0, so 21,
210,1 BAABBA
Hence solution is
t
tyBA
ttx 1
21,1
21
(Ans).
Nov-Dec 2010 34. Write the formula for P.I. for the method of variation of parameters.
Ans: - Let equation is rybaDD )( 2 , then P.I is given by
Wrdxyy
Wrdxyyy p
.. 12
21 where
'' 21
21
yyyy
W .
35. Solve )2sin(8)2( 222 xxeyD x . Ans: - )2sin(8)2( 222 xxeyD x ---------------------(1)
Its characteristics equation is 2,20)2( 2 DD
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So, xxh BxeAeyFC 22.. -----------------------(2)
Now, )2sin(8)2(
1.. 222 xxe
DyIP x
p
222
22 )2(
12sin)2(
1)2(
18 xD
xD
eD
y xp
222
2
21
1412sin
441
)2(218 x
Dx
DDe
Dxy x
p
2
2
222 ..........
43
221
412sin
4421
218 xDDx
Dexy x
p
..........00
462
41
22cos
41
28 22
2
xxxexy xp
3422cos4 222 xxxexy xp
So, Solution 3422cos4 22222 xxxexBxeAeyyy xxxph (Ans).
36. Solve, by the method of variation of parameters: 2
3
2
2
96xey
dxdy
dxyd x
.
Ans: - 2
3
2
2
96xey
dxdy
dxyd x
Its Homogeneous equation is 0962
2
ydxdy
dxyd
Its symbolic form is 0)96( 2 yDD
Its characteristics equation is 0)96( 2 DD 3,3 D
Hence, Homogeneous solution is 213)( ByAyeBxAy x
h
xx xeyey 32
31 ,
xxxxxxx
xx
exexeexeee
xeeyyyy
W 6666333
33
21
21 3333''
Wrdxyy
Wrdxyyy p
.. 12
21
dxxe
eexedx
xe
exeey
x
x
xx
x
x
xx
p 2
3
6
33
2
3
6
33 ..
dxx
xedxx
ey xxp 2
33 11
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xxexey x
ex
p1log 33 x
ex
p exey 33 log
So, Solution xe
xxxph exeBxeAeyyy 3333 log (Ans).
37. Solve: xydxdyx
dxydx elog2
22 .
Ans: - xydxdyx
dxydx elog2
22 ---------(1)
Let logzx e z x and 2
22' , '( ' 1)dy d yx D y x D D y
dx dx where ' dD
dz
Putting all the values in (1) we get zyDDzyDDD ]12[1)1( 2 -----------(2)
Its auxiliary equation is 1,10122 mmm
So, xBxAxBzeAeyFC ezz
h log..
Now, zD
yIP p 2)1(1..
zDDzDy p .........)321()1( 22
xzzy ep log22........0002
So, general solution is xxBxAxyyy eeph log2log (Ans).
April -May, 2011
38. Find the particular integral of Solve 3
3. 4 sin 2d y dy xdx dx
Ans. 3
3. 4 sin 2d y dy xdx dx
P.I will be 3
1 sin 24
PI xD D
3 2
1 1sin 2 sin 24 4
x xD D D D
3 2
1 1sin 2 sin 24 2 4
x xD D D
3
1 1sin 2 sin 24 0
x xD D D
By differentiating 3 4D D as 1, 0in f D can not be
f D
CSVTU II SEMESTER SOLUTION ALL UNITS
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3 2
1 1sin 2 sin 24 3 4
x x xD D D
3 2
1 1sin 2 sin 24 3( 2 ) 4
x x xD D
3
1 sin 2 sin 24 8
xx xD D
3
1 sin 2sin 24 8
x xxD D
P.I of3
3. 4 sin 2d y dy xdx dx
will be sin 28
x x
39. Solve 2
2 3 2xd y dy ex y e
dx dx
Ans: - Its symbolic form is 2 3 2xeD D y e .
Its Auxiliary equation is 2 3 2 0D D 1, 2D
Hence, Complementary function is 2x xcy Ae Be
Now, 2
1 1 1 13 2 ( 2)( 1) ( 2) ( 1)
x x xe e ePI e e eD D D D D D
2 2 21 12 2
x x x xx x e x e x x x e x x ePI e e e dx e e e e e e dx e e e dxD D
2 xx ePI e e
So, Solution 2 2x xc p
xx ey y y Ae Be e e (Ans)
40. Solve the differential equation: 2
22 5 4 . log .
d y dyx x y x x
dx dx
(Ans). Let
zx e and d Ddz
. Then 2
22, ( 1)dy d yx Dy x D D y
dx dx
Thus the given differential equation reduces to the following fotm:
2
( 1) 5 4
( 4 4)
z
z
D D D y ze
D D y ze
Which is the linear differential equation with constant coefficient s, for which the auxiliary equation is:
2 4 4 0D D 2, 2D
C.F= 2 21 2 1 2 logzc c z e c c x x
P.I Will be
CSVTU II SEMESTER SOLUTION ALL UNITS
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1( )
zzef D
i.e
21.
( 4 4)zP I ze
D D
2 2 2
1.( 4 4) ( 1) 4( 1) 4 ( 6 9)
z zz e eP I ze z z
D D D D D D
26. (1 )9 9 9
ze D DP I z
6.9 9
zeP I z
2. log9 3xP I x
Complete Ans is C.F +P.I
21 2 logc c x x +
2log9 3x x
41. Solve sin , cos ,dx dyy t x tdt dt
Given that x=2 and y=0 when t=0.
Ans: Given simultaneous differential equation are
sindx y t
dt
-------------(1)
cosdy x t
dt
------------(2)
From equation (2) we get cosdyx tdt
---------(3)
Differentiating (3) with respect to t we get
2
2 sindx d y tdt dt
---------(4)
From (1) and (4) we get
2
2sin sind yt y tdt
2
2 2sind y y tdt
2( 1) 2sinD y t -----------(5)
Its Auxiliary equation is 2 1 0 1,1m m
So, . . t thC F y Ae Be
CSVTU II SEMESTER SOLUTION ALL UNITS
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And 2
1. . ( 2sin )1pP I y t
D
2
1( 2sin ) sin1 1py t t
So, h py y y
sint ty Ae Be t -------(6) Putting the value of y in (2) we get
cos cost tAe Be t x t
t tx Ae Be ---------(7)
Given that 2, 0, 0x y when t
So, equation (6) and (7) becomes 0 A B and 2 A B
1, 1A B So, solution t tx e e and sint ty e e t (Ans)
Nov-Dec 2011
42. Explain Cauchy’s homogeneous linear differential equation.
Ans: - A Differential equation of the form
Xyadx
ydxadx
ydxadx
ydx nn
nn
n
nn
n
nn
........2
22
21
11
1 is known as Cauchy’s
Linear differential equation.
43. Solve the differential equation2
2 2 sinxd y dy y xe xdx dx
.
Ans: 2
2 2 sinxd y dy y xe xdx dx
Its auxiliary equation is 2 22 1 0 ( 1) 0 1,1m m m m
So, . . x xhC F y Ae Bxe
Now, 2
1. . sin( 1)
xpP I y xe x
D
2 2
1 1sin sin( 1 1)
x xpy e x x e x x
D D
1 sinx
py e x xdxD
1 cos sinxpy e x x x
D
cos sinxpy e x x x dx
CSVTU II SEMESTER SOLUTION ALL UNITS
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sin cos cosxpy e x x x x
sin 2cosxpy e x x x
So, general solution h py y y
sin 2cosx x xy Ae Bxe e x x x (Ans)
44. Solve the differential equation 2
22 log .sin(log )
d y dyx y x xdx dx
zx e and d Ddz
. Then 2
22, ( 1)dy d yx Dy x D D y
dx dx
Thus the given differential equation reduces to the following fotm:
2
( 1) 1 (sin )
( 1) sin
D D D y z z
D y z z
Which is the linear differential equation with constant coefficient s, for which the auxiliary equation is:
2 1 0D ,D i i
C.F= 1 2 1 2cos sin cos log sin logc z c z c x c x
P.I Will be 1 sin( )
z zf D
i.e
2 2
1 1. sin .( 1) ( 1)
izP I z z I P zeD D
221. sin .
( 1) 1)
izeP I z z I P zD D i
2
1. sin .( 1) 2
izeP I z z I P zD D i
21 1 1. sin .
( 1) 2 12
izP I z z I Pe zDD ii
1
2
1 1. sin . 1( 1) 2 2
DizP I z z I Pe zD i i
2
1 1. sin . 1( 1) 2 2
DizP I z z I Pe zD i i
CSVTU II SEMESTER SOLUTION ALL UNITS
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2
1 1 1. sin .( 1) 2 2
izP I z z I Pe zD i i
2
1. sin .( 1) 2 2
i iizP I z z I Pe zD
2
1 1. sin .( 1) 2 4
iizP I z z I Pe zD
2
1 1. sin . cos sin( 1) 2 4
iP I z z I P z i z zD
2
1 cos sin sin. sin . cos( 1) 2 4 2 4
i z z z i zP I z z I P z zD
2
1 cos sin sin 1. sin . cos( 1) 4 2 4 2
z z z zP I z z I P i z zD
2
1 sin 1. sin cos( 1) 4 2
zP I z z z zD
2
1 sin log 1. sin log cos log( 1) 4 2
xP I z z x xD
Ans.= C.F+P.I
1 2cos log sin logc x c x +sin log 1 log cos log
4 2x x x
45. Solve the following simultaneous equation 2dx ydt
, 2dy zdt ,
2 .dz xdt
(Ans.) The given equation is:
2dx ydt
………..(1), 2 ..........(2)dy zdt
,
2 ..........(3)dz xdt
Differentiating (1)w . r. “t.”, we get 2
2 2 2(2 )d x dy zdt dt
using(2)
Differentiating again w. r. t. “t”,, we get 3
3 4 4(2 )d x dz xdt dt
3
3 2
21 2 3
( 8) 0,
8 0 ( 2)() 2 4) 0
2, 1 3
cos( 3 )12
t t
dD x whereDdt
D or D D D
D i
x c e c e t cdxydt
CSVTU II SEMESTER SOLUTION ALL UNITS
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21 2 3 2 3
21 2 3 3
21 2 3
1 2 cos( 3 ) 3 sin( 3 )2
2 2cos cos( 3 ) sin sin( 3 )3 3
2cos 33
t t t
t t
t t
c e c e t c c e t c
c e c e t c t c
c e c e t c
From(2)….. 21 2 3 2 3
21 2 3
12
1 2 22 cos( 3 ) 3 sin( 3 )2 3 3
4cos( 3 )3
t t t
t t
dyzdt
c e c e t c c e t c
c e c e t c
April -May, 2012
46. Solve 3
3 0d y ydx
Ans: - Its symbolic form is Solve 013 yD .
Its Auxiliary equation is 0)1)(1(01 23 DDDD
231,
231,1 iiD
So, solution 23
23221 sincos..x
xxx eCCeCyFC (Ans).
47. Solve 2
22 5 4 logd y dyx x y x x
dx dx
(Ans). Let
zx e and d Ddz
. Then 2
22, ( 1)dy d yx Dy x D D y
dx dx
Thus the given differential equation reduces to the following fotm:
2
( 1) 5 4
( 4 4)
z
z
D D D y ze
D D y ze
Which is the linear differential equation with constant coefficient s, for which the auxiliary equation is:
2 4 4 0D D 2, 2D
C.F= 2 21 2 1 2 logzc c z e c c x x
P.I Will be 1( )
zzef D
i.e
CSVTU II SEMESTER SOLUTION ALL UNITS
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21.
( 4 4)zP I ze
D D
2 2 2
1.( 4 4) ( 1) 4( 1) 4 ( 6 9)
z zz e eP I ze z z
D D D D D D
26. (1 )9 9 9
ze D DP I z
6.9 9
zeP I z
2. log9 3xP I x
Complete Ans is C.F +P.I
21 2 logc c x x +
2log9 3x x
48. Solve :
3 2
3. 22 sin 2xd y d y dy e xdx dx dx
Sol. It’s symbolic form is xeyDDD x 2sin2 23
Its Auxiliary equation is 0)12(02 223 DDDDDD
1,1,0 D
So, xxc exCCeCyFC 32
01..
xc exCCCy 321 ---------------------(1)
Now, xeDDD
yIP xp 2sin
21.. 23
xDDD
eDDD
y xp 2sin
21
21
2323
xDDDD
eDD
xy xp 2sin
2.1
1431
222
xDD
eD
xy xp 2sin
)4(2)4.(1
4612
xD
exyx
p 2sin83
14)1(6
2
xD
Dexyx
p 2sin64983
2 2
2
xDexyx
p 2sin64)4(983
2
2
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
10
2sin82cos62
2 xxexyx
p
-------------------(2)
So, solution 10
2sin82cos62
2
321xxexexCCCyyy
xx
pc
(Ans).
49. Solve the equations simultaneously
5 2 , 2 0, 0dx dyx t t x y x y
dt dt when t=0
Ans: - + 5푥 − 2푦 = 푡, + 2푥 + 푦 = 0 Its symbolic form is (퐷 + 5)푥 − 2푦 = 푡------------------(1) 2푥 + (퐷 + 1)푦 = 0-------------------(2) Now, eq(1) X (D+1) + eq(2) X 2 we get ⇒ [(퐷 + 1)(퐷 + 5) + 4]푥 = (퐷 + 1)푡 + 0 ⇒ (퐷 + 6퐷 + 9)푥 = 1 + 푡, which is linear differential equation. Its auxiliary equation is (퐷 + 6퐷 + 9) = 0 ⇒ 퐷 = −3,−3 So, 퐶.퐹. = (퐶 + 퐶 푡)푒
Now, 푃. 퐼. = ( ) (1 + 푡) = 1 + (1 + 푡) = 1− 2 + 3 + … … … … . (1 + 푡)
⇒ 푃. 퐼. = 1 + 푡 − = 푡 + = +
So, 푥 = (퐶 + 퐶 푡)푒 + + ---------------------------------(3) Now, from (1) 2푦 = (퐷 + 5)푥 − 푡 = 퐷푥 + 5푥 − 푡 ⇒ 2푦 = 5(퐶 + 퐶 푡)푒 + + − 3(퐶 + 퐶 푡)푒 + 퐶 푒 + − 푡
⇒ 푦 = (2퐶 + 퐶 ) + 푡퐶 푒 − + ---------------(4) Given that at 푡 = 0, 푥 = 푦 = 0, so equation (3) and (4) becomes 퐶 + = 0 ⇒ 퐶 = − and
(2퐶 + 퐶 ) + = 0 ⇒ 퐶 = − − 2퐶 == − + = − = − So, Solution is
푥 = (− − 푡)푒 + + and 푦 = − + + 푡푒 − +
⇒ 푥 = − {(1 + 6푡)푒 − (3푡 + 1)} and 푦 = − {(4 + 6푡)푒 + (6푡 − 4)}
Nov -Dec, 2012 50. Write the formula for P.I. for the method of variation of parameters.
Ans: - Let equation is rybaDD )( 2 , then P.I is given by
Wrdxyy
Wrdxyyy p
.. 12
21 where
'' 21
21
yyyy
W .
51. Solve 2 1 sin cosD y x x
Ans: 2 1 sin cosD y x x x
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2( 1) 0D therefore 1D
1 2. . x xcC F y C e C e
2
1. . sin cos1
P I x x xD
2 2
1 1. . sin cos1 1
P I x x xD D
2 2
1 1. . . cos1 1
ixP I I P xe xD D
2 2
1. . . cos11
ixeP I I P x xDD i
22
1. . . cos1 11 2 1
ixeP I I P x xD Di
22
1. . . cos1 11 2 1
ixeP I I P x xD Di
2
1. . . cos22 2
ixeP I I P x xD Di
2
1. . . cos2
2 12
ixeP I I P x xDDi
12 1. . . 1 cos2 2 2
ixe DP I I P Di x x
1. . . 1 cos2 2
ixeP I I P Di x x
1. . . cos2 2
ixeP I I P x i x
cos sin 1. . . cos2 2
x i xP I I P x i x
cos cos sin sin 1. . . cos2 2
x x i x ix x xP I I P x
By selecting only imaginary part from above we will have
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
cos sin 1. . cos2 2
x x xP I x
Y = C.F + P.I
Y= 1 2. . x xcC F y C e C e +
cos sin 1 cos2 2
x x xx
52. Solve the equation : - 2
22(1 ) (1 ) sin 2 log(1 )d y dyx x y x
dx dx .
Ans: 2
22(1 ) (1 ) sin 2 log(1 )d y dyx x y x
dx dx ……(1) is a Legendre’s equation.
Now let us put 1 tx e we get
2
22(1 ) ( 1) , (1 )d y dyx D D y x Dy
dx dx where
dDdt
So, equation (1) becomes ( 1) sin 2D D y Dy y t
2( 1) sin 2D y t which is linear differential equation with constant coefficients.
Its auxiliary equation is 2( 1) 0D D i
So, 1 2. . cos sinC F c t c t ……………..(2)
Now, 2 2
1 sin 2 sin 2. . sin 21 2 1 3
t tP I tD
So, solution 1 2sin 2. . . . cos sin
3ty C F P I c t c t (Ans)
53. Solve the simultaneously equation:
22 3 , 3 2 tdx dxx y t x y edt dt
Sol. The given equation can be expressed as:
2
( 2) 3 ............(1)3 ( 2) ......(2)t
D x y tx D e
To eliminate y , operating equation (1) by (D+2) and equation (2) by 3 then adding , we get 2 2
2 2
( 2) 9 ( 2) 3( 4 5) 1 2 3
t
t
D x x D t eD D x t e
…………………..(3)
The root of auxiliary equation of the equation corresponding homogeneous equation 2( 4 5) 0D D x
of the equation (3) is given by 2( 4 5) 0, 5,1
D Dor D
Hence the complementary function of equation (3) is:
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
51 2. . t tC F c e c e
The particular integral of equation (3) is
22
22 2
12
2
2
1. . 1 2 3( 4 5)
1 1(1 2 ) 3( 4 5) ( 4 5)
1 4 31 (1 2 )5 5 5 4 8 5
1 4 31 (1 2 )5 5 71 8 3(1 2 )
5 5 7
t
t
t
t
t
P I t eD D
t eD D D D
D eD t
eD t
et
Hence the general solution of equation (3) is:
25
1 2
. . . .1 8 3(1 2 )
5 5 7
tt t
x C F P Iex c e c e t
…………………….(4)
25
1 22 6, 5
5 8
tt tdx eNow c e c e
dt
2 25 5
1 2 1 21 1 2 6 1 8 3, 2 5 2 (1 2 )3 3 5 8 5 5 7
t tt t t tdx e eNow y x t c e c e c e c e t t
dt
2 25
1 21 3 2 8 6 33 23 5 5 5 8 7
t tt t t e ey c e c e t
25
1 21 2 153 2 13 5 8
tt t t ey c e c e t
Unit III – Multiple Integral
(May-Jun-2006)
1. Evaluate by changing the order of integration.
Ans 0 0
2xx yxe dydx
0 0
/ .2
xyx dydxxe
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
0 0
2xx yxe dydx
by changing the order of the equation we will have 0
2
y
xyxe dydx
0 0
2 2
y y
x xy yxe dydx xe dx dy
0 0
2
2 2
2
2
2
22
2 2 2
y
y
t
y y
x xyy yxe dydx e dy
y
xxyxe dx taking ty
xdx ydt
xx yy y yyxdx dt xe dx e dt e
y
0 0 0 0
2 20
2 2 2y
x xy y yy y y yxe dydx e dy e dy e dy
yy
0 0
21 1. 1/ 22 1y
x yey yxe dydx y e
(Ans).
2. Find the volume common to the cylinder 2 2 2 2 2 2,x y a x z a .
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Ans:
2 2 2 2 2 2 2 2
2 2 2 2 0 0 0
Volume 8a a x a x a a x a x
a a x a x
dxdydz dxdydz
2 2 2 2
2 22 2
00 0 0 0
Volume 8 8a a x a a x
a xz dxdy a x dxdy
2 2
2 2 2 2 2 2 2 20
0 0 0
Volume 8 . 8 . 8 ( )a a a
a xa x y dx a x a x dx a x dx
3 3 3 3
2 2
0
2 16Volume 8 8 83 3 3 3
ax a a aa x a
(Ans).
3. Prove that .
Sol.
1
10
1 !log
1
nn
n
nmx x dxm
1
0
log nmx x dx ……………………………………………….1
Let log tx t x e tdx e dt
1 becomes 0
0
1 1n nm t m te t dt e t dt
……………………………….2
1,0,)1(
!.)1().(log 1
1
0
mn
mndxxx n
nnm
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
1
1dpm t p dt
m
2 becomes 0
11 1
nppe dpm m
0
11 11
nn
p ne p dpm
=
0
1 1 11 11
nn npe p dp
m
0
1 11 11 11 1 11 1
n nn nnpe p dp n
m m
1 1 !1 1 1 11 1
nn
n nn nm m
Hence Proved
(Nov –Dec 2006)
4. Evaluate the following integral by changing of order of integration 2
4 2 20
a a
ax
y dxdyy a x
.
Ans:
Here region of integration is x varies from 0 to a, y varies from ax to a.
After changing order y varies from 0 to a, x varies from 0 to2y
a.
2 /2 2
4 2 2 4 2 20 0 0
y aa a a
ax
y dxdy y dydxy a x y a x
putting 2a as common we get
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2
2
4 2 2 220 0 0 2
2a a a
ax
y y dydxay dxdy ay a x y x
a
2
2 21
24 2 2 220 0 0 02
0
22
sina a a a
ax
yay y dxay dxdy y xa dy dy
a yy a x y x aa
2
2 2 21 1 1
2 24 2 20 0 0
0
2
sin sin sin 0a a a a
ax
ya y
y dxdy y x y ady dya ay yy a x
a a
2 2 2
4 2 20 0 0
02 2
a a a a
ax
y dxdy y ydy dya ay a x
2 2 2
3
04 2 20 02 6 6
a a aa
ax
y dxdy y ady ya ay a x
(Ans)
5. Find the volume common to gas cylinder 2 2 2 2 2 2,x y a x z a .
Ans:
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 2 2 2 2 2 2 2
2 2 2 2 0 0 0
Volume 8a a x a x a a x a x
a a x a x
dxdydz dxdydz
2 2 2 2
2 22 2
00 0 0 0
Volume 8 8a a x a a x
a xz dxdy a x dxdy
2 2
2 2 2 2 2 2 2 20
0 0 0
Volume 8 . 8 . 8 ( )a a a
a xa x y dx a x a x dx a x dx
3 3 3 3
2 2
0
2 16Volume 8 8 83 3 3 3
ax a a aa x a
(Ans).
6. Define Beta function and show that ( ) ( )( , )( )m nm nm n
.
Ans: The Beta function is defined as 1 1
1 1
0 0
( , ) (1 ) , , 0(1 )
nm n
m nxB m n x x dx dx m nx
.
We know that 1
0
( ) .x nn e x dx
By putting x az dx adz we get
1
0
( ) .( )az nn e az adz
1
0
( ) .n az nn a e z dz
Putting z = x we get
1
0
( ) .n ax nn a e x dx
Now, taking a = z we get
1
0
( ) .zx n nn e x z dx
By multiplying 1z me z both side we get
1 (1 ) 1 1
0
( ) .z m z x n m nn e z e x z dx
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Now taking integration with respect to z both side from 0z we get
1 1 (1 ) 1
0 0 0
( ) .z m n z x m nn e z dz x e z dzdx
Now, let (1 )1 1
y dyz x y z dzx x
we get
1
1
0 0
.( ) ( )(1 )
y m nn
m n
e y dyn m x dxx
1
1
0 0
( ) ( ) .(1 )
ny m n
m n
xn m e y dy dxx
1
0
( ) ( ) ( )(1 )
n
m nxm n m n dxx
1
0
( ) ( ) ( ) ( ). ( , )(1 )
n
m nxm n m n dx m n B m nx
( ) ( )( , ) .( )m nB m nm n
(proved)
(May-Jun-2007)
7. Change the order of integration in 2
1 2
0
x
x
I xydxdy
and hence evaluate the same.
Ans: From 2
1 2
0
x
x
I xydxdy
, region of integration is 2 , 2 , 0, 1y x y x x x
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
By changing of order of integration we get two regions
In one region limit of x is 0,x x y and limit of y is 0, 1y y
In another region limit of x is 0, 2x x y and limit of y is 1, 2y y
So, 2
21 2 1 2
0 0 0 1 0
y yx
x
I xydxdy xydydx xydydx
21 2
0 0 1 0
y y
I xydx dy xydx dy
21 22 2
0 10 02 2
y yx y x yI dy dy
1 22 2
0 1
0 (2 ) 02 2
y y yI dy dy
1 22 3
2
0 1
2 22 2y yI dy y y dy
1 23 3 4
2
0 1
26 3 8y y yI y
1 16 16 2 10 4 16 3 8 3 8
I
4 96 128 48 24 16 3
24I
164 153 11
24 24I
(Ans)
8. Find the volume bounded by the cylinder 2 2 4x y and the planes 4, 0y z z
Ans:
From the figure 4z y is to be integrated over the circle 2 2 4x y in the xy plane.
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
To cover the shaded half of the circle, x varies from 0 to 24 y
So, 2 24 42 2
2 0 2 0
Volume 2 2 (4 )y y
zdxdy y dxdy
2
24
02
Volume 2 (4 ) yy x dy
2
2
2
Volume 2 (4 ) 4y y dy
2 2
2 2
2 2
Volume 8 4 2 4y dy y y dy
22
1
2
4 4Volume 8 sin 02 2 2
y y y
{second term is zero because of odd function}
1 1Volume 8 0 0 2sin 1 2sin ( 1)
1Volume 8 4sin 1 8 4 162 (Ans).
9. Solve 2 2 /2( )nxy x y dxdy over the positive quadrant of 2 2 4x y supposing n+3>0. Ans:
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 2
2 2 /2 2 2 /2
0 0
( ) ( )a a x
n nxy x y dxdy xy x y dxdy
2 2
32 2 2
2 2 /2
0
0
( )( ) 322
a xn
an
y
x x yxy x y dxdy dxn
3
2 2 /2 2 2 2 2
0
1( ) ( 0)3
a nnxy x y dxdy x x a x dx
n
3 322 2
2 2 /2
0 0
( )3 3 2
n n aan a a xxy x y dxdy xdx
n n
73
222 2 /2( )
3 2 2( 3)
nn
naa axy x y dxdy
n n
(Ans)
(Nov –Dec 2007)
10. Define Beta Function. Ans. The Beta function ,m n is defined as
1
0
1 1, 1m nm n x x dx for m>0 and n>0 It is also known as first Eularian Integral
11. Evaluate .
Ans:
2 2 2c b a
x y z dxdydzc ab
3 32 2 2 2 22 2
3 3
ac b c bz ax z y z dxdy ax ay dxdyc cab b
3 3 3 3
2 2 2 2 22 2 4 42 43 3 3 3
bc b a c cay a y ab a bx y z dxdydz ax y dx abx dxc a c cbb
3 3 3 3 3 3
2 2 2 4 4 4 8 8 83 3 3 3 3 3
cc b a abx ab x a bx abc ab c a bcx y z dxdydzc a cb
2 2 2 2 2 283
c b a abcx y z dxdydz a b cc ab
(Ans)
c
c
b
b
a
a
dxdydzzyx )( 222
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
12. Prove that .
Sol.
1
10
1 !log
1
nn
n
nmx x dxm
1
0
log nmx x dx ……………………………………………….1
Let log tx t x e tdx e dt
1 becomes 0
0
1 1n nm t m te t dt e t dt
……………………………….2
1
1dpm t p dt
m
2 becomes 0
11 1
nppe dpm m
0
11 11
nn
p ne p dpm
=
0
1 1 11 11
nn npe p dp
m
0
1 11 11 11 1 11 1
n nn nnpe p dp n
m m
1 1 !1 1 1 11 1
nn
n nn nm m
Hence Proved
13. Find the area enclosed by the parabolas 푦 = 4푥 − 푥 , 푦 = 푥 .
Ans:
1,0,)1(
!.)1().(log 1
1
0
mnm
ndxxx n
nnm
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
23 4
0
Areax x
x
dxdy
2
3 3 34 2 2
0 0 0
Area (4 ) (3 )x x
xy dx x x x dx x x dx
32 3
0
27 27 27 9Area 3 0 02 3 2 3 6 2x x
(Ans)
(May-Jun-2008)
14. Define Gamma Function.
Ans. Gamma function is defined as 1
0
, 0x nn e x dx n
15. Evaluate the integral: 1
loglog
1 1
xye ezdxdydz
Sol. 1
loglog
1 1
xye ezdxdydz
1 1 1
log log loglog log 0 11
1 1 1 1
xx
x xy y ye e e eezdz dxdy z z z dxdy e x e dx dy
1 1 1
log log loglog 1 1
1 1 1
xx x x x x
y ye e e e yzdxdydz e x e dx dy e x e e x dy
1
1 1
1 1
loglog log 2 log 2 1
1 1log
log log 2 log 11 1
x
x
ye e ezdxdydz y y y y e e dy
ye e ezdxdydz y y y y e dy
2
2 2
1
loglog log log 1
4 11 1
xy
eye e yzdxdydz y y y y y e
2 2
1
log 5log 1 0 5 / 4 0 12 41 1
xye e e ezdxdydz e e e e
2
1
loglog 2 13 / 4
41 1
xye e ezdxdydz e
Ans.
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
16. Given 1
0 1 sin( )
nx dxx n
Show that 1
sin( )n n
n
Hence evaluate 40 1
dyy
Sol. Given that
1
0 1 sin( )
nx dxx n
We know that ( , ) m nm nm n
1
0 1
n
m nx m ndx
m nx
1 1let m n m n
1
0
1 111
n
m nx n ndx n nx
1
0
1sin( )1
nx dx n nnx
Therefore 1sin( )
n nn
Now for
40 1
dyy
……………………………………….1
Let 4y =x
Y= 1/4x
Y= 3/414
dy x dx
3/4 1/4 1
40 0 0
1 1 1 1 111 4 1 4 1 4 sin 4 4 2 2sin
4 2
dy x dx x dxy x x
(Ans)
17. Find the area enclosed by the parabolas 푦 = 4푎푥,푥 = 4푎푦 .
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Area is given by 4
0
2
2/4
a axdxdy
x a
4 4
0 0
2 22/42/4
a aax axdxdy y dxx a
x a
4 42
0 0
22 / 4
2/4
a aaxdxdy ax x a dx
x a
44 3 2 2 2
0 0
2 3/2 1 32 16 1623 / 2 4 3 3 3 32/4
aa ax x x a a adxdy a
ax a
(Ans)
(Nov –Dec 2008)
18. Write only the value of .
Ans By using Legendre duplication formula
1 22 12 2
m m mm
14
1 1 1 1 12 214 4 2 4 22 122 2
19. Evaluate .
Ans:
43.
41
c
c
b
b
a
a
dxdydzzyx )( 222
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 2 2c b a
x y z dxdydzc ab
3 32 2 2 2 22 2
3 3
ac b c bz ax z y z dxdy ax ay dxdyc cab b
3 3 3 3
2 2 2 2 22 2 4 42 43 3 3 3
bc b a c cay a y ab a bx y z dxdydz ax y dx abx dxc a c cbb
3 3 3 3 3 3
2 2 2 4 4 4 8 8 83 3 3 3 3 3
cc b a abx ab x a bx abc ab c a bcx y z dxdydzc a cb
2 2 2 2 2 283
c b a abcx y z dxdydz a b cc ab
(Ans)
20. Change the order of integration 0
y
x
e dxdyy
and hence evaluate it.
Ans 0
y
x
e dxdyy
0
y
x
e dxdyy
by changing the order of the equation we will have 0 0
y ye dxdyy
0 0 0 0
y yy ye edxdy dx dyy y
0 0 0 0
yy y ye edxdy x dyy y
0 0 0 0
y y yye edxdy y dy e dy
y y
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
0
0 0 0 0
1y y y
y ye edxdy y dy e dy ey y
(Ans).
21. Find the area of the loop of the curve 푎푦 = 푥 (푎 − 푥) . Ans.
Area of the loop of the curve 푎푦 = 푥 (푎 − 푥) .will be given by 0 0
a xx aadxdy
0 0 0 0
a x a xx xa aa adxdy dy dx
……………………………..1
0 0 0 0 0 0
a x a x a xx x xa aa a a adxdy dy dx y dx
0 0 0 0 0 00 0
a x a x a xx x xa aa a a aa a xdxdy dy dx y dx x dxa
0 0 0 0
a xx aa a a xdxdy x dxa
…………………………..2
Take 2sinx a
Becomes /2 /2
2 2 2 4 2
0 0 0 0
sin cos 2 sin cos 2 sin cos
a xx aadxdy a a d a a d
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
/2 22 4 2 2
0 0 0
3 1 12 sin cos 2
6 4 2 2 16
a xx aa adxdy a a d a
(Ans).
(May-Jun-2009)
22. Write the relation between Beta and Gamma function.
Ans: ( ). ( )( , )( )m nB m nm n
23. Evaluate the following integral by changing of order of integration 2
4 2 20
a a
ax
y dxdyy a x
.
Ans:
Here region of integration is x varies from 0 to a, y varies from ax to a.
After changing order y varies from 0 to a, x varies from 0 to2y
a.
2 /2 2
4 2 2 4 2 20 0 0
y aa a a
ax
y dxdy y dydxy a x y a x
putting 2a as common we get
2
2
4 2 2 220 0 0 2
2a a a
ax
y y dydxay dxdy ay a x y x
a
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2
2 21
24 2 2 220 0 0 02
0
22
sina a a a
ax
yay y dxay dxdy y xa dy dy
a yy a x y x aa
2
2 2 21 1 1
2 24 2 20 0 0
0
2
sin sin sin 0a a a a
ax
ya y
y dxdy y x y ady dya ay yy a x
a a
2 2 2
4 2 20 0 0
02 2
a a a a
ax
y dxdy y ydy dya ay a x
2 2 2
3
04 2 20 02 6 6
a a aa
ax
y dxdy y ady ya ay a x
(Ans)
24. Evaluate 2 2( )
0 0
x ye dxdy
by changing to polar coordinates. Hence show that 2
0 2xe dx
.
Ans: 2 2 2
/2( )
0 0 0 0
x y r
r
e dxdy e rdrd
2 2 2
/2( )
0 0 0 0
1 22
x y r
r
e dxdy re dr d
2 2 2
/2( )
00 0 0
12
x y re dxdy e d
2 2
/2( )
0 0 0
1 (0 1)2
x ye dxdy d
2 2/2
/2( )0
0 0 0
1 1 12 2 2 2 4
x ye dxdy d
(Ans)
Now, 2 2( )
0 0 4x ye dxdy
2 2( ) ( )
0 0 4x ye dx e dy
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 2( ) ( )
0 0 4x xe dx e dx
2
2
( )
0 4xe dx
2
0 2xe dx
(Proved)
25. Find by double integration, the area lying between the parabolas 24y x x and the line y x .
Ans:
23 4
0
Areax x
x
dxdy
2
3 3 34 2 2
0 0 0
Area (4 ) (3 )x x
xy dx x x x dx x x dx
32 3
0
27 27 27 9Area 3 0 02 3 2 3 6 2x x
(Ans)
(Nov –Dec 2009)
26. Find the value of 2
0
axe x dx
.
Ans: - ∫ 푒 푥 푑푥∞ Let 푎푥 = 푡 ⇒ 푎푑푥 = 푑푡 ⇒ 푑푥 = 푑푡
= ∫ 푒 푡 푑푥∞ = Γ(3) = × 2 = (Ans).
27. Change the order of integration and evaluate 43
0 1
( )y
x y dxdy
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Ans: - ∫ ∫ (푥 + 푦)푑푥 푑푦 = ∫ ∫ (푥 + 푦)푑푦 푑푥 = ∫ 푥푦 + 푑푥
= ∫ 푥(4 − 푥 ) + ( ) 푑푥
= ∫ 4푥 − 푥 + 8− 4푥 + 푑푥
= 2푥 − + 8푥 − +
= 2(4 − 1) − (16 − 1) + 8(2 − 1) − (8 − 1) + (32 − 1)
= 6− + 8− + = = (Ans).
28. Find the volume bounded by the plane xy, the cylinder 2 2 1x y and plane 3x y z .
Ans: - Volume = ∭푑푥푑푦푑푧
= ∫ ∫ ∫ 푑푥푑푦푑푧√√ = ∫ ∫ 푧] 푑푥푑푦√
√
= ∫ ∫ (3 − 푥 − 푦)푑푥푑푦 = ∫ 3푦 − 푥푦 −√
√푑푥√
√
= ∫ (3− 푥)(√1− 푥 + √1− 푥 ) − ( )푑푥
= ∫ 2(3− 푥)√1− 푥 푑푥 = ∫ 6√1 − 푥 − 푥√1− 푥 푑푥
= 6 √ + 6 푠푖푛 푥 + ( ) /
= 6(0 − 0) + 3 + + (0 − 0) = 3휋 (Ans).
29. Prove that: 1 12
4 40 0 4 21 1
x dx dxx x
.
Ans: - ∫ √ Let 푥 = 푠푖푛푡 ⇒ 2푥푑푥 = 푐표푠푡푑푡 ⇒ 푑푥 = =
√푑푡
⇒ ∫ √= ∫ /
√푑푡 = ∫ √푠푖푛푡푑푡/ =
Γ Γ
Γ=
Γ
Γ√
⇒ ∫ √=
Γ
Γ√ ----------------------------------(1)
∫ √ Let 푥 = 푡푎푛푡 ⇒ 2푥푑푥 = 푠푒푐 푡푑푡 ⇒ 푑푥 = =
√푑푡
⇒ ∫ √= ∫ /
√푑푡 = ∫
√푑푡/ = ∫ √
푑푡/
⇒ ∫ √=
√∫ √
푑푡/ Let 2푡 = 휃 ⇒ 2푑푡 = 푑휃 ⇒ 푑푡 =
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
⇒ ∫ √=
√∫ √
/ =√
Γ Γ
Γ=
√
Γ
Γ√ --------------------(2)
Now, ∫ √× ∫ √
=Γ
Γ√ ×
√
Γ
Γ√
⇒ ∫ √× ∫ √
=√
Γ
Γ휋 =
√
Γ
Γ휋 =
√ (Ans).
(May-Jun-2010) 30. Compute, 훽 , .
Ans: -
9 7 7 5 3 1 5 3 1. . . . . . . .9 7 52 2 2 2 2 2 2 2 2,9 72 2 7.6.5.4.3.2.1 20482 2
(Ans).
31. Show that : ∫ √= √ ∫ √
=.
.
Ans: -
1
061 x
dx putting ddxxx cossin
31sinsin 3/23/13
2/
0
1)2/1(21)6/1(22/
0
3/22/
0
3/21
06
cossin31sin
31
cos
cossin31
1
dd
d
xdx
3/2
.6/161
3/222/1.6/1
31
1
1
06
xdx
--------------------(1)
By Legendre’s Duplication method nnn n 222
112
---------(2)
Now, putting 61
n in eqn (2) we get 3/12
3/26/1 3/2
3/2
3/126/13/2
So, equation (1) becomes
3/23/23/12
61
3/23/12
3/261
1
3/23/21
06
xdx
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
233/2
2
23/21
06 )3/sin(/
3/1.261
3/23/13/13/12
61
1
x
dx as
nrn
rnr
sin1.
3/7
333/21
06 2
3/1.4.
3.3/1.261
1
xdx
(Proved)--------(I)
Again
1
0312
3
xdx
putting ddxx cossin32sin 3/13/2
2/
0
1)2/1(21)3/1(22/
0
3/12/
0
3/11
03
cossin3
1cossin3
1cos
cossin32
23
123
dd
d
xdx
6/5
.3/132
16/52
2/1.3/13
1
123 1
03
xdx
-----------------(3)
By Legendre’s Duplication method nnn n 222
112
---------(4)
Now, putting 31
n in eqn (4) we get 3/22
6/53/1 3/2
3/1
3/226/53/2
So, equation (3) becomes
3/2
3/132
1
3/13/22
3/132
112
3 2
3/53/2
1
03
xdx
)3/sin(/
3/132
13/23/13/13/1
321
123 3
3/5
2
3/5
1
03
xdx
as
nrn
rnr
sin1.
3/7
33
3/5
1
03 2
3/1233/1
321
123
xdx
(Proved)--------------(II)
From (I) and (II)
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
3/7
3
1
03
1
06 2
31
123
1
xdx
xdx
(Proved)
32. Change the order of following integration: ∫ ∫ 푥푦 푑푥 푑푦 and evaluate.
Ans: From 2
1 2
0
x
x
I xydxdy
, region of integration is 2 , 2 , 0, 1y x y x x x
By changing of order of integration we get two regions
In one region limit of x is 0,x x y and limit of y is 0, 1y y
In another region limit of x is 0, 2x x y and limit of y is 1, 2y y
So, 2
21 2 1 2
0 0 0 1 0
y yx
x
I xydxdy xydydx xydydx
21 2
0 0 1 0
y y
I xydx dy xydx dy
21 22 2
0 10 02 2
y yx y x yI dy dy
1 22 2
0 1
0 (2 ) 02 2
y y yI dy dy
1 22 3
2
0 1
2 22 2y yI dy y y dy
1 23 3 4
2
0 1
26 3 8y y yI y
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
1 16 16 2 10 4 16 3 8 3 8
I
83
249
24155164
243162448128964
I (Ans).
33. Find the volume enclosed by the cylinders
x2 + y2 = 2ax and z2 = 2ax.
Ans: - 2 2 2 2 2 2 22 2 0 ( )x y ax x y ax x a y a
At 2 20, ( ) 0,2y x a a x a
Volume = 2
2
2 2 2
0 22
a ax x ax
axax x
dxdydz
Volume = 2
2
2 22
20 2
a ax xax
axax x
z dxdy
Volume = 2
2
2
2
2 2 22
20 02
2 2 2 2a ax x a
ax x
ax xax x
axdxdy a x y dx
Volume = 2
2
0
2 2 .2 2a
a x ax x dx
Volume = 2
2
0
4 2 . 2a
a x ax x dx Let 22 sin 4 sin cosx a dx a d
Volume = /2
0
4 2 2 sin .2 sin cos .4 sin cosa a a a d
Volume = /2 3
3 3 2 3
0
2 12864 sin .cos 64 .5 3 15
aa d a
(Ans).
(Nov –Dec 2010)
34. Write the relation between beta and gamma function. ( ) ( )( , ) .( )m nB m nm n
35. Change the order of integration and evaluate : 2
4 2
0 /4
a ax
x a
I dydx
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2
4 2
0 /4
a ax
x a
I dydx By changing the order of the integration we will have figure below and the
solution will be 2 2
24 2 4
0 0/4 /4
aya ax a
x a y a
dydx dydx
=
24
0 2/4
ayadydx
y a =
4
0
24 22/40 2/4
aaya aydydx x dyy ay a
4 42
0 0
22 / 4
2/4
a aaydxdy ay y a dy
y a
44 3 2 2 2
0 0
2 3/2 1 32 16 1623 / 2 4 3 3 3 32/4
aa ay
y y a a adxdy aa
y a
(Ans)
36. Evaluate :
2 22 11 1
0 0 0
x yx
I xyzdydxdz
Sol.
2 22 22 2 111 1 1 1 2
0 0 0 0 0 02
x yx yx x xyzI xyzdydxdz dydx
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 22 2 2 211 1 1 1
0 0 0 0 0
12
x yx x xy x yI xyzdydxdz dydx
2 22 2 3 311 1 1 1
0 0 0 0 0 2
x yx x xy x y y xI xyzdydxdz dydx
2
2 22
12 3 2 4
11 1 1
0 0 0 0
0
2 2 42
x
x yxxy x y y x
I xyzdydxdz dx
2 222 2 4
2 3 2 211 1 1
0 0 0 0
1 1 1
4 4 8
x yx x x x x x xI xyzdydxdz dx
2 22 22 3 2 211 1 1
0 0 0 0
1 1 14 4 8
x yx x x x x x xI xyzdydxdz dx
2 22 3 3 5 5 311 1 1
0 0 0 0
24 4 8
x yx x x x x x x xI xyzdydxdz dx
2 22
12 4 4 6 2 6 4
11 1
0 0 0
0
22 4 4 6 2 6 4
4 4 8
x yxx x x x x x x
I xyzdydxdz
2 22 11 1
0 0 0
1 1 1 1 1 1 18 16 16 24 16 48 16
x yx
I xyzdydxdz
2 22 11 1
0 0 0
1 1 1 1 1 1 1 18 16 16 24 16 48 16 48
x yx
I xyzdydxdz
(Ans)
37. Given 1
0 1 sin( )
nx dxx n
Show that 1
sin( )n n
n
Hence evaluate 40 1
dyy
Sol. Given that
1
0 1 sin( )
nx dxx n
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
We know that ( , ) m nm nm n
1
0 1
n
m nx m ndx
m nx
1 1let m n m n
1
0
1 111
n
m nx n ndx n nx
1
0
1sin( )1
nx dx n nnx
Therefore 1sin( )
n nn
Now for
40 1
dyy
……………………………………….1
Let 4y =x
Y= 1/4x
Y= 3/414
dy x dx
3/4 1/4 1
40 0 0
1 1 1 1 111 4 1 4 1 4 sin 4 4 2 2sin
4 2
dy x dx x dxy x x
(Ans)
(May-Jun-2011)
38. Evaluate 3
2
Ans. We know that
( 1) ( )n n n
( 1)( ) nnn
3 13 2
322
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
13 2
322
………………………..1
As we know that
( 1)( ) nnn
Therefore we will have
1 111 2 2
1 122 2
…………………..2
From 1 and 2 we will have
121 1
3 2 23 3 12
2 2 2
143 2
2 3
Ans
39. Evaluate the following integral by changing of order of integration
2
4 2 20
a a
ax
y dxdyy a x
.
Ans:
Here region of integration is x varies from 0 to a, y varies from ax to a.
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
After changing order y varies from 0 to a, x varies from 0 to2y
a.
2 /2 2
4 2 2 4 2 20 0 0
y aa a a
ax
y dxdy y dydxy a x y a x
putting 2a as common we get
2
2
4 2 2 220 0 0 2
2a a a
ax
y y dydxay dxdy ay a x y x
a
2
2 21
24 2 2 220 0 0 02
0
22
sina a a a
ax
yay y dxay dxdy y xa dy dy
a yy a x y x aa
2
2 2 21 1 1
2 24 2 20 0 0
0
2
sin sin sin 0a a a a
ax
ya y
y dxdy y x y ady dya ay yy a x
a a
2 2 2
4 2 20 0 0
02 2
a a a a
ax
y dxdy y ydy dya ay a x
2 2 2
3
04 2 20 02 6 6
a a aa
ax
y dxdy y ady ya ay a x
(Ans)
40. State and prove relation between beta and gamma function. Ans: The Beta function is defined as
1 11 1
0 0
( , ) (1 ) , , 0(1 )
nm n
m nxB m n x x dx dx m nx
.
We know that 1
0
( ) .x nn e x dx
By putting x az dx adz we get
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
1
0
( ) .( )az nn e az adz
1
0
( ) .n az nn a e z dz
Putting z = x we get
1
0
( ) .n ax nn a e x dx
Now, taking a = z we get
1
0
( ) .zx n nn e x z dx
By multiplying 1z me z both side we get
1 (1 ) 1 1
0
( ) .z m z x n m nn e z e x z dx
Now taking integration with respect to z both side from 0z we get
1 1 (1 ) 1
0 0 0
( ) .z m n z x m nn e z dz x e z dzdx
Now, let (1 )1 1
y dyz x y z dzx x
we get
1
1
0 0
.( ) ( )(1 )
y m nn
m n
e y dyn m x dxx
1
1
0 0
( ) ( ) .(1 )
ny m n
m n
xn m e y dy dxx
1
0
( ) ( ) ( )(1 )
n
m nxm n m n dxx
1
0
( ) ( ) ( ) ( ). ( , )(1 )
n
m nxm n m n dx m n B m nx
( ) ( )( , ) .( )m nB m nm n
(proved)
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
41. Find the volume common to gas cylinder 2 2 2 2 2 2,x y a x z a .
Ans:
2 2 2 2 2 2 2 2
2 2 2 2 0 0 0
Volume 8a a x a x a a x a x
a a x a x
dxdydz dxdydz
2 2 2 2
2 22 2
00 0 0 0
Volume 8 8a a x a a x
a xz dxdy a x dxdy
2 2
2 2 2 2 2 2 2 20
0 0 0
Volume 8 . 8 . 8 ( )a a a
a xa x y dx a x a x dx a x dx
3 3 3 3
2 2
0
2 16Volume 8 8 83 3 3 3
ax a a aa x a
(Ans).
(Nov –Dec 2011)
42. Write the relation between Beta and Gamma function
Ans: - The relation between beta & gamma function is ( ) ( )( , )( )m nm nm n
.
43. Change the order of integration 0
y
x
e dxdyy
and hence evaluate it.
Ans 0
y
x
e dxdyy
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
0
y
x
e dxdyy
by changing the order of the equation we will have 0 0
y ye dxdyy
0 0 0 0
y yy ye edxdy dx dyy y
0 0 0 0
yy y ye edxdy x dyy y
0 0 0 0
y y yye edxdy y dy e dy
y y
0
0 0 0 0
1y y y
y ye edxdy y dy e dy ey y
(Ans).
44. Find the volume common to the cylinder 2 2 2 2 2 2,x y a x z a .
Ans:
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 2 2 2 2 2 2 2
2 2 2 2 0 0 0
Volume 8a a x a x a a x a x
a a x a x
dxdydz dxdydz
2 2 2 2
2 22 2
00 0 0 0
Volume 8 8a a x a a x
a xz dxdy a x dxdy
2 2
2 2 2 2 2 2 2 20
0 0 0
Volume 8 . 8 . 8 ( )a a a
a xa x y dx a x a x dx a x dx
3 3 3 3
2 2
0
2 16Volume 8 8 83 3 3 3
ax a a aa x a
(Ans).
45. Prove that 2
1 2 12 2 1n
nnn
Ans.
2 11 2 1 12 2 2
nnn
2 1 2 1 2 1 2 1 2 3 2 31 1 ..........2 2 2 2 2 2 2
n n n n n nn
2 1 2 1 2 3 2 5 2 71 1 11 ............2 2 2 2 2 2 2 2
n n n n nn
2 1 . 2 3 . 2 5 . 2 7 ................112 2
n n n nn n
2 . 2 1 . 2 2 . 2 3 . 2 4 . 2 5 . 2 6 . 2 7 ................112 2 .2 . 2 2 . 2 4 . 2 6 ......................2
n n n n n n n nn n n n n n
1 2 !2 2 .2 . 1 . 2 . 3 ......................1
nn n n n n n
1 2 !22 2 . !
nn n n
1 2 122 2 . 1
nn n n
hence Proved
(May-Jun-2012)
46. Evaluate the following Integral:
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
2 3 2
0 1xy dxdy
3 23 22 3 2 22
0 1 0 01 0
26 26 52 / 33 3 6
xy x xxy dy dx dx dx
47. Evaluate the following integral by changing the order of integration:
0
y
x
e dydxy
Ans 0
y
x
e dxdyy
0
y
x
e dxdyy
by changing the order of the equation we will have 0 0
y ye dxdyy
0 0 0 0
y yy ye edxdy dx dyy y
0 0 0 0
yy y ye edxdy x dyy y
0 0 0 0
y y yye edxdy y dy e dy
y y
0
0 0 0 0
1y y y
y ye edxdy y dy e dy ey y
(Ans).
48. Given 1
0 1 sin
nx dxx n
, show that ( ) (1 )sin
n nn
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Sol. Given that
1
0 1 sin( )
nx dxx n
We know that ( , ) m nm nm n
1
0 1
n
m nx m ndx
m nx
1 1let m n m n
1
0
1 111
n
m nx n ndx n nx
1
0
1sin( )1
nx dx n nnx
Therefore 1sin( )
n nn
Now for
40 1
dyy
……………………………………….1
Let 4y =x
Y= 1/4x
Y= 3/414
dy x dx
3/4 1/4 1
40 0 0
1 1 1 1 111 4 1 4 1 4 sin 4 4 2 2sin
4 2
dy x dx x dxy x x
(Ans)
49. Find the area included between the parabola: 24y x x and the line y x Ans:
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
23 4
0
Areax x
x
dxdy
2
3 3 34 2 2
0 0 0
Area (4 ) (3 )x x
xy dx x x x dx x x dx
32 3
0
27 27 27 9Area 3 0 02 3 2 3 6 2x x
(Ans)
(Nov-Dec2012)
50. Define Beta and Gamma Function.
Ans:
Beta function
The Beta function is defined as 1 1
1 1
0 0
( , ) (1 ) , , 0(1 )
nm n
m nxB m n x x dx dx m nx
.
Gamma Function: - The Gamma function )(n is defined as dxxen nx
0
1)( for 0n .
It is also know as Second Eulerian Integral
51. Evaluate0
y
x
e dydxy
Ans 0
y
x
e dxdyy
0
y
x
e dxdyy
by changing the order of the equation we will have 0 0
y ye dxdyy
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
0 0 0 0
y yy ye edxdy dx dyy y
0 0 0 0
yy y ye edxdy x dyy y
0 0 0 0
y y yye edxdy y dy e dy
y y
0
0 0 0 0
1y y y
y ye edxdy y dy e dy ey y
(Ans).
52. Prove that
Hence evaluate 1 3
0(log )x x dx
Sol.
1
10
1 !log
1
nn
n
nmx x dxm
1
0
log nmx x dx ……………………………………………….1
Let log tx t x e tdx e dt
1 becomes 0
0
1 1n nm t m te t dt e t dt
……………………………….2
1
1dpm t p dt
m
2 becomes 0
11 1
nppe dpm m
0
11 11
nn
p ne p dpm
=
0
1 1 11 11
nn npe p dp
m
0
1 11 11 11 1 11 1
n nn nnpe p dp n
m m
1 1 !1 1 1 11 1
nn
n nn nm m
Hence Proved
1,0,)1(
!.)1().(log 1
1
0
mnm
ndxxx n
nnm
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
As we know that
1
10
1 !log
1
nn
n
nmx x dxm
there fore
31 3
3 10
1 3! 6 3(log )16 81 1
x x dx
=
53. Find the volume of the cylinder 2 2 4,x y and the planes 4y z and z=0
Ans:
From the figure 4z y is to be integrated over the circle 2 2 4x y in the xy plane.
To cover the shaded half of the circle, x varies from 0 to 24 y
So, 2 24 42 2
2 0 2 0
Volume 2 2 (4 )y y
zdxdy y dxdy
2
24
02
Volume 2 (4 ) yy x dy
2
2
2
Volume 2 (4 ) 4y y dy
2 2
2 2
2 2
Volume 8 4 2 4y dy y y dy
22
1
2
4 4Volume 8 sin 02 2 2
y y y
{second term is zero because of odd function}
1 1Volume 8 0 0 2sin 1 2sin ( 1)
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
1Volume 8 4sin 1 8 4 162 (Ans).
APPLIED MATHS – II
300214 B.E.(Second Semester)
Unit IV – Vector Calculus
(May-Jun-2006)
1. Show that 2 2( 1)n nr n n r where 푟 = 푥푖 + 푦푗 + 푧푘.
Ans: - 222ˆˆˆ zyxkzjyixRr
2222222nn
n zyxzyxr
2 2 2 2 2 2 2 2 22 2 2ˆ ˆˆ ˆ ˆ ˆ
n n nn n n
nx y z x y z x y zr r rr i j k i j k
x y z x y z
1 1 12 2 2 2 2 2 2 2 22 2 2ˆˆ ˆ.2 .2 .22
n n nn nr i x y z x j x y z y k x y z z
2
2 2 2 2 ˆˆ ˆn
nr n x y z ix jy kz
2n nr nr R
…………………………………1
2 2n nr nr R
2 2ˆ ˆ ˆn nr i j k nr Rx y z
2 2 2 2n n n nr nr R i nr R j nr R kx y z
2 2 2 2n n n nr nr x nr y nr z
x y z because ˆˆ ˆR xi yj zk
2 2 2 2n n n nr n r x r y r z
x y z
2 2 2 2 2 2 2n n n n n n nr n r x x r r y y r r z z r
x x y y z z
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 2 3 2 3 2 3.1 2 .1 2 .1 2
r r rn n n n n n nr n r x n r r y n r r z n rx y z
2 2 3 2 3 2 3.1 2 .1 2 .1 2yx zn n n n n n nr n r x n r r y n r r z n r
r r r
2 2 2 4 2 2 4 2 2 4.1 2 .1 2 .1 2n n n n n n nr n r x n r r y n r r z n r
2 2 4 2 2 23 2n n nr n r n r x y z
2 2 4 23 2n n nr n r x n r r
2 2 23 2n n nr n r n r
2 2 3 2n nr nr n
2 2 1n nr nr n
2 21n nr n n r hence proved
2. Verify Stoke’s theorem for the vector field on the upper half surface of , bounded by its projection on the xy-plane.
Ans: Stoke’s theorem states that ˆ. ( ).C S
F dr curl F nds
1. Here C is unit circle 2 2 1, 0x y z
2 2 ˆ ˆˆ ˆ ˆ ˆ. (2 ) .F dr x y i yz j y zk idx jdy kdz
2 2. (2 )F dr x y dx yz dy y zdz
Here C is z = 0, dz = 0 and cos , sinx y
. (2 )F dr x y dx
So, 2
0
. (2 ) (2cos sin )( sin )C C
F dr x y dx d
2
0
1 cos 2. sin 22C
F dr d
2
0
cos 2 sin 2.2 2 4C
F dr
1 1 2 0 0 0.2 2 4C
F dr
-------(1)
2. Now, 2 2 ˆˆ ˆ(2 )F x y i yz j y zk
2 2 ˆˆ ˆ(2 )F x y i yz j y zk
2 2 2 1x y z
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 2
ˆˆ ˆ
( )
2
i j k
curl Fx y z
x y yz y z
ˆ ˆˆ ˆ( ) ( 2 2 ) (0 0) (0 1)curl F i yz yz j k k
By using spherical co-ordinate we have
ˆˆ ˆˆ sin cos sin sin cosn i j k
So, ˆ( ). coscurl F n
Hence /2 2
0 0
ˆ( ). cos .sinS
curl F nds d d
/2 2
0 0
ˆ( ). cos .sinS
curl F nds d d
/22
2
00
sinˆ( ).2S
curl F nds
1 0ˆ( ). (2 0)2S
curl F nds
------------(2)
From (1) and (2) it verifies Stoke’s theorem. (Ans)
3. Use divergence theorem to evaluate where and S is the
surface of the sphere .
Ans: By divergence theorem .S V
F ds divFdv
3 3 3. ( ) ( ) ( )S V
F ds x y z dvx y z
2 2 2. 3S V
F ds x y z dxdydz
2 2 2 2 2. 3S V
F ds a dxdydz x y z a given
3 3
2 4 4. 33 3S V
a aF ds a volume of sphere dxdydz
5. 4S
F ds a Ans
S
dsF
kzjyixF ˆˆˆ 333
2222 azyx
CSVTU II SEMESTER SOLUTION ALL UNITS
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Nov-Dec, 2006
4. Verify Green’s theorem for: 2 2( )C
xy y dx x dy where C is bounded by 2,y x y x .
Ans:
By Green’s theorem 2 11 2
C S
F FF dx F dy dxdyx y
(1) Here 2 2ˆ ˆ( )F xy y i x j
So, 2 12 , 2F Fx x yx y
2
12 1
0
2 2x
S x
F F dxdy x x y dxdyx y
2
122 1
0
x
xS
F F dxdy xy y dxx y
1
2 2 3 42 1
0S
F F dxdy x x x x dxx y
11 5 4
4 32 1
0 05 4S
F F x xdxdy x x dxx y
2 1 1 1 15 4 20S
F F dxdyx y
----------------(1)
(2) Now we have to find line integral along two curves. Along 2 , 2y x dy xdx .
So, 1 1
2 21 2 ( )
C C
F dx F dy xy y dx x dy
CSVTU II SEMESTER SOLUTION ALL UNITS
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1
12 4 2
1 20
( . ) .2C
F dx F dy x x x dx x xdx
1
11 4 53 4
1 20 0
3(3 )4 5C
x xF dx F dy x x dx
1
1 23 1 194 5 20C
F dx F dy
Along ,y x dy dx
So, 2 2
2 21 2 ( )
C C
F dx F dy xy y dx x dy
2
02 2
1 21
( . ) .C
F dx F dy x x x dx x dx
2
002 3
1 2 11
3 0 1 1C
F dx F dy x dx x
Now, 1 2
1 2 1 2 1 219 1120 20C C C
F dx F dy F dx F dy F dx F dy ----------(2)
From (1) and (2) it verifies Green’s theorem.
5. Evaluate .S
F dS
where 2 2 ˆˆ ˆ4 2F xi y j z k
and S is the surface bounding the region
2 2 4x y , 0, 3z z . Ans:
CSVTU II SEMESTER SOLUTION ALL UNITS
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By divergence theorem .S V
F ds divFdv
2 2. (4 ) ( 2 ) ( )S V
F ds x y z dvx y z
. 4 4 2S V
F ds y z dxdydz
2
2
2 4 3
2 04
. 4 4 2x
S x
F ds y z dxdydz
2
2
2 432
02 4
. 4 4x
zS x
F ds z yz z dxdy
2 2
2 2
2 4 2 4
2 24 4
. 12 12 9 21 12x x
S x x
F ds y dxdy y dxdy
2
2
242
42
. 21 6x
xS
F ds y y dx
2
2
2
. 42 4 0S
F ds x dx
2
2
2
. 42 4S
F ds x dx
22
1
2
4 4. 42 sin2 2 2S
x x xF ds
1 1. 42 0 0 2sin 1 2sin ( 1)S
F ds
1. 42 4sin 1 42 4 842S
F ds (Ans).
CSVTU II SEMESTER SOLUTION ALL UNITS
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6. Evaluate C
F dR
, where ˆˆ ˆ2F yi zj xk
and the curve cos , sin , 2cosx t y t z t
from 0 / 2t to .
Ans: Here ˆˆ ˆ2F yi zj xk
So,
ˆˆ ˆˆˆ ˆ2 ( ) (2 ) (2 )
i j kF dR y z x i zdz xdy j ydz xdx k ydy zdx
dx dy dz
Now, ˆˆ ˆ( ) (2 ) (2 )C C
F dR i zdz xdy j ydz xdx k ydy zdx
2 2/2
0
ˆ ˆ(4sin cos cos ) ( 4sin sin cos )ˆ(2sin cos 2sin cos )C
i t tdt tdt j tdt t tdtF dR
k t tdt t tdt
/2
2 2
0
ˆ ˆ(4sin cos cos ) (4sin sin cos )C
F dR i t t t j t t t dt
/2
0
1 cos 2 sin 2ˆ ˆ2sin 2 2(1 cos 2 )2 2C
t tF dR i t j t dt
/2
0
sin 2cos 22ˆ ˆcos 2 2 sin 2
2 4C
tt tF dR i t j t t
0 0 0 1 12ˆ ˆ1 1 2 0 0 02 2 4C
F dR i j
1ˆ ˆ24 2C
F dR i j
(Ans)
May-June, 2007
7. Find the work done in moving a particle in the force field: 2 ˆˆ ˆ3 (2 )F x i xz y j zk
, along:
(a) The straight line from (0, 0, 0) to (2, 1, 3). Ans: The equation of straight line joining (0, 0, 0) and (2, 1, 3) is
0 0 0 2 , , 3
2 0 1 0 3 0x y z t x t y t z t
where 0t to 1t
So, Work done .C
F dr
CSVTU II SEMESTER SOLUTION ALL UNITS
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1
2
0
Work done 3(2 ) .2 (4 (3 )) 3 .3t dt t t t dt t dt
1 1
2 2 2
0 0
Work done 24 12 8 36 8t t t dt t t dt
13 2
0Work done 12 4 12 4 0 0 16t t
(b) The curve defined by 2 34 ,3 8x y x z from 0, 2x x . Ans: 2 34 ,3 8x y x z
Let 2 33,4 8t tx t y z where 0, 2t t
So, Work done .C
F dr
1 4 2 3 2
2
0
3 3 9Work done 3 .4 4 2 8 8t t t t tt dt dt dt
1 15 3 5 3 5
2 2
0 0
3 27 51Work done 3 38 8 64 8 64t t t t tt dt t dt
24 6
3
0
17 16 17 64 1 17Work done 8 8 1632 128 32 128 2 2t tt
(Ans)
8. Evaluate divergence and curl of 2 2 3 ˆˆ ˆ3 5F x i xy j xyz k
at the point (1, 2, 3).
Ans: 2 2 3 ˆˆ ˆ3 5F x i xy j xyz k
2( ) 6 10 3div F x xy xyz
(1,2,3)( ) 6 20 54 80div F
(Ans).
Now,
2 2 3
ˆˆ ˆ
( )
3 5
i j k
curl Fx y zx xy xyz
3 3 2ˆˆ ˆ( ) ( 0) ( 0) (5 0)curl F i xz j yz k y
3 3 2 ˆˆ ˆ( ) 5curl F xz i yz j y k
(1,2,3)ˆˆ ˆ( ) 9 54 20curl F i j k
(Ans).
9. Use divergence theorem to evaluate .S
F dS
where 2 2 ˆˆ ˆ4 2F xi y j z k
and S is the
surface bounding the region 2 2 4x y , 0, 3z z . Ans:
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
By divergence theorem .S V
F ds divFdv
2 2. (4 ) ( 2 ) ( )S V
F ds x y z dvx y z
. 4 4 2S V
F ds y z dxdydz
2
2
2 4 3
2 04
. 4 4 2x
S x
F ds y z dxdydz
2
2
2 432
02 4
. 4 4x
zS x
F ds z yz z dxdy
2 2
2 2
2 4 2 4
2 24 4
. 12 12 9 21 12x x
S x x
F ds y dxdy y dxdy
2
2
242
42
. 21 6x
xS
F ds y y dx
2
2
2
. 42 4 0S
F ds x dx
2
2
2
. 42 4S
F ds x dx
22
1
2
4 4. 42 sin2 2 2S
x x xF ds
CSVTU II SEMESTER SOLUTION ALL UNITS
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1 1. 42 0 0 2sin 1 2sin ( 1)S
F ds
1. 42 4sin 1 42 4 842S
F ds (Ans).
Nov - Dec, 2007 10. Define Gradient.
Ans. Gradient of a scalar point function : -Let f be a scalar point function then vector point function f is known as gradient of f. Its geometrical interpretation is grad(f) is the normal to the
surface f = const.So, ˆˆ ˆ( ) f f fgrad f f i j kx y z
11. Find the values of a and b such that the surfaces xabyzax )2(2 and 44 32 zyx cut orthogonally at (1, -1, 2).
Ans: - Here 0)2(2 xabyzaxF and 044 32 zyxG
So, kbyjbziaaxF ˆˆˆ)22( and kzjxixyG ˆ3ˆ4ˆ8 22
kbjbiaF ˆˆ2ˆ)2()2,1,1( kjiG ˆ12ˆ4ˆ8)2,1,1(
As surfaces are orthogonal, so 01281680 bbaGF
04201684 baab ---------------(1)
As point (1, -1, 2) lies on the surface, so it satisfies surface F.
10220)2(2 bbaba
Putting in (1) we get 2/50412 aa
So, 1,2/5 ba (Ans).
12. If 2 2 2 ˆˆ ˆ( ) 2F x y i xyj y xy k
then find dif F and Curl F . Ans: If 2 2 2 ˆˆ ˆ( ) 2F x y i xyj y xy k
then dif F will be defined as
2 2 2ˆ ˆˆ ˆ ˆ ˆ. . ( ) 2f f fF i j k x y i xyj y xy kx y z
. (2 2F x x
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If 2 2 2 ˆˆ ˆ( ) 2F x y i xyj y xy k
then Curl F will be defined as
2 2 2ˆ ˆˆ ˆ ˆ ˆ( ) 2f f fF i j k x y i xyj y xy kx y z
2 2 2
ˆˆ ˆ
( ) (2 )
i j k
Fx y z
x y xy y xy
ˆˆ ˆ2 2 2F y x i y j y y k
ˆ ˆ2F y x i y j
Ans.
13. Verify Stoke’s theorem for the vector field 2 2 ˆ ˆ( ) 2F x y i xyj
taken around the rectangle bounded by , 0x a y to y b
Ans: Stoke’s theorem states that ˆ. ( ).C S
F dr curl F nds
……………………………1
Here C is Rectangle bounded by ,x a y b
2 2 ˆˆ ˆ ˆ ˆ. ( ) 2 .F dr x y i xyj idx jdy kdz
2 2. ( ) 2F dr x y dx xydy
2 2. ( ) 2F dr x y dx xydy
So, 2 2. ( ) 2C C DA AB BC CD
F dr x y dx xydy
0
2 2 2
0
2 2a b a
DA AB BC CD a a b
x dx aydy x b dx aydy
CSVTU II SEMESTER SOLUTION ALL UNITS
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0
2 2 2
0 0
4 2 2a b a b
DA AB BC CD a a b
x dx aydy x b dx aydy aydy
3 2 3
2
0
43 2 3
a b a
DA AB BC CD a a
x ay x b x
3 3
2 22 22 23 3DA AB BC CD
a aab ab
24DA AB BC CD
ab …………………………………………2
ˆ( ).S
curl F nds
( )curl F F
=
2 2
ˆˆ ˆ
( ) ( 2 ) 0
i j k
Fx y z
x y xy
2 2 4F y y k yk
0
ˆ( ). ( )b a
S a
curl F nds curl F Nds
0 0 0
( ) 4 4b a b a b a
a a a
curl F Nds yk kds y dxdy
0 0 0
( ) 4 4b a b a b
a
aa a
curl F Nds y dxdy yx dy
0 0 0
( ) 4 4 2b a b a b
a a
curl F Nds y dxdy ay dy
2 20
0 0 0
( ) 4 4 2 4 4b a b a b
b
a a
curl F Nds y dxdy ay dy a y ab
………………………………..3
From 1 , 2, and 3rd we can Verify the Stokes’s theorem
(May-Jun-2008)
14. Define gradient.
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Ans. Gradient of a scalar point function : -Let f be a scalar point function then vector point function f is known as gradient of f. Its geometrical interpretation is grad(f) is the normal to the
surface f = const.So, ˆˆ ˆ( ) f f fgrad f f i j kx y z
15. Find the values of a and b such that the surfaces and cut orthogonally at (1, -1, 2).
Ans: - Here 0)2(2 xabyzaxF and 044 32 zyxG
So, kbyjbziaaxF ˆˆˆ)22( and kzjxixyG ˆ3ˆ4ˆ8 22
kbjbiaF ˆˆ2ˆ)2()2,1,1( kjiG ˆ12ˆ4ˆ8)2,1,1(
As surfaces are orthogonal, so 01281680 bbaGF 04201684 baab ---------------(1) As point (1, -1, 2) lies on the surface, so it satisfies surface F. 10220)2(2 bbaba
Putting in (1) we get 2/50412 aa So, 1,2/5 ba (Ans).
16. Prove that : 2 2( ) ''( ) '( )f r f r f r
r
Ans: - 222ˆˆˆ zyxkzjyixRr
2 ( ) . ( )f r f r
( )f r =
ˆ ˆ ˆ ( )i j k f rx y z
ˆ ˆ ˆ( ) ( ) ( ) ( )f r i f r j f r k f rx y z
/ / /ˆ( ) ( )r r r
f r f r i f j f kx y z
/ ˆ( ) ( )yx z
f r f r i j kr r r
/( )
( )f r R
f rr
2
/( )( ) . ( ) .
f r Rf r f r
r
xabyzax )2(2 44 32 zyx
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 / /( ) ( ) . . ( )
R Rf r f r f r
r r
2 / /ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) . ( )
R Rf r f r i j k i j k f r
x y z r r x y z
2 / /ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) . ( )
2 2 2 2 2 2 2 2 2x x x R
f r f r i j k i j k i j k f rx y z r x y zx y z x y z x y z
2 / // ˆ ˆ ˆ( ) ( ) ( )
2 2 2 2 2 2 2 2 2yx z R r r r
f r f r f r i j kx y z r x y zx y z x y z x y z
2 / // ˆ ˆ ˆ( ) ( ) ( )
2 2 2x R r r r
f r f r f r i j kx r x y zx y z
2
2 2 2 2 2 2/ // ˆ ˆ ˆ( ) ( ) ( )22 2 2
x y z x x x y z yR x zx xf r f r f r i j kr r r r
x y z
2
22 2 22 2 22/ //( ) ( ) ( )22 2 2
xx y z x
x y z R Rf r f r f r
r rx y z
2
22 2 2 2/ //( ) ( ) ( )
2 2 2 2 2 2
x y z x Rf r f r f r
rx y z x y z
2
221/ //( ) ( ) ( )3/22 2 2 2 2 2
x Rf r f r f r
rx y z x y z
2
22 2 23/ //( ) ( ) ( )3/22 2 2 2 2 2
x y z Rf r f r f r
rx y z x y z
CSVTU II SEMESTER SOLUTION ALL UNITS
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2
223/ //( ) ( ) ( )3/22 2 2 2
r Rf r f r f r
rx y z r
2
23 1/ //( ) ( ) ( )
2 2 2
Rf r f r f r
r rx y z
2
23 1/ //( ) ( ) ( )
2 2 2 2 2 2
Rf r f r f r
rx y z x y z
2
22/ //( ) ( ) ( )
2 2 2
Rf r f r f r
rx y z
222
2/ // 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) ( ).( )2 2 2
rf r f r f r R xi yj zk xi yj zk x y z r
rx y z
2 2/ //( ) ( ) ( ) Prf r f r f r Hence oved
r
17. Evaluate the line integral 2 2 2( ) ( )
C
x xy dx x y dy where C is square formed by the
lines 푦 = ±1, 푥 = ± 1. Ans. Here C is Square bounded by 1, 1x y
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
So, 2 2 2. ( ) ( )C C DA AB BC CD
F dr x xy dx x y dy
02 2 2
0
2 2a b a
DA AB BC CD a a b
x dx aydy x b dx aydy
1 1 1 1
2 2 2 2
1 1 1 1
1 1DA AB BC CD
x x dx y dy x x dx y dy
1 1 1 1 1 1
2 2 2 2 2 2
1 1 1 1 1 1
1 1 1 1DA AB BC CD
x x dx y dy x x dx y dy y dy y dy
1 1
2 2
1 1DA AB BC CD
x x dx x x dx
1 1 1 13 2 3 2
1 1 1 13 2 3 2DA AB BC CD
x x x x
2 2 03 3DA AB BC CD
2 2 2. ( ) ( )C C DA AB BC CD
F dr x xy dx x y dy
= 0
(Nov –Dec 2008)
18. Evaluate 푑푖푣 (3푥 횤̂ + 5푥푦 횥̂ + 5푥푦푧 푘) at (1, 2, 3).
CSVTU II SEMESTER SOLUTION ALL UNITS
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Ans: - 2 2 3 2 2 3ˆˆ ˆ3 5 5 3 5 5div x i xy j xyz k x xy xyzx y z
2 2 3 2ˆˆ ˆ3 5 5 6 10 15div x i xy j xyz k x xy xyz
2 2 3
(1,2,3)ˆˆ ˆ3 5 5 6 20 270 296div x i xy j xyz k (Ans).
19. What is the directional derivative of 2 3xy yz at the point (2, -1, 1) in the direction of
normal to the surface at (-1, 2, 1)? Ans: Let 2 3xy yz
The gradient of above is and its directional derivative is in the direction of unit vector normal to the surface
(2, 1,1)ˆˆ ˆi j k
x y z
2 3(2, 1,1)
ˆˆ ˆi j k xy yzx y z
2 3 2(2, 1,1)
ˆˆ ˆ2 3y i yx z j yz k
(2, 1,1)ˆˆ ˆ3 3i j k
Normal vector to the surface 2log 4x z y is given by 2log 4x z y
2 2ˆˆ ˆlog 4 log 4x z y i j k x z yx y z
2 ˆˆlog 4 4x z y j k
Now directional derivative of 2 3xy yz in the direction of ˆˆ4 j k will be
ˆˆ4ˆˆ ˆ3 3 .17
j ki j k
Which is equal to 1517
20. If 푓 = (푥 + 푦 + 푧 ) , find 푑푖푣 푔푟푎푑(푓) and determine 푛 if 푑푖푣 푔푟푎푑(푓) =
0.
Ans: - 222ˆˆˆ zyxkzjyixRr
2 2 2 2n nx y z r
( ) .div gradf f
4log 2 yzx
4log 2 yzx
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 2 2 2n nf x y z r
2 2 2ˆˆ ˆ
n n nr r rf i j k
x y z
2 1 2 1 2 1 ˆˆ ˆ2 2 2r r rn n nf n r i n r j n r kx y z
2 1 ˆˆ ˆ2 r r rnf n r i j kx y z
2 1 ˆˆ ˆ2 x y znf n r i j kr r r
2 12 Rnf n rr
2 12 Rnf n rr
2 1ˆˆ ˆ 2 Rnf i j k n rx y z r
2 1ˆˆ ˆ2 Rnf n i j k rx y z r
2 1 2 1ˆ ˆˆ ˆ ˆ ˆ2 x y z Rn nf n r i j k i j k rx r y r z r r x y z
2 1 2 1ˆ2 2 2 2
r x x r r y y r r z z r Rn nx x x x x xf n r i rr xr r r
2 1 2 2 ˆ2 2 12 2 2
x y zr x r y r z R rn nr r rf n r n r ir xr r r
2 1 2 2 ˆ2 2 12 2 2
x y zr x r y r z R xn nr r rf n r n r ir rr r r
2 2 2 2 2 22 1 2 22 2 13 3 3
r x r y r z R Rn nf n r n rr rr r r
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22 2 2 2 2 22 1 2 22 2 13 3 3
y z x z y x Rn nf n r n rrr r r
2 2 2 222 1 2 22 2 13
x y z Rn nf n r n rrr
2
2 2 2 22 .2 2 1 .1 1Rn nf n r n rr
2 22 .2 2 1nf n r n
2 22 . 2 1nf n r n
0f
2 22 . 2 1 0 . 2 1 0 1/ 2nn r n n n
21. Verify Green’s theorem for: where C is bounded by
. Ans:
By Green’s theorem 2 11 2
C S
F FF dx F dy dxdyx y
Here 2 2ˆ ˆ( )F xy y i x j
So, 2 12 , 2F Fx x yx y
2
12 1
0
2 2x
S x
F F dxdy x x y dxdyx y
2
122 1
0
x
xS
F F dxdy xy y dxx y
2 2( )C
xy y dx x dy 2,y x y x
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1
2 2 3 42 1
0S
F F dxdy x x x x dxx y
11 5 4
4 32 1
0 05 4S
F F x xdxdy x x dxx y
2 1 1 1 15 4 20S
F F dxdyx y
----------------(1)
Now we have to find line integral along two curves. Along 2 , 2y x dy xdx .
So, 1 1
2 21 2 ( )
C C
F dx F dy xy y dx x dy
1
12 4 2
1 20
( . ) .2C
F dx F dy x x x dx x xdx
1
11 4 53 4
1 20 0
3(3 )4 5C
x xF dx F dy x x dx
1
1 23 1 194 5 20C
F dx F dy
Along ,y x dy dx
So, 2 2
2 21 2 ( )
C C
F dx F dy xy y dx x dy
2
02 2
1 21
( . ) .C
F dx F dy x x x dx x dx
2
002 3
1 2 11
3 0 1 1C
F dx F dy x dx x
Now, 1 2
1 2 1 2 1 219 1120 20C C C
F dx F dy F dx F dy F dx F dy ---------- (2)
From (1) and (2) it verifies Green’s theorem.
May-June 2009
22. State the Green’s theorem in the plane. Ans: If C be the regular closed curve in xy plane bounding any region R and 1( , )F x y and
2 ( , )F x y be continuous functions on C and its interior, having continuous partial derivatives
1Fy
and 2Fx
in R, then 2 11 2
C R
F FF dx F dy dxdyx y
, the line integral being
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taken along the boundary C of R such that R is on the left as one advances in the direction of integration.
23. Prove that 5 3
1 3( )( )A R B R A BA Br r r
where A and B are constants vectors
ˆˆ ˆR xi yj zk and 2 2 2r x y z .
Ans: 2 2 2r x y z
3/22 2 2 2 2 2
1 1 1 1 ˆˆ ˆ2 2 22
xi yj kr x y z x y z
3 32 2 2
1 1 1 RRr r rx y z
3
1 B RBr r
3
1 B RBr r
3 3
1 1 1B B R B Rr r r
-----------(1)
Where ˆ ˆˆ ˆ ˆ ˆB R ai cj ck xi yj zk ax by cz
ˆˆ ˆB R ax by cz ai cj ck B
So equation (1) becomes
3 4
1 3BB B R rr r r
2 2 2
3 4
1 3B RBB x y z
r r r
3 4 2 2 2
1 1 ˆˆ ˆ3 2 2 22
B RBB xi yj kr r r x y z
3 5
1 3B RBB R
r r r
3 5
1 3A R B RA BA B
r r r
CSVTU II SEMESTER SOLUTION ALL UNITS
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5 3
1 3A R B R A BA B
r r r
(Proved)
24. A vector is given by 2 2 ˆ ˆ( ) (2 )F x y x i xy y j
. Show that the field is irrotational and find its scalar potential. Hence evaluate the line integral from (1, 2) to (2, 1).
Ans: Here 2 2 ˆ ˆ( ) (2 )F x y x i xy y j
2 2
ˆˆ ˆ
( ) (2 ) 0
i j k
Fx y z
x y x xy y
ˆˆ ˆ(0 0) (0 0) ( 2 2 ) 0F i j k y y
So, 2 2 ˆ ˆ( ) (2 )F x y x i xy y j
is irrotational and can be expressed as the gradient of a scalar potential.
Hence, F
2 2 ˆ ˆ ˆ ˆ( ) (2 )x y x i xy y j i jx y
2 2( )..........(1) (2 )..........(2)x y x xy yx y
Now, integrating (1) we get
3 2
2 ( )............(3)3 2x xy x f y
Now, differentiating (3) with respect to y we get
2 '( )yx f yy
(2 ) 2 '( )xy y yx f y
2
'( ) ( )2yf y y f y
So, 3 2 2
2
3 2 2x x yy x (Ans)
Again .F dR
from (1, 2) to (2, 1) (1,2) (2,1)
1 1 4 8 4 1. 4 23 2 2 3 2 2
F dR
2 24 3 12 16 12 12 3 31 13 44 22.6 6 6 6 6 3
F dR
(Ans)
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25. Verify Stoke’s theorem for the vector field 2 2 ˆˆ ˆ(2 )F x y i yz j y zk
on the upper half surface of 2 2 2 1x y z , bounded by its projection on the xy-plane.
Ans: Stoke’s theorem states that ˆ. ( ).C S
F dr curl F nds
Here C is unit circle 2 2 1, 0x y z
2 2 ˆ ˆˆ ˆ ˆ ˆ. (2 ) .F dr x y i yz j y zk idx jdy kdz
2 2. (2 )F dr x y dx yz dy y zdz
Here C is z = 0, dz = 0 and cos , sinx y
. (2 )F dr x y dx
So, 2
0
. (2 ) (2cos sin )( sin )C C
F dr x y dx d
2
0
1 cos 2. sin 22C
F dr d
2
0
cos 2 sin 2.2 2 4C
F dr
1 1 2 0 0 0.2 2 4C
F dr
-------(1)
Now, 2 2 ˆˆ ˆ(2 )F x y i yz j y zk
2 2
ˆˆ ˆ
( )
2
i j k
curl Fx y z
x y yz y z
ˆ ˆˆ ˆ( ) ( 2 2 ) (0 0) (0 1)curl F i yz yz j k k
By using spherical co-ordinate we have
ˆˆ ˆˆ sin cos sin sin cosn i j k
So, ˆ( ). coscurl F n
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Hence /2 2
0 0
ˆ( ). cos .sinS
curl F nds d d
/2 2
0 0
ˆ( ). cos .sinS
curl F nds d d
/22
2
00
sinˆ( ).2S
curl F nds
1 0ˆ( ). (2 0)2S
curl F nds
------------(2)
From (1) and (2) it verifies Stoke’s theorem. (Ans)
Nov-Dec, 2009
26. Find 2 2 3 ˆˆ ˆ3 5 3div x i xy j xyz k at the point (1,2,3) .
Ans: - 2 2 3 2 2 3ˆˆ ˆ3 5 3 3 5 3div x i xy j xyz k x xy xyzx y z
2 2 3 2ˆˆ ˆ3 5 3 6 10 9div x i xy j xyz k x xy xyz
2 2 3
(1,2,3)ˆˆ ˆ3 5 3 6 20 162 188div x i xy j xyz k (Ans).
27. Verify Green’s theorem for: 2 2( )C
xy y dx x dy where C is bounded by 2,y x y x .
Ans: -
By Green’s theorem 2 11 2
C S
F FF dx F dy dxdyx y
(3) Here 2 2ˆ ˆ( )F xy y i x j
So, 2 12 , 2F Fx x yx y
CSVTU II SEMESTER SOLUTION ALL UNITS
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2
12 1
0
2 2x
S x
F F dxdy x x y dxdyx y
2
122 1
0
x
xS
F F dxdy xy y dxx y
1
2 2 3 42 1
0S
F F dxdy x x x x dxx y
11 5 4
4 32 1
0 05 4S
F F x xdxdy x x dxx y
2 1 1 1 15 4 20S
F F dxdyx y
----------------(1)
(4) Now we have to find line integral along two curves. Along 2 , 2y x dy xdx .
So, 1 1
2 21 2 ( )
C C
F dx F dy xy y dx x dy
1
12 4 2
1 20
( . ) .2C
F dx F dy x x x dx x xdx
1
11 4 53 4
1 20 0
3(3 )4 5C
x xF dx F dy x x dx
1
1 23 1 194 5 20C
F dx F dy
Along ,y x dy dx
So, 2 2
2 21 2 ( )
C C
F dx F dy xy y dx x dy
2
02 2
1 21
( . ) .C
F dx F dy x x x dx x dx
2
002 3
1 2 11
3 0 1 1C
F dx F dy x dx x
Now, 1 2
1 2 1 2 1 219 1120 20C C C
F dx F dy F dx F dy F dx F dy ----------(2)
From (1) and (2) it verifies Green’s theorem. OR
28. Evaluate ˆ.S
F ndS
where 2 2 ˆˆ ˆ4 2F xi y j z k
and S is the surface bounding the region
2 2 4x y , 0, 3z z . Ans: -
CSVTU II SEMESTER SOLUTION ALL UNITS
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By divergence theorem .S V
F ds divFdv
2 2. (4 ) ( 2 ) ( )S V
F ds x y z dvx y z
. 4 4 2S V
F ds y z dxdydz
2
2
2 4 3
2 04
. 4 4 2x
S x
F ds y z dxdydz
2
2
2 432
02 4
. 4 4x
zS x
F ds z yz z dxdy
2 2
2 2
2 4 2 4
2 24 4
. 12 12 9 21 12x x
S x x
F ds y dxdy y dxdy
2
2
242
42
. 21 6x
xS
F ds y y dx
2
2
2
. 42 4 0S
F ds x dx
2
2
2
. 42 4S
F ds x dx
2
21
2
4 4. 42 sin2 2 2S
x x xF ds
1 1. 42 0 0 2sin 1 2sin ( 1)S
F ds 1. 42 4sin 1 42 4 842S
F ds
(Ans). 29. Find the directional derivative of 2 3( , , )f x y z xy yz at (2, 1,1) in the direction of
vector ˆˆ ˆ2 2i j k .
Ans: - 2 3( , , )f x y z xy yz
2 3[ ( , , )] [ ]grad f x y z grad xy yz
2 3 2 ˆˆ ˆ(2 ) 3f y i xy z j yz k (2, 1,1)ˆˆ ˆ3 3f i j k
Given that ˆˆ ˆ2 2a i j k
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So, directional derivative of 2 3( , , )f x y z xy yz at (2,-1,1) in the direction of
ˆˆ ˆ2 2a i j k
is (2, 1,1)
ˆˆ ˆ2 2 1 6 6 11ˆˆ ˆ3 3ˆˆ ˆ 3 32 2
a i j kf i j ka i j k
(Ans).
May-June 2010
30. Explain in brief the directional derivatives.
It is the maximum rate of change of a scalar point function in the direction of a vector.
So, directional derivative of f at a point P in the direction of vector a is given by a fa
31. Evaluate C
dszxy )( 2 , where C is the arc of the helix tztytx ,sin,cos , which joins
the points )0,0,1( and ),0,1( .
Ans: - Here z t , so 0 0z t
2 2 2
2 2sin cos 1 2dx dy dzds dt t t dt dtdt dt dt
Now, 2 2
0
( ) (sin .cos ) 2C
xy z ds t t t dt
3
2 2
0 0
sin 2 cos 2( ) 2 22 4 3C
t t txy z ds t dt
3 3
2 1 1 2( ) 2 04 3 4 3C
xy z ds
(Ans).
32. Evaluate S
dSF
, where kzjyixF ˆˆ2ˆ4 22
and S is the surface bounding the region
3,0,422 zzyx . Ans:
CSVTU II SEMESTER SOLUTION ALL UNITS
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By divergence theorem .S V
F ds divFdv
2 2. (4 ) ( 2 ) ( )S V
F ds x y z dvx y z
. 4 4 2S V
F ds y z dxdydz 2
2
2 4 3
2 04
. 4 4 2x
S x
F ds y z dxdydz
2
2
2 432
02 4
. 4 4x
zS x
F ds z yz z dxdy
2 2
2 2
2 4 2 4
2 24 4
. 12 12 9 21 12x x
S x x
F ds y dxdy y dxdy
2
2
242
42
. 21 6x
xS
F ds y y dx
2
2
2
. 42 4 0S
F ds x dx
2
2
2
. 42 4S
F ds x dx
2
21
2
4 4. 42 sin2 2 2S
x x xF ds
1 1. 42 0 0 2sin 1 2sin ( 1)S
F ds
1. 42 4sin 1 42 4 842S
F ds (Ans).
33. If kxjiyF ˆˆ3ˆ2 2
and S is the surface of the parabolic cylinder xy 82 in the first
octant bounded by the planes 6,4 zy , show that 132S
dSNF
.
Ans: -
CSVTU II SEMESTER SOLUTION ALL UNITS
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Here Let 28 yx , jyigrad ˆ2ˆ8)(
16
ˆˆ4464
ˆ2ˆ8ˆ2ˆ8
ˆ2ˆ8)()(ˆ
22
yjyi
yjyi
jyijyi
gradgradn
164ˆ
16
ˆˆ4ˆˆ22
yi
yjyiin
16
1116
3816
ˆˆ4ˆˆ3ˆ2ˆ222
2
yy
yyy
yjyikxjiynF
dydzy
y
dydzy
yin
dydzNFdSNFS
4
0
6
02
4
0
6
02
4
0
6
0 411
16416
11ˆˆ
.132684
1124
11 60
4
0
2
zydSNF
S
(Proved).
(Nov –Dec 2010)
34. State the Green’s theorem in the plane. Ans: If C be the regular closed curve in xy plane bounding any region R and 1( , )F x y and
2 ( , )F x y be continuous functions on C and its interior, having continuous partial derivatives 1Fy
and 2Fx
in R, then 2 11 2
C R
F FF dx F dy dxdyx y
, the line integral being taken along
the boundary C of R such that R is on the left as one advances in the direction of integration Or
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Ans: - Green’s theorem in the plane: If xyyxyx ,),,(),,( be continuous in a region
E of the xy plane bounded by a closed curve C, then dxdyyx
dydxEC
)..( .
35. Find the values of a and b such that the surfaces xabyzax )2(2 and 44 32 zyx cut orthogonally at (1, -1, 2).
Ans: - Here 0)2(2 xabyzaxF and 044 32 zyxG
So, kbyjbziaaxF ˆˆˆ)22( and kzjxixyG ˆ3ˆ4ˆ8 22
kbjbiaF ˆˆ2ˆ)2()2,1,1( kjiG ˆ12ˆ4ˆ8)2,1,1(
As surfaces are orthogonal, so 01281680 bbaGF 04201684 baab ---------------(1) As point (1, -1, 2) lies on the surface, so it satisfies surface F. 10220)2(2 bbaba
Putting in (1) we get 2/50412 aa So, 1,2/5 ba (Ans).
36. Using Green’s theorem, evaluate C
xdydxxy cos)sin( , where C is the plane triangle
enclosed by the lines
xyxy 2,2
,0
Ans: -
For
C
xdydxxy cos)sin(
Here xxy
xxy sin,1cos,sin
CSVTU II SEMESTER SOLUTION ALL UNITS
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So, by Green’s theorem dxdyyx
xdydxxyEC
cos)sin(
2/
0
/2
0
)1sin(cos)sin( x
x
xy
yC
dxdyxxdydxxy
2/
0
/20)1sin(cos)sin(
x
x
x
C
dxyxxdydxxy
2/
0
2/
0
)1(sin22)1sin(cos)sin(
x
x
x
xC
dxxxdxxxxdydxxy
2/
0
2
2sin)cos(2cos)sin(
xxxxxxdydxxyC
000
81
20
22cos)sin(
2C
xdydxxy
24
18
2cos)sin(2
C
xdydxxy (Ans).
37. Find the work done in moving a particle in the force field kzjyxzixF ˆˆ)2(ˆ3 2
along:
i. The straight line from (0, 0, 0) to (2, 1, 3). ii. The curved defined by zxyx 83,4 32 from 0x to 2x .
Ans: - The equation of straight line joining (0, 0, 0) and (2, 1, 3) is
0 0 0 2 , , 3
2 0 1 0 3 0x y z t x t y t z t
where 0t to 1t
So, Work done .C
F dr
1
2
0
Work done 3(2 ) .2 (4 (3 )) 3 .3t dt t t t dt t dt
1 1
2 2 2
0 0
Work done 24 12 8 36 8t t t dt t t dt
13 2
0Work done 12 4 12 4 0 0 16t t
i. The curved defined by zxyx 83,4 32 from 0x to 2x . Ans: - 2 34 ,3 8x y x z
Let 2 33,4 8t tx t y z where 0, 2t t
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So, Work done .C
F dr
1 4 2 3 2
2
0
3 3 9Work done 3 .4 4 2 8 8t t t t tt dt dt dt
1 15 3 5 3 5
2 2
0 0
3 27 51Work done 3 38 8 64 8 64t t t t tt dt t dt
24 6
3
0
17 16 17 64 1 17Work done 8 8 1632 128 32 128 2 2t tt
(Ans)
(May-Jun-2011)
38. What are irrational and solenoidal vectors?
Irrotational Field (Conservative): -If F is irrotational field, then
1. 0
F and RdF
is path independent.
2. 0C
RdF
i.e. circulation along every closed surface is zero.
3. F , where is a scalar function.
Solenoidal Field (Incompressible): - If F is solenoidal field, then
1. 0 F .
2. Flux dsNF
across every closed surface is zero.
3. VF
, where V
is a Vector function.
39. The temperature of points in space is given by 푇(푥, 푦, 푧) = 푥 + 푦 − 푧. A mosquito located at (1,1,2) desires of fly in such a direction that it will get warm as soon as possible. In what direction should it move?
Ans:The temperature of points in space is given by 푇(푥, 푦, 푧) = 푥 + 푦 − 푧. A mosquito located at (1,1,2) desires of fly in such a direction that it will get warm as soon as possible. In direction of directional derivative of surface 푇(푥, 푦, 푧) = 푥 + 푦 − 푧.
The direction will be given by T
(1,1,2)ˆˆ ˆT i j k T
x y z
(1,1,2)2 2ˆˆ ˆT i j k
x y zx y z
CSVTU II SEMESTER SOLUTION ALL UNITS
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(1,1,22 2
)2 2 2 2ˆˆ ˆT i j k
x yx y z x y z x y
zz
(1,1,2)ˆˆ ˆ2 2 2T xi yj zk
(1,1,2)ˆˆ ˆ2 2 4T i j k
A mosquito located at (1,1,2) desires of fly in direction along the vector ˆˆ ˆ2 2 4i j k
40. Verify Green’s theorem for ∫ [(3푥 − 8푦 )푑푥 + (4푦 − 6푥푦)푑푦] where C is the boundary of the region bounded by x=0, y=0 and x+y=1.
Ans: -
By Green’s theorem 2 11 2
C S
F FF dx F dy dxdyx y
(5) Here 2(3 8 ) (4 6 )ˆ ˆF i x jx y y y
So, 2 16 , 16F Fx yx y
1
2 1
0 0
16 16
S
xF F dxdy y y dxdyx y
1
2 1
0 0
110
S
xF F dxdy y dxdyx y
2 1
00
1 125S
xF F dxdy y dxx y
2 1
0
1 25 1S
F F dxdy x dxx y
2 1
0
125 1 2
S
F F dxdy x x dxx y
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1
2 1
0
3 25 5 53S
F F xdxdy x xx y
2 1 5 55 53 3S
F F dxdyx y
----------------(1)
Verification of Green’s theorem by line integral 2(3 8 ) ( 6 ). 4x y dx y xy dyF dr
So, 2(3. 8 ) (4 6 )C C OA AB BO
x y dx yF r xyd dy
1
0
20
1
(3 8 ) (4 6 )3 4OA AB BO AB
x yxd dx y xy dyx ydy
220
1
(3 8 ) (4 6 ) (3 8 1 ) (4 1 6 1 )( )AB
x y dx y xy dy x x dx x x x dx
1 0 0
0 1
2
1
(3 8 1 ) (4 1 6 1 ) 4(3 )OA AB BO
x x dxdx x x x ydyx dx
1 002 2
2
10 1
3 8 1 2 4 1 12
63 42OA AB BO
x x dx x x x dx yx x
0
2 2
1
19 8 83 22
4 4 6 6OA AB BO
xx dx x x dxx
0
2 2
1
19 8 8 4 4 6 2632OA AB BO
x x x dxx x
0 02 3
0
11 1
29 13 22 2
4 123OA AB BO
x xx
3 29 14 22 2
1 13
2OA AB BO
3 29 14 22 2
1 13
2OA AB BO
=
53
…………………………….2
From 1 and 2 Greens theorem verified
41. Apply divergence theorem to evaluate ˆ.S
F ndS
where 2 2 ˆˆ ˆ4 2F xi y j z k
and S is the
surface bounding the region 2 2 4x y , 0, 3z z .
Ans: -
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
By divergence theorem .S V
F ds divFdv
2 2. (4 ) ( 2 ) ( )S V
F ds x y z dvx y z
. 4 4 2S V
F ds y z dxdydz
2
2
2 4 3
2 04
. 4 4 2x
S x
F ds y z dxdydz
2
2
2 432
02 4
. 4 4x
zS x
F ds z yz z dxdy
2 2
2 2
2 4 2 4
2 24 4
. 12 12 9 21 12x x
S x x
F ds y dxdy y dxdy
2
2
242
42
. 21 6x
xS
F ds y y dx
2
2
2
. 42 4 0S
F ds x dx
2
2
2
. 42 4S
F ds x dx
2
21
2
4 4. 42 sin2 2 2S
x x xF ds
1 1. 42 0 0 2sin 1 2sin ( 1)S
F ds 1. 42 4sin 1 42 4 84
2S
F ds (Ans).
(Nov –Dec 2011)
42. State the Gauss Divergent theorem
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
Ans: Gauss Divergence Theorem: - If F is a continuously differentiable vector function in
region E bounded by closed surface S, then ( )S E
F Nds div F dv
Where N
is the unit
normal vector at any point of S and kfjfifF ˆˆˆ321
31 21 2 3( )
S E
ff ff dydz f dzdx f dxdy dxdydzx y z
43. Prove that : 2 2( ) ''( ) '( )f r f r f r
r
Ans: - 222ˆˆˆ zyxkzjyixRr
2 ( ) . ( )f r f r
( )f r =
ˆ ˆ ˆ ( )i j k f rx y z
ˆ ˆ ˆ( ) ( ) ( ) ( )f r i f r j f r k f rx y z
/ / /ˆ( ) ( )r r r
f r f r i f j f kx y z
/ ˆ( ) ( )yx z
f r f r i j kr r r
/( )
( )f r R
f rr
2
/( )( ) . ( ) .
f r Rf r f r
r
2 / /( ) ( ) . . ( )
R Rf r f r f r
r r
2 / /ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) . ( )
R Rf r f r i j k i j k f r
x y z r r x y z
2 / /ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) . ( )
2 2 2 2 2 2 2 2 2x x x R
f r f r i j k i j k i j k f rx y z r x y zx y z x y z x y z
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 / // ˆ ˆ ˆ( ) ( ) ( )
2 2 2 2 2 2 2 2 2yx z R r r r
f r f r f r i j kx y z r x y zx y z x y z x y z
2 / // ˆ ˆ ˆ( ) ( ) ( )
2 2 2x R r r r
f r f r f r i j kx r x y zx y z
2
2 2 2 2 2 2/ // ˆ ˆ ˆ( ) ( ) ( )22 2 2
x y z x x x y z yR x zx xf r f r f r i j kr r r r
x y z
2
22 2 22 2 22/ //( ) ( ) ( )22 2 2
xx y z x
x y z R Rf r f r f r
r rx y z
2
22 2 2 2/ //( ) ( ) ( )
2 2 2 2 2 2
x y z x Rf r f r f r
rx y z x y z
2
221/ //( ) ( ) ( )3/22 2 2 2 2 2
x Rf r f r f r
rx y z x y z
2
22 2 23/ //( ) ( ) ( )3/22 2 2 2 2 2
x y z Rf r f r f r
rx y z x y z
2
223/ //( ) ( ) ( )3/22 2 2 2
r Rf r f r f r
rx y z r
2
23 1/ //( ) ( ) ( )
2 2 2
Rf r f r f r
r rx y z
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2
23 1/ //( ) ( ) ( )
2 2 2 2 2 2
Rf r f r f r
rx y z x y z
2
22/ //( ) ( ) ( )
2 2 2
Rf r f r f r
rx y z
222
2/ // 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) ( ).( )2 2 2
rf r f r f r R xi yj zk xi yj zk x y z r
rx y z
2 2/ //( ) ( ) ( ) Prf r f r f r Hence oved
r
44. Evaluate ∮ 퐹⃗ 푑푟⃗ by Stoke’s theorem, where 퐹⃗ = 푦 횤⃗ + 푥 횥⃗ − (푥 + 푧)푘⃗ and c is the boundary of the triangle with vertices at (0,0,0), (1,0,0) and (1,1,0).
Ans: We know that from Stoke’s theorem ˆ. ( ).C S
F dr curl F nds
As ˆ. ( ).
C S
F dr curl F nds
Therefore ( ) ( )curl F F
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 2
ˆˆ ˆ
ˆˆ 2 2
i j k
F j x y kx y z
y x x z
ˆˆ 2 2F j x y k
ˆ ˆˆˆ( ). 2 2S S
curl F nds j x y k kdxdy
ˆ( ). 2 2S S
curl F nds x y dxdy
1 1
0 0
ˆ( ). 2 2x
S
curl F nds x y dxdy
11 2
0 0
ˆ( ). 22
x
S
ycurl F nds xy dx
21
0
1ˆ( ). 2 1
2S
xcurl F nds x x dx
212
0
1 2ˆ( ). 2 2
2S
x xcurl F nds x x dx
1
2 2
0
1ˆ( ). 4 4 1 22S
curl F nds x x x x dx
1
2
0
1ˆ( ). 6 5 12S
curl F nds x x dx
12 3
0
1 3 5ˆ( ).2 2 3S
x xcurl F nds x
1
0
1 3 5 1 9 10 6 13ˆ( ). 12 2 3 2 2 4S
curl F nds
Therefore ˆ. ( ).C S
F dr curl F nds
= 134
Ans.
45. Evaluate ∬ 푦 푧 횤⃗ + 푧 푥 횥⃗ + 푧 푦 푘⃗ . 푛 푑푠 where S is the point if the sphere 푥 + 푦 + 푧 = 1 above XY-plane and bounded by this plane. Ans: By divergence theorem .
S V
F ds divFdv
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2 2 2 2 2 2. ( )S V
y z i z xF ds i j k dvx
zy
kz
j y
2 2 2 2 2 2
V
dvx y z
y z z x z y
22
V
zy dv
2
2 22 112
1
1 021
x yxdxdzy ydz
x
2 2
211 2 210
1 21
x x ydxdz y y
x
22 2
2111
1 21
xx y dxdy
x
y
2 42 2
211
1 21
xx y dxdyy y
x
2 4 2 42 2 2 2
0
2 21 11 12
1 121
x xx y dxdy x y dxdyy y
x
y y
2 4 2 4
0
2 2 2
2 21 11 12
1 121
1
x xx y dxdy x y dxdyy y
x
y
3 5
2
0
2112
3 511
xyyx dx
22 2 2 22
1 1 1 1 12
3 511
x x x xx dx
2 22 2 22 21 11 1 11 1 423 3 5 15 31 1
x x xx x dx dx
22 2
0
18 1 115
x x dx
sin cos ,x dx d then
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22 2 6
0 0
1 /28 81 1 cos15 15
x x dx d
By using reduction formula we will have
2 2 2 2 2 2( ).12S V
y z i z x j z yF ds i j k dvy z
kx
April -May, 2012
46. State Gauss divergence theorem.
Ans: Gauss Divergence Theorem: - If F is a continuously differentiable vector function in
region E bounded by closed surface S, then ( )S E
F Nds div F dv
Where N
is the unit
normal vector at any point of S and kfjfifF ˆˆˆ321
31 21 2 3( )
S E
ff ff dydz f dzdx f dxdy dxdydzx y z
47. Find the directional derivative of 2 24x yz xz at the point (1, 2,1) in the direction of the
vector 2 2I J k . Ans: The directional derivative of 2 24x yz xz at the point (1, 2,1) in the direction of
the vector 2 2I J k is given by
2 24A i j k x yz xz Ax y z
where A is 2 2I J k ………………..1
2 24i j k x yz xzx y z
2 2 22 4 8xyz z i x z j x y xz k
4 4 2 8 6(1, 2,1) i j k j k …………………..2
2 2 2 2 2 22 2 39
I J k I J k I J kAI J k
…………………………3
From 1,2 and 3 we will have directional derivative of 2 24x yz xz at the point (1, 2,1) in
the direction of the vector 2 2I J k
2 24A i j k x yz xz Ax y z
= 2 26 .
3I J kj k
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A = 6 .j k2 2 13 / 3
3I J k
48. Verify Green theorem for 2 2 ,c
xy y dx x dy where C is bounded by 2,y x y x
Ans: -
By Green’s theorem 2 11 2
C S
F FF dx F dy dxdyx y
(6) Here 2 2ˆ ˆ( )F xy y i x j
So, 2 12 , 2F Fx x yx y
2
12 1
0
2 2x
S x
F F dxdy x x y dxdyx y
2
122 1
0
x
xS
F F dxdy xy y dxx y
1
2 2 3 42 1
0S
F F dxdy x x x x dxx y
11 5 4
4 32 1
0 05 4S
F F x xdxdy x x dxx y
2 1 1 1 15 4 20S
F F dxdyx y
----------------(1)
Now we have to find line integral along two curves. Along 2 , 2y x dy xdx .
So, 1 1
2 21 2 ( )
C C
F dx F dy xy y dx x dy
1
12 4 2
1 20
( . ) .2C
F dx F dy x x x dx x xdx
1
11 4 53 4
1 20 0
3(3 )4 5C
x xF dx F dy x x dx
1
1 23 1 194 5 20C
F dx F dy
Along ,y x dy dx
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So, 2 2
2 21 2 ( )
C C
F dx F dy xy y dx x dy
2
02 2
1 21
( . ) .C
F dx F dy x x x dx x dx
2
002 3
1 2 11
3 0 1 1C
F dx F dy x dx x
Now, 1 2
1 2 1 2 1 219 1120 20C C C
F dx F dy F dx F dy F dx F dy ----------(2)
From (1) and (2) it verifies Green’s theorem. 49. Find the work done in moving a particle in the force field 23 (2 )F x I xz y J zK , along
the straight line from (0,0,0) to (2,1,3). Ans: The equation of straight line joining (0, 0, 0) and (2, 1, 3) is
0 0 0 2 , , 3
2 0 1 0 3 0x y z t x t y t z t
where 0t to 1t
So, Work done .C
F dr
1
2
0
Work done 3(2 ) .2 (4 (3 )) 3 .3t dt t t t dt t dt
1 1
2 2 2
0 0
Work done 24 12 8 36 8t t t dt t t dt
13 2
0Work done 12 4 12 4 0 0 16t t
Nov-Dec, 2012
50. Evaluate divF and curl of F at point (1,2,3) where F is 2 2 2 ˆˆ ˆF x yzi xy zj xyz k
Ans: 2 2 2 ˆˆ ˆF x yzi xy zj xyz k
( ) 6 2 2 8 2div F xyz xy xyz xyz xy
(1,2,3)( ) 8 2 52div F xyz xy
(Ans).
Now,
2 2 2
ˆˆ ˆ
( )
i j k
curl Fx y z
x yz xy z xyz
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2 2 2 2 2 2ˆˆ ˆ( ) ( ) ( ) ( )curl F i xz xy j yz yx k yz zx
(1,2,3)ˆˆ ˆ( ) 16 15curl F i j k
(Ans).
51. Find the directional derivative of 2 24x yz xz at (1, 2,1, ) in the direction of vectorˆˆ ˆ2 2i j k .
Ans: The directional derivative of 2 24x yz xz at the point (1, 2,1) in the direction of
the vector 2 2I J k is given by
2 24A i j k x yz xz Ax y z
where A is 2 2I J k ………………..1
2 24i j k x yz xzx y z
2 2 22 4 8xyz z i x z j x y xz k
4 4 2 8 6(1, 2,1) i j k j k …………………..2
2 2 2 2 2 22 2 39
I J k I J k I J kAI J k
…………………………3
From 1,2 and 3 we will have directional derivative of 2 24x yz xz at the point (1, 2,1) in
the direction of the vector 2 2I J k
2 24A i j k x yz xz Ax y z
= 2 26 .
3I J kj k
A = 6 .j k2 2 13 / 3
3I J k
52. Find the value of a if the vector 2 2 2 2 22 2ax z yz i xy z x j xyz x y k has zero
divergence find the curl of the above vector which has zero divergence.
Ans: 2 2 2 2 22 2F ax z yz i xy z x j xyz x y k
it is given that divF is Zero
2 2 2 2 2( ) 2 2 0div F ax z yz i xy z x j xyz x y k
( ) 2 2 2 0div F ax xy xy
( ) 2 4 0div F ax xy
2a y Curl of above vector whose divergence is equal to zero is given by
CSVTU II SEMESTER SOLUTION ALL UNITS
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2 2 2 2 2
ˆˆ ˆ
( )
2 2 2
i j k
curl Fx y z
yx z yz xy z x xyz x y
2 2 2 2 2 2 ˆˆ ˆ( ) 2 4 2 2 4 2 2curl F xz x y xz i yz xy yx y j y z x z z k
2 2 2 2 2 2 ˆˆ ˆ( ) 4 4 2 4 2 2curl F xz x y i yz y xy yx j y z x z z k
53. Evaluate ∫ 퐹⃗.푑푠 where 퐹⃗ = 4x횤̂ - 2y2횥̂ + z2푘 and s is the surface bounding the region x2 + y2 = 4, z = 0 and z = 3.
Ans: -
By divergence theorem .S V
F ds divFdv
2 2. (4 ) ( 2 ) ( )S V
F ds x y z dvx y z
. 4 4 2S V
F ds y z dxdydz
2
2
2 4 3
2 04
. 4 4 2x
S x
F ds y z dxdydz
2
2
2 432
02 4
. 4 4x
zS x
F ds z yz z dxdy
2 2
2 2
2 4 2 4
2 24 4
. 12 12 9 21 12x x
S x x
F ds y dxdy y dxdy
2
2
242
42
. 21 6x
xS
F ds y y dx
2
2
2
. 42 4 0S
F ds x dx
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2
2
2
. 42 4S
F ds x dx
2
21
2
4 4. 42 sin2 2 2S
x x xF ds
1 1. 42 0 0 2sin 1 2sin ( 1)S
F ds 1. 42 4sin 1 42 4 842S
F ds
(Ans).
Unit V – Theory Of Equations April -May, 2006
1. If O, A, B, C are the four point on a straight line such that the distance of A, B and C from
O are the roots of the equation 3 23 3 0ax bx cx d . If B is the middle point of AC show that 3 33 2 0a d abc b Ans . Let us assume , , are the roots of the equation 3 23 3 0ax bx cx d and
, ,OA OB OC
According to the given condition 2 --------------(1) According to the property of roots
Now, 3ba
--------------(2)
3ca
------------(3)
da
-------------(4)
From 2 and 3 we will have 33 b ba a
-------------(5)
3 3c ca a
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From 1 2
2 3 32 2c b ca a a
2
2
3 2c ba a
-------------(6)
again from 4 we have
d
a
da
From 5 and 6 we have 2
2
3 2b c b da a a a
By solving above we will have
2
2
3 2ca bb da
Hence we will have 3 33 2 0a d abc b
2. Solve by Cardan’s method of equation 3 23 3 0x x . Ans: 3 23 3 0x x --------------(1)
Here we can remove second term of equation (1) by diminishing its roots by
3 1
3 1bhna
We can diminished each root by 1 by synthetic division method
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Hence the transformed equation is 3 3 1 0y y ---------(2)where 1y x
Let 1/3 1/3y p q be the solution of equation (2).
3 1/3 1/3 1/3 1/33y p q p q p q
3 1/3 1/33y p q p q y
3 1/3 1/33 ( ) 0y p q y p q -------------(3)
By comparing equation (2) and (3) we get
1/3 1/3 1, ( ) 1p q p q
1, ( ) 1pq p q
So, p and q are the roots of the equation 2 1 0t t
1 1 4 1 3
2 2it
.
So, let1 3 1 3,
2 2i ip q
.
So roots of given cubic equation (1) are
(i) 1/3 1/3
1/3 1/3 1 3 1 32 2i ip q
1/3 1/32 2 2 2cos sin cos sin
3 3 3 3i i
2 2 2 2cos sin cos sin9 9 9 9
i i
22cos9
(ii) 1/3 1/3
1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q
4/3 4/3
1 3 1 32 2i i
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4/3 4/32 2 2 2cos sin cos sin
3 3 3 3i i
8 8 8 8cos sin cos sin9 9 9 9
i i
82 cos9
(iii) 1/3 1/3
2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq
1/3 1/32 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin
3 3 3 3 3 3 3 3i i i i
2 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin3 3 9 9 3 3 9 9
i i i i
2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3
i i
4 4 4 4cos sin cos sin9 9 9 9
i i
4 4 142cos 2cos 2 2cos9 9 9
So, roots of equation (2) are 2 8 142cos , 2 cos , 2 cos9 9 9
Now, roots of given equation (1) are 2 8 141 2cos , 1 2 cos , 1 2cos9 9 9
(Ans)
3. Solve the equation by Ferrari’s method: 4 3 212 41 18 72 0x x x x . Ans: 4 3 212 41 18 72 0x x x x
Let 4 3 2 2 2 2( ) 12 41 18 72 ( 6 ) ( ) 0f x x x x x x x mx n ----------(1)
4 3 2 4 2 2 3 2 2 2 212 41 18 72 36 12 12 2 2x x x x x x x x x m x mnx n
4 3 2 4 3 2 2 2 212 41 18 72 12 (36 2 ) ( 12 2 )x x x x x x m x mn x n
Equating the coefficients we get 2 2 2(36 2 ) 41, ( 12 2 ) 18, 72m mn n
2 2 22 5, 9 6 , 72m mn n
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222 5 72 9 6
3 2 22 144 5 360 81 36 108 0 3 22 41 252 441 0 2( 3)(2 35 147) 0
2123,7,
By taking 3 1, 9m n equation (1) becomes
4 3 2 2 2 212 41 18 72 ( 6 3) ( 9) 0x x x x x x x
2 2( 6 3 9)( 6 3 9) 0x x x x x x
2 2( 5 6)( 7 12) 0x x x x ( 6)( 1)( 3)( 4) 0x x x x
1,3, 4,6x (Ans)
Nov-Dec, 2006 4. Solve by Cardan’s method of equation 3 23 3 0x x .
Ans: 3 23 3 0x x --------------(1)
Here we can remove second term of equation (1) by diminishing its roots by
3 1
3 1bhna
We can diminished each root by 1 by synthetic division method
Hence the transformed equation is 3 3 1 0y y ---------(2)where 1y x
Let 1/3 1/3y p q be the solution of equation (2).
3 1/3 1/3 1/3 1/33y p q p q p q
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3 1/3 1/33y p q p q y
3 1/3 1/33 ( ) 0y p q y p q -------------(3)
By comparing equation (2) and (3) we get
1/3 1/3 1, ( ) 1p q p q
1, ( ) 1pq p q
So, p and q are the roots of the equation 2 1 0t t
1 1 4 1 3
2 2it
.
So, let1 3 1 3,
2 2i ip q
.
So roots of given cubic equation (1) are
(i) 1/3 1/3
1/3 1/3 1 3 1 32 2i ip q
1/3 1/32 2 2 2cos sin cos sin
3 3 3 3i i
2 2 2 2cos sin cos sin9 9 9 9
i i
22cos9
(ii) 1/3 1/3
1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q
4/3 4/3
1 3 1 32 2i i
4/3 4/32 2 2 2cos sin cos sin
3 3 3 3i i
8 8 8 8cos sin cos sin9 9 9 9
i i
82 cos9
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(iii) 1/3 1/3
2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq
1/3 1/32 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin
3 3 3 3 3 3 3 3i i i i
2 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin3 3 9 9 3 3 9 9
i i i i
2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3
i i
4 4 4 4cos sin cos sin9 9 9 9
i i
4 4 142cos 2cos 2 2cos9 9 9
So, roots of equation (2) are 2 8 142cos , 2 cos , 2 cos9 9 9
Now, roots of given equation (1) are 2 8 141 2cos , 1 2 cos , 1 2cos9 9 9
(Ans)
5. Solve the equation 5 4 3 26 43 43 6 0x x x x x . Ans: 5 4 3 26 43 43 6 0x x x x x
The given equation is a reciprocal equation of odd degree having coefficient of terms equidistance from beginning and end.
So, 1x is its root.
4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x
Dividing by 2x we get
22
5 66 5 38 0x xx x
22
1 16 5 38 0x xx x
21 16 2 5 38 0x xx x
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26 2 5 38 0t t where 1x tx
26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2
t t t
1 10 1 53 2
x and xx x
2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x
10 100 36 5 25 166 4
x and x
10 8 5 36 4
x and x
1 13, 2,2 2
x and x
So, roots are 1 12, 1, , , 32 2
x (Ans)
6. If , , are the roots of the equation 3 0x qx r . Find the equation whose roots are
, , .
Ans: Given that , , are the roots of the equation 3 0x qx r .
Now, 0 --------------(1)
q ------------(2)
r -------------(3)
Now 2 2 2 2( ) 2( ) 0 q q ----------(4)
Let , ,A B C
A B C
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2 2 2 2 2 2
A B C
2 2 2 2 2 2
A B C
2 2 2 2 2 2
A B C
( ) ( ) ( )A B C
A B C
as 0
3 3A B C
------------------(5)
AB BC CA
2 2 2 2 2 2 2 2 2 2 2 2
AB BC CA
2 2 2 2 2 2 2 2 2 2 2 2
AB BC CA
2 2 2 2 2 2q q q q q qAB BC CA
r r r
2 2 2 2 2 2q q q q q qAB BC CA
r r r
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2( ) ( ) ( )q q q q q qAB BC CA
r r r
2 3 2 2 2 2 3 2 2 2 2 3 2 2 2q q q q q q q q qAB BC CA
r r r
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2 3 2 2 2 2 3 2 2 2 2 3 2 2 2q q q q q q q q qAB BC CA
r
2 3 3 3 3 3 3 3( )( ) ( )q q qAB BC CA
r
2 3 2( ) ( ) 0 ( ) ( ) 3 ( )q q q q rAB BC CA
r
3 4 2 3( )( )
q q q q r qAB BC CA
r r
3 4 2
2
2 3q q q rAB BC CAr
-----------(6)
ABC
2 2 2 2 2 2
ABC
2 2 2q q qABC
3 2
2 2 2
( ) ( )q q qABC
3 2 3 2
2 2
0q q r q q rABCr r
Equation whose roots are A, B, C is
3 2( ) ( ) 0x A B C x AB BC CA x ABC
3 4 2 3 2
3 22 2
2 33 0q q q r q q rx x xr r
2 3 2 2 3 4 2 3 23 (2 3 ) ( ) 0r x r x q q q r x q q r (Ans)
7. Solve by Cardan’s method of equation 3 23 3 0x x .
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Ans: 3 23 3 0x x --------------(1)
Here we can remove second term of equation (1) by diminishing its roots by
3 1
3 1bhna
We can diminished each root by 1 by synthetic division method
Hence the transformed equation is 3 3 1 0y y ---------(2)where 1y x
Let 1/3 1/3y p q be the solution of equation (2).
3 1/3 1/3 1/3 1/33y p q p q p q
3 1/3 1/33y p q p q y
3 1/3 1/33 ( ) 0y p q y p q -------------(3)
By comparing equation (2) and (3) we get
1/3 1/3 1, ( ) 1p q p q
1, ( ) 1pq p q
So, p and q are the roots of the equation 2 1 0t t
1 1 4 1 3
2 2it
.
So, let1 3 1 3,
2 2i ip q
.
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So roots of given cubic equation (1) are
(i) 1/3 1/3
1/3 1/3 1 3 1 32 2i ip q
1/3 1/32 2 2 2cos sin cos sin
3 3 3 3i i
2 2 2 2cos sin cos sin9 9 9 9
i i
22cos9
(ii) 1/3 1/3
1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q
4/3 4/3
1 3 1 32 2i i
4/3 4/32 2 2 2cos sin cos sin
3 3 3 3i i
8 8 8 8cos sin cos sin9 9 9 9
i i
82 cos9
(iii) 1/3 1/3
2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq
1/3 1/32 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin
3 3 3 3 3 3 3 3i i i i
2 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin3 3 9 9 3 3 9 9
i i i i
2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3
i i
4 4 4 4cos sin cos sin9 9 9 9
i i
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4 4 142cos 2cos 2 2cos9 9 9
So, roots of equation (2) are 2 8 142cos , 2 cos , 2 cos9 9 9
Now, roots of given equation (1) are 2 8 141 2cos , 1 2 cos , 1 2cos9 9 9
(Ans)
8. Solve the equation 5 4 3 26 43 43 6 0x x x x x .
Ans: 5 4 3 26 43 43 6 0x x x x x
The given equation is a reciprocal equation of odd degree having coefficient of terms equidistance from beginning and end.
So, 1x is its root.
4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x
Dividing by 2x we get
22
5 66 5 38 0x xx x
22
1 16 5 38 0x xx x
21 16 2 5 38 0x xx x
26 2 5 38 0t t where 1x tx
26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2
t t t
1 10 1 53 2
x and xx x
2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x
10 100 36 5 25 166 4
x and x
10 8 5 36 4
x and x
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1 13, 2,2 2
x and x
So, roots are 1 12, 1, , , 32 2
x (Ans)
9. If , , are the roots of the equation 3 0x qx r . Find the equation whose roots are
, , .
Ans: Given that , , are the roots of the equation 3 0x qx r .
Now, 0 --------------(1)
q ------------(2)
r -------------(3)
Now 2 2 2 2( ) 2( ) 0 q q ----------(4)
Let , ,A B C
A B C
2 2 2 2 2 2
A B C
2 2 2 2 2 2
A B C
2 2 2 2 2 2
A B C
( ) ( ) ( )A B C
A B C
as 0
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3 3A B C
------------------(5)
AB BC CA
2 2 2 2 2 2 2 2 2 2 2 2
AB BC CA
2 2 2 2 2 2 2 2 2 2 2 2
AB BC CA
2 2 2 2 2 2q q q q q qAB BC CA
r r r
2 2 2 2 2 2q q q q q qAB BC CA
r r r
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2( ) ( ) ( )q q q q q qAB BC CA
r r r
2 3 2 2 2 2 3 2 2 2 2 3 2 2 2q q q q q q q q qAB BC CA
r r r
2 3 2 2 2 2 3 2 2 2 2 3 2 2 2q q q q q q q q qAB BC CA
r
2 3 3 3 3 3 3 3( )( ) ( )q q qAB BC CA
r
2 3 2( ) ( ) 0 ( ) ( ) 3 ( )q q q q rAB BC CA
r
3 4 2 3( )( )
q q q q r qAB BC CA
r r
3 4 2
2
2 3q q q rAB BC CAr
-----------(6)
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ABC
2 2 2 2 2 2
ABC
2 2 2q q qABC
3 2
2 2 2
( ) ( )q q qABC
3 2 3 2
2 2
0q q r q q rABCr r
Equation whose roots are A, B, C is
3 2( ) ( ) 0x A B C x AB BC CA x ABC
3 4 2 3 2
3 22 2
2 33 0q q q r q q rx x xr r
2 3 2 2 3 4 2 3 23 (2 3 ) ( ) 0r x r x q q q r x q q r (Ans)
May-June, 2007
10. Solve the equation by Ferrari’s method: 4 3 212 41 18 72 0x x x x . Ans: 4 3 212 41 18 72 0x x x x
Let 4 3 2 2 2 2( ) 12 41 18 72 ( 6 ) ( ) 0f x x x x x x x mx n ----------(1)
4 3 2 4 2 2 3 2 2 2 212 41 18 72 36 12 12 2 2x x x x x x x x x m x mnx n
4 3 2 4 3 2 2 2 212 41 18 72 12 (36 2 ) ( 12 2 )x x x x x x m x mn x n
Equating the coefficients we get 2 2 2(36 2 ) 41, ( 12 2 ) 18, 72m mn n
2 2 22 5, 9 6 , 72m mn n
222 5 72 9 6
3 2 22 144 5 360 81 36 108 0
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3 22 41 252 441 0 2( 3)(2 35 147) 0
2123,7,
By taking 3 1, 9m n equation (1) becomes
4 3 2 2 2 212 41 18 72 ( 6 3) ( 9) 0x x x x x x x
2 2( 6 3 9)( 6 3 9) 0x x x x x x
2 2( 5 6)( 7 12) 0x x x x ( 6)( 1)( 3)( 4) 0x x x x
1,3, 4,6x (Ans)
11. Solve by Cardan’s method of equation 3 15 126 0x x . Ans: 3 15 126 0x x
Here the equation having terms involving 2x term missing. Let 1/3 1/3x p q be the solution of equation (1).
3 1/3 1/3 1/3 1/33x p q p q p q
3 1/3 1/33x p q p q x
3 1/3 1/33 ( ) 0x p q x p q -------------(2) By comparing equation(1) and (2) we get 1/3 1/3 5, ( ) 126p q p q 125, ( ) 126pq p q
So, p and q are the roots of the equation 2 126 125 0t t ( 1)( 125) 0 1,125t t t . So, let p = 1 and q = 125.
So roots of given cubic equation (1) are (i) 1/3 1/3 1 5 6p q
(ii) 1/3 2 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iwp w q i
(iii) 2 1/3 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iw p wq i
So, roots are 6, 3 2 3, 3 2 3i i (Ans).
12. Solve 3 24 20 48 0x x x given that the roots , are connected by the relation
2 0 . Ans: Let , , are the roots of 3 24 20 48 0x x x . Given that 2 0 2 Now, 4 --------------(1)
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20 ------------(2) 48 -------------(3) From (1) 2 4 4 ----------(4) From (3) 2 . .(4 ) 48
3 24 24
3 24 24 0
2( 2)( 6 12) 0
6 36 482, 2, 3 3
2i
So, 2 4, 6 So, roots are -4, 2, 6.
Nov-Dec, 2007 13. Find the number of real roots the equation 3 2 x x 4x 4 0
Ans: number of real roots the equation 3 2 x x 4x 4 0 are 3 , two positive roots and 1
negative roots.
14. Find the condition that the equation 3 2x x 4x 4 0p had roots , , which satisfy
1 0 .
Ans: Let , , are the roots of 3 2 0x px qx r . Given that 1 0 1 --------------(1)
Now, p --------------(2) 4 ------------(3) 4 -------------(4) From (1) and (4) 4 ----------(4) Putting the value of 4 in the equation 3 2 0x px qx r we will have
64 16 16 4 0p
84 2116 4
p
15. If , , are roots of the equation 3 0x qx r Find the equation whose roots are
2 2 2, ,
Ans: Let , , are the roots of 3 0x qx r …………………(1) Now, 0 --------------(2)
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q ------------(3) r -------------(4)
Let 2y
2 2 4
2 2 4
2 2 40 r
2 2 4rxx
2 4ry xx
3 4 0x yx r ……………………………..(5)
Subtracting 4 from 1 we will have
3 0q y x r
3rx
q y
………………..(6)
Put the value of x in the equation 1 we will have 3
3 3 0r rq rq y q y
2 3327 3 0r rq q y r q y
3 2 2 3 3 2 227 3 2 3 3 0r rq q r rq r q r r q rq
3 3 3 2 2 3 4 3 2 227 3 3 6 3 3 0r rq r q r q rq r r q r q 3 3 3 2 2 427 4 6 9 0r rq r q r q r
16. Solve by Cardan’s methods: 3 15 126 0x x
Ans: 3 15 126 0x x
Here the equation having terms involving 2x term missing. Let 1/3 1/3x p q be the solution of equation (1).
3 1/3 1/3 1/3 1/33x p q p q p q
3 1/3 1/33x p q p q x
3 1/3 1/33 ( ) 0x p q x p q -------------(2) By comparing equation(1) and (2) we get
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1/3 1/3 5, ( ) 126p q p q 125, ( ) 126pq p q
So, p and q are the roots of the equation 2 126 125 0t t ( 1)( 125) 0 1,125t t t . So, let p = 1 and q = 125.
So roots of given cubic equation (1) are (i) 1/3 1/3 1 5 6p q
(ii) 1/3 2 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iwp w q i
(iii) 2 1/3 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iw p wq i
So, roots are 6, 3 2 3, 3 2 3i i (Ans).
April -May 2008
17. Form the equation whose roots are 1, 2, 3.
Ans .The equation will be 1 2 3x x x 3 21 2 3 6 11 6x x x x x x
18. Solve the equation 2x2+x2-7x-6 = 0 when the difference of two roots is 3.
Ans: Let , , are the roots of 3 22x x 7x 6 0 . Given that 3 -------------- (1) Now, 1/ 2 -------------- (2) 7 / 2 ------------ (3) 3 ------------- (4) From (1) and (2) 2 7 / 2 ------------- (5) From (4) and (1) . .(3 ) 3 ------------- (6)
From (5) and (6) 7 2 .(3 ) 3
2
From
3 24 24
3 24 24 0
2( 2)( 6 12) 0
6 36 482, 2, 3 3
2i
So, 2 4, 6
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So, roots are -4, 2, 6.
19. If 휶,휷,휸 are the roots of the cubic 3 2 0x px qx r find the equation whose roots are
1 1 1, ,
Ans. As , , are the roots of the cubic 3 2 0x px qx r .
Here , ,p q r
Let root of the new equation is y, and according to the given condition 1 1 1ry y
x
1 1r ry xx y
and x is the root of the equation 3 2 0x px qx r
3 21 1 1 0r r rp q r
y y y
3 2 2 31 1 1 0r p r y q r y ry
3 2 2 2 31 3 3 1 2 1 0r r r p r r y q r y ry
3 2 2 3 21 1 2 1 3 3 0ry q r y p r r y r r r
Required equation will be
3 2 2 3 21 1 2 1 3 3 0ry q r y p r r y r r r
20. Solve by Cardan’s methods: 3 29 +6x -1=0 x Ans: 3 29 +6x -1=0 x
Here the equation having terms involving 2x term. We can have another equation 1= yx
And by putting the value of y in the equation 3 29 +6x -1=0 x we will have 3 2
31 19 +6 -1=0 y 6 9 0yy y
Here in the 3y 6 9 0y ……………………………………………(2)
equation term involving 2y is missing .
Let y u v be the solution of equation (2).
3 3 3 3y u v uv u v
3 3 3 3y u v uvy
3 3 33 0y uvy u v -------------(2)
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By comparing equation(1) and (2) we get 3 32, ( ) 9uv u v ------------(3)
3 3 3 3 3 32, ( ) 9 8, ( ) 9uv u v u v u v
So, 3u and 3v are the roots of the equation 2 9 8 0t t 2 9 8 0 ( 8) 1t t t t .
So, let 3u = 8 and 22, 2 , 2u w w therefore from 3 we will have 21, ,v w w
So roots of given cubic equation (1) are (i) 2 1 3u v
(ii) 2 1 3 1 3 3 32 22 2 2i i iu v w w
(iii) 2 1 3 1 3 3 32 22 2 2i i iu v w w
So, roots of the equation 3y 6 9 0y are 3 3 3 33, ,
2 2i i
And roots of the equation 3 29 +6x -1=0 x are
1 11 3 3 3 3, ,3 2 2
i i
as
1= yx
Nov -Dec 2008
21. From the equation whose roots are reciprocal of the roots of the equation5 3 22 4 3 7 6 0x x x x
Ans: The equation whose roots are reciprocal of the roots of the equation5 3 22 4 3 7 6 0x x x x is y
1yx
by putting the value of y in the above equation we will have
5 3 22 1/ 4 1/ 3 1/ 7 1/ 6 0x x x x
4 35 26 7 3 4 2 0x x x x 22. If 1 2 3, ,r r r are the root of the equation 3 22 3 1 0x x kx find constant k if sum of
two roots is 1.
Let 1 2 3, ,r r r are the roots of 3 22 3 1 0x x kx .
Given that 1 2 1r r -------------- (1)
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Now, 1 2 3 3 / 2r r r -------------- (2)
1 2 2 3 1 3 / 2r r r r r r k ------------ (3)
1 2 3 1/ 2r r r ------------- (4)
From (1) and (2) 3 1/ 2r ------------- (5)
From (5) and (4) 1 2 1r r ------------- (6)
From (5) and (6)we can construct a quadratic equation that is 2 1 0x x
By solving equation 2 1 0x x we will have 2 4 1 3
2 2b b ac ix
a
1 2 31 3 1 3
2, , 1 / 2
2i ir r r
23. Find the equation whose roots are the roots of 풙ퟒ + 풙ퟑ − ퟑ풙ퟐ − 풙 + ퟐ = ퟎ.each diminished by 3. Ans. The above equation can be solved by using synthetic division then we will have
3y x Let , , , are the roots of 풙ퟒ + 풙ퟑ − ퟑ풙ퟐ − 풙 + ퟐ = ퟎ.
Now, 1 -------------- (1) As roots are diminished by 3 therefore new root will be 3, 3, 3, 3y
3 3 3 3 1 12
3 3 3 3 13
3 1 1 3 1 2
0 3 12 27 78
1 4 9 26 80
0 3 21 90
1 7 30 116
0 3 30
1 10 60
0 3
1 13
Hence the new equation is 4 3 213 60 116 80x x x x
24. Solve by Cardan’s method of equation 3 15 126 0x x .
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Ans: 3 15 126 0x x
Here the equation having terms involving 2x term missing. Let 1/3 1/3x p q be the solution of equation (1).
3 1/3 1/3 1/3 1/33x p q p q p q
3 1/3 1/33x p q p q x
3 1/3 1/33 ( ) 0x p q x p q -------------(2) By comparing equation(1) and (2) we get 1/3 1/3 5, ( ) 126p q p q 125, ( ) 126pq p q
So, p and q are the roots of the equation 2 126 125 0t t ( 1)( 125) 0 1,125t t t . So, let p = 1 and q = 125.
So roots of given cubic equation (1) are (i) 1/3 1/3 1 5 6p q
(ii) 1/3 2 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iwp w q i
(iii) 2 1/3 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iw p wq i
So, roots are 6, 3 2 3, 3 2 3i i (Ans).
May-June 2009
25. Write the relation between roots and coefficients of the equation.
1 20 1 2 1.......... 0n n n
n na x a x a x a x a
Ans: Let 1 2, ,........., n are roots of 1 20 1 2 1.......... 0n n n
n na x a x a x a x a .
Then 1
1 0i
i n
aa
, 2 2
0
( 1)i ji j
aa
3 3
0
( 1)i j ki j k
aa
Similarly, 1 20
......... ( 1)n nn
aa
26. If , , are the roots of the cubic 3 0x px q . Show that i. 5 5 5 5
ii. 2 5 3 43 5
Ans: As , , are the roots of the cubic 3 0x px q .--------(1)
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So, 0, ,p q
Now, 22 2 0 2 2p p .
2 2 p -----------(2)
As 3 30 3 0 3( ) 0q
3 3q -------------(3)
Let us multiply equation (1) by x we get 4 2 0x px qx
4 2 0p q
4 2 2( 2 ) (0) 2p q p p q p
4 22 p ---------------(3)
(i) 5 5 5 5
Ans: Now by multiplying (1) by 2x we get 5 3 2 0x px qx
5 3 2 0p q
5 3 2 ( 3 ) ( 2 ) 5p q p q q p pq ----------(4)
5 5( )q p
5 5
5 5 5 5 (Proved)
(ii) 2 5 3 43 5
Ans: 2 5 23 3.( 2 ).(5 ) 30p pq p q -------------(5)
3 4 2 25 5( 3 )(2 ) 30q p p q ------------(6)
From (5) and (6) 2 5 3 43 5 (Proved)
27. Show that the equation 4 3 210 23 6 15 0x x x x can be transformed into reciprocal
equation by diminishing the roots by 2. Hence solve the equation. Ans: Let us diminished the roots of 4 3 210 23 6 15 0x x x x --------------(1) by 2. Putting 2 2y x x y By Using synthetic division method
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The transformed equation is 4 3 22 13 2 1 0y y y y ------------(2) Equation (2) is a reciprocal equation of even degree having coefficients of terms equidistant from the beginning and end equal.
Dividing equation (2) by 2y we get 22
2 12 13 0y yy y
22
1 12 13 0y yy y
Now putting 2 22
1 1 2y z y zy y
2 2 2 13 0z z 2 2 15 0 ( 5)( 3) 0z z z z 3,5z
1 3yy
1 5yy
2 1 3y y 2 1 5y y
2 3 1 0y y 2 5 1 0y y
3 9 42
y
5 25 42
y
3 52
y
5 212
y
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3 5 3 5 5 21 5 21, , ,2 2 2 2
y
So, solution for 2x y
3 5 3 5 5 21 5 212, 2, 2, 22 2 2 2
x
1 5 1 5 9 21 9 21, , ,2 2 2 2
x (Ans)
28. Solve by Cardan’s method, the equation 3 227 54 198 73 0x x x .
Ans: 3 227 54 198 73 0x x x
3 2 22 732 03 27
x x x ------------(1)
Let us compare this equation with 3 2 0ax bx cx d
Then 22 731, 2, ,3 27
a b c d
Putting 2 23 3
z ax b x x z in (1) we get
3 22 2 22 2 732 0
3 3 3 3 27z z z
3 2 24 8 8 8 22 44 732 2 03 27 3 9 3 9 27
z z z z z z
3 6 7 0z z ------------------------(2) Now comparing with 3 3 0z Hz G Where 2, 7H G
3 21 1 14 7 32 49 7 9 82 2 2
p G H G
So roots of given cubic equation (2) are
(i) 1/31/3
22 12
Hpp
(ii) 1/3 2 2 21/3
22 2 2( ) 32
Hwp w w w w w wwp w
1/31/3
1 3 4 3 3 3 1 3 32 32 2 2
H i i iwpwp
(iii) 2 1/3 2 2 22 1/3 2
22 2 2( ) 32
Hw p w w w w w ww p w
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2 1/32 1/3
1 3 4 3 3 3 1 3 32 32 2 2
H i i iw pw p
So, roots of equation (2) are 1 3 3 1 3 31, ,
2 2i i
.
Now, roots of given equation (1) are 2 1 3 3 2 1 3 3 21 , ,3 2 3 2 3
i i
1 3 9 3 4 3 9 3 4, ,3 6 6
i i
1 7 9 3 7 9 3, ,3 6 6
i i (Ans).
Nov-Dec, 2009
29. Write the relation between roots and coefficients of the equation. 1 2
0 1 2 1.......... 0n n nn na x a x a x a x a
Ans: Let 1 2, ,........., n are roots of 1 20 1 2 1.......... 0n n n
n na x a x a x a x a .
Then 1
1 0i
i n
aa
, 2 2
0
( 1)i ji j
aa
3 3
0
( 1)i j ki j k
aa
Similarly, 1 20
......... ( 1)n nn
aa
30. If ,, are the roots of the equation 3 0x px q , then show that i. 5 5 5 5
ii. 2 5 3 43 5 .
Ans: As , , are the roots of the cubic 3 0x px q .--------(1)
So, 0, ,p q
Now, 22 2 0 2 2p p .
2 2 p -----------(2)
As 3 30 3 0 3( ) 0q
3 3q -------------(3)
Let us multiply equation (1) by x we get 4 2 0x px qx
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4 2 0p q
4 2 2( 2 ) (0) 2p q p p q p
4 22 p ---------------(3)
(i) 5 5 5 5
Ans: Now by multiplying (1) by 2x we get 5 3 2 0x px qx
5 3 2 0p q
5 3 2 ( 3 ) ( 2 ) 5p q p q q p pq ----------(4)
5 5( )q p
5 5
5 5 5 5 (Proved)
(ii) 2 5 3 43 5
Ans: 2 5 23 3.( 2 ).(5 ) 30p pq p q -------------(5)
3 4 2 25 5( 3 )(2 ) 30q p p q ------------(6)
From (5) and (6) 2 5 3 43 5 (Proved)
31. Show that the equation 4 3 210 23 6 15 0x x x x can be transformed into reciprocal equation by diminishing the roots by 2. Hence solve the equation
Ans: Let us diminished the roots of 4 3 210 23 6 15 0x x x x --------------(1) by 2. Putting 2 2y x x y By Using synthetic division method
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The transformed equation is 4 3 22 13 2 1 0y y y y ------------(2) Equation (2) is a reciprocal equation of even degree having coefficients of terms equidistant from the beginning and end equal.
Dividing equation (2) by 2y we get 22
2 12 13 0y yy y
22
1 12 13 0y yy y
Now putting 2 22
1 1 2y z y zy y
2 2 2 13 0z z 2 2 15 0 ( 5)( 3) 0z z z z 3,5z
1 3yy
1 5yy
2 1 3y y 2 1 5y y
2 3 1 0y y 2 5 1 0y y
3 9 42
y
5 25 42
y
3 52
y
5 212
y
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3 5 3 5 5 21 5 21, , ,2 2 2 2
y
So, solution for 2x y
3 5 3 5 5 21 5 212, 2, 2, 22 2 2 2
x
1 5 1 5 9 21 9 21, , ,2 2 2 2
x (Ans)
32. Solve by Cardan’s method, the equation 3 227 54 198 73 0x x x .
Ans: 3 227 54 198 73 0x x x
3 2 22 732 03 27
x x x ------------(1)
Let us compare this equation with 3 2 0ax bx cx d
Then 22 731, 2, ,3 27
a b c d
Putting 2 23 3
z ax b x x z in (1) we get
3 22 2 22 2 732 0
3 3 3 3 27z z z
3 2 24 8 8 8 22 44 732 2 03 27 3 9 3 9 27
z z z z z z
3 6 7 0z z ------------------------(2) Now comparing with 3 3 0z Hz G Where 2, 7H G
3 21 1 14 7 32 49 7 9 82 2 2
p G H G
So roots of given cubic equation (2) are
(i) 1/31/3
22 12
Hpp
(ii) 1/3 2 2 21/3
22 2 2( ) 32
Hwp w w w w w wwp w
1/31/3
1 3 4 3 3 3 1 3 32 32 2 2
H i i iwpwp
(iii) 2 1/3 2 2 22 1/3 2
22 2 2( ) 32
Hw p w w w w w ww p w
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2 1/32 1/3
1 3 4 3 3 3 1 3 32 32 2 2
H i i iw pw p
So, roots of equation (2) are 1 3 3 1 3 31, ,
2 2i i
.
Now, roots of given equation (1) are 2 1 3 3 2 1 3 3 21 , ,3 2 3 2 3
i i
1 3 9 3 4 3 9 3 4, ,3 6 6
i i
1 7 9 3 7 9 3, ,3 6 6
i i (Ans).
May-June 2010
33. If ,, are the roots of the equation 08126 23 xxx , find an equation whose roots are 2,2,2 .
Ans: - It is same as to find an equation whose roots are diminished by 2 of 08126 23 xxx 0)2( 3 x -----------------(1) Let 22 yxxy ,
so eqn (1) becomes 03 y . Then the equation is 3 0x .
34. If the roots of the equation 023 rqxpxx are in H.P., then prove that
02927 32 qpqrr .
Ans: - Let ,, are the roots of the equation 023 rqxpxx . Here , ,p q r Given that roots are in HP.
So, 1 1 2
2 2 3
3 33 r
q
As is a root of the given eqn so it must satisfy the equation.
Hence 3 2 0p q r
3 2 3 2 3
3
3 3 3 27 9 20 0r r r r pqr rqp q rq q q q
3 2 327 9 2 0r pqr rq
2 327 9 2 0r pqr q (Proved).
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35. If ,, are the roots of the equation 03 rqxx , form an equation whose roots are:
(i)
,, .
(ii) 2 2 2, ,
Ans: - Here 0, ,q r
i. , ,
Let 22 2 2 2 2 2 2, , , , , ,r r r r ry x
x y
Now, 03 rqxx 3 6 4 2 2 22x qx r x qx q x r
3 22 22r r rq q r
y y y
3 22 2
3 22r r rq q ry y y
2 2 2 32r qry q y ry
3 2 2 22 0ry q y qry r
New eqn is 3 2 2 22 0rx q x qrx r
(ii) 222 ,,
Ans: - Here 0, ,q r
Let 2 2 2 2 2 2
1 1 1, , , ,y
1,1,110,10,10 y as 0
yx
xy 11
So, eqn 03 rqxx becomes 01011 323
ryqyr
yq
y
0123 qyry
New eqn is 0123 qxrx
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36. Solve by Cardan’s method: 01928 23 xx Ans: - 01928 23 xx ------------------(1)
Let 1 1y xx y
, then eqn (1) becomes
3 9 28 0y y ---------------(2)
Here the equation having terms involving 2y term missing. Let y u v be the solution of equation (2).
3 3 3 3 ( )y u v uv u v 3 3 3 3y u v uvy
3 3 33 0y uvy u v -------------(3)
By comparing equation(2) and (3) we get 3 33, 28uv u v 3 3 3 327, 28u v u v
So, 3 3,u v are the roots of the equation 2 28 27 0t t
( 1)( 27) 0 1, 27t t t . So, let 3 31, 27u v .
21, ,u w w , then 23 3, 3 , 3v w wu
Then 2 21 3, 3 , 3y u v w w w w 24,1 2 ,1 2y u v w w
1 3 1 34,1 2 ,1 22 2i iy u v
4, 2 3,2 3y i i
1 1 2 3 2 3, ,
4 7 7i ix
y
So, roots are 1 2 3 2 3, ,
4 7 7i ix
. (Ans).
Nov-Dec 2010
37. State the intermediate property.
Ans Intermediate property of roots says that for the equation ( ) 0f x for two numbers
let a and b such that, )(af and )(bf have different sign then equation 0)( xf has at least one root lies between a and b.
38. If ,, are the roots of the equation 3 0x px q , then show that i. 5 5 5 5
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ii. 2 5 3 43 5 .
Ans: As , , are the roots of the cubic 3 0x px q .--------(1)
So, 0, ,p q
Now, 22 2 0 2 2p p .
2 2 p -----------(2)
As 3 30 3 0 3( ) 0q
3 3q -------------(3)
Let us multiply equation (1) by x we get 4 2 0x px qx
4 2 0p q
4 2 2( 2 ) (0) 2p q p p q p
4 22 p ---------------(3)
(i) 5 5 5 5
Ans: Now by multiplying (1) by 2x we get 5 3 2 0x px qx
5 3 2 0p q
5 3 2 ( 3 ) ( 2 ) 5p q p q q p pq ----------(4)
5 5( )q p
5 5
5 5 5 5 (Proved)
(ii) 2 5 3 43 5
Ans: 2 5 23 3.( 2 ).(5 ) 30p pq p q -------------(5)
3 4 2 25 5( 3 )(2 ) 30q p p q ------------(6)
From (5) and (6) 2 5 3 43 5 (Proved)
39. Solve: 5 4 3 26 41 97 97 41 6 0x x x x x . Ans: 5 4 3 26 43 43 6 0x x x x x
The given equation is a reciprocal equation of odd degree having coefficient of terms equidistance from beginning and end.
So, 1x is its root.
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4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x
Dividing by 2x we get
22
5 66 5 38 0x xx x
22
1 16 5 38 0x xx x
21 16 2 5 38 0x xx x
26 2 5 38 0t t where 1x tx
26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2
t t t
1 10 1 53 2
x and xx x
2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x
10 100 36 5 25 166 4
x and x
10 8 5 36 4
x and x
1 13, 2,2 2
x and x
So, roots are 1 12, 1, , , 32 2
x (Ans)
40. Solve by Ferrari’s method: 031052 234 xxxx .
Ans: - 031052 234 xxxx ---------------------------(1) Equation (1) can be written as by combining 푥 and 푥 into perefect square
0)()( 222 nmxxx ----------------------------(2)
0)22()12(2 222234 nmnxmxxx ----------(3) 퐹푟표푚 (1)푎푛푑 (3)
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33
5102262512
2222
22
nnmnmnmm
--------------------(4)
From (4) 222 )(. mnnm 22 )5()3).(62(
07452 23 Then 1 is one of the root. So, 246)1(22 mm (one of the value)
245 nmn So, (2) can be written as 0)22()1( 222 xxx
0)221)(221( 22 xxxxxx
0)13)(3( 22 xxxx
0)13(,0)3( 22 xxxx
2493,
21211
xx
253,
2131
xx (Ans)
April -May-2011
41. What is geometrical meaning of root of the equation f(x)=0
Ans. For any value of x which satisfies 0)( xf are called as solution of 0)( xf or root of
0)( xf .
Example: - For 062 x
a. If 2x , it is not satisfies 062 x . So, 2x is not a solution of 062 x . If 3x , it is not satisfies 062 x . So, 3x is a solution of 062 x .
Geometrically for the curve f(x) =0 the roots are the curve of , , ,................x a x b x c which touches the curve at a, b, and c respectively.
42. Solve the following equations : 6x5-41x4+97x3-97x2+41x-6=0
Ans: 5 4 3 26 43 43 6 0x x x x x
The given equation is a reciprocal equation of odd degree having coefficient of terms equidistance from beginning and end.
So, 1x is its root.
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4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x
Dividing by 2x we get
22
5 66 5 38 0x xx x
22
1 16 5 38 0x xx x
21 16 2 5 38 0x xx x
26 2 5 38 0t t where 1x tx
26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2
t t t
1 10 1 53 2
x and xx x
2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x
10 100 36 5 25 166 4
x and x
10 8 5 36 4
x and x
1 13, 2,2 2
x and x
So, roots are 1 12, 1, , , 32 2
x (Ans)
43. Find the equation of squared difference of the roots of the cubic 3 2x 6x 7x 2 0.
Ans: Let , , are the roots of 3 2x 6x 7x 2 0. …………………(1) Now, 6 --------------(2) 7 ------------(3) 2 -------------(4)
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As sum of the roots is -6 therefore to remove the 2nd term we will make a new equation whose roots will be increased by 2
Here by removing the term containing 2x We will have
2 1 6 7 2
0 2 8 2
1 4 1 4
0 2 4
1 2 5
0 2
1 0
The transformed equation will be 3-5y 4 0.y and roots of it will be 2, 2, 2 …(5)
Let 2, 2, 2a b c
Now, 0a b c --------------(6) 5ab bc ca ------------(7) 4abc -------------(8)
2 2
2 2
2 2
b c
c a
a b
And we are here to find 2 2 2, , which is equivalent to
2 2 2, ,a b b c c a if z be the root of the equation squared differences we will have
2 2 4z b c b c bc
2 2 4z b c b c bc from equation 6 and 8 we will have
3
2 2 2 4 164 abc az b c b c bc aa a
3 16aza
3 16yzy
as y is the root of the equation 3-5y 4 0.y and equivalent to a, b, and c
3 16 0y zy ………………………(9)
Subtracting the equation 9 from equation 8 we will have 125 12 0
5z y y
z
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Putting the value of y in the equation 8 we will have
312 12-5 4 0.
5 5z z
3 2-30z 225 68 0.z z and roots of the obtained equation are 2 2 2, ,a b b c c a
which is equivalent to 2 2 2, , .
Hence 3 2-30z 225 68 0.z z is the required equation.
44. Solve by Cardan’s methods: x3-3x+1=0.
Ans: 3 3 1 0x x --------------(1)
Hence the transformed equation is 3 3 1 0x x ---------(2)where
Let 1/3 1/3x p q be the solution of equation (2).
3 1/3 1/3 1/3 1/33x p q p q p q
3 1/3 1/33x p q p q x
3 1/3 1/33 ( ) 0y p q y p q -------------(3)
By comparing equation (2) and (3) we get
1/3 1/3 1, ( ) 1p q p q
1, ( ) 1pq p q
So, p and q are the roots of the equation 2 1 0t t
1 1 4 1 3
2 2it
.
So, let1 3 1 3,
2 2i ip q
.
So roots of given cubic equation (1) are
(i) 1/3 1/3
1/3 1/3 1 3 1 32 2i ip q
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1/3 1/32 2 2 2cos sin cos sin
3 3 3 3i i
2 2 2 2cos sin cos sin9 9 9 9
i i
22cos9
(ii) 1/3 1/3
1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q
4/3 4/3
1 3 1 32 2i i
4/3 4/32 2 2 2cos sin cos sin
3 3 3 3i i
8 8 8 8cos sin cos sin9 9 9 9
i i
82 cos9
(iii) 1/3 1/3
2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq
1/3 1/32 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin
3 3 3 3 3 3 3 3i i i i
2 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin3 3 9 9 3 3 9 9
i i i i
2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3
i i
4 4 4 4cos sin cos sin9 9 9 9
i i
4 4 142cos 2cos 2 2cos9 9 9
So, roots of equation (2) are 2 8 142cos , 2 cos , 2 cos9 9 9
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Dec-Jan-2011
45. Explain Descartes’ rule of sign of equation. Ans Descarte’s Rule of Sign:- a. Maximum number of positive real root of 0)( xf is same as the number of sign changes
in )(xf . b. Maximum number of negative real root of 0)( xf is same as the number of sign changes
in )( xf . c. If an equation having degree n such that at most p number of positive roots, at most q
number of negative roots, then equation has atleast )( qpn number of imaginary roots.
46. If O, A, B, C are the four point on a straight line such that the distance of A, B and C from
O are the roots of the equation 3 3 0ax cx d . If B is the middle point of AC show that 3 33 2 0a d abc b
Ans . Let us assume , , are the roots of the equation 3 23 3 0ax bx cx d and , ,OA OB OC
According to the given condition 2 --------------(1) According to the property of roots
Now, 3ba
--------------(2)
3ca
------------(3)
da
-------------(4)
From 2 and 3 we will have 33 b ba a
-------------(5)
3 3c ca a
From 1 2
2 3 32 2c b ca a a
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2
2
3 2c ba a
-------------(6)
again from 4 we have
d
a
da
From 5 and 6 we have 2
2
3 2b c b da a a a
By solving above we will have
2
2
3 2ca bb da
Hence we will have 3 33 2 0a d abc b
47. Solve by Cardan’s methods : X3 – 3X2 + 3 =0.
Ans: 3 23 3 0x x --------------(1)
Here we can remove second term of equation (1) by diminishing its roots by
3 1
3 1bhna
We can diminished each root by 1 by synthetic division method
Hence the transformed equation is 3 3 1 0y y ---------(2)where 1y x
Let 1/3 1/3y p q be the solution of equation (2).
3 1/3 1/3 1/3 1/33y p q p q p q
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3 1/3 1/33y p q p q y
3 1/3 1/33 ( ) 0y p q y p q -------------(3)
By comparing equation (2) and (3) we get
1/3 1/3 1, ( ) 1p q p q
1, ( ) 1pq p q
So, p and q are the roots of the equation 2 1 0t t
1 1 4 1 3
2 2it
.
So, let1 3 1 3,
2 2i ip q
.
So roots of given cubic equation (1) are
(i) 1/3 1/3
1/3 1/3 1 3 1 32 2i ip q
1/3 1/32 2 2 2cos sin cos sin
3 3 3 3i i
2 2 2 2cos sin cos sin9 9 9 9
i i
22cos9
(ii) 1/3 1/3
1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q
4/3 4/3
1 3 1 32 2i i
4/3 4/32 2 2 2cos sin cos sin
3 3 3 3i i
8 8 8 8cos sin cos sin9 9 9 9
i i
82 cos9
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(iii) 1/3 1/3
2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq
1/3 1/32 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin
3 3 3 3 3 3 3 3i i i i
2 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin3 3 9 9 3 3 9 9
i i i i
2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3
i i
4 4 4 4cos sin cos sin9 9 9 9
i i
4 4 142cos 2cos 2 2cos9 9 9
So, roots of equation (2) are 2 8 142cos , 2 cos , 2 cos9 9 9
Now, roots of given equation (1) are 2 8 141 2cos , 1 2 cos , 1 2cos9 9 9
(Ans)
48. Solve by Ferrari’s method: 4 3 212 41 18 72 0x x x x .
Ans: 4 3 212 41 18 72 0x x x x
Let 4 3 2 2 2 2( ) 12 41 18 72 ( 6 ) ( ) 0f x x x x x x x mx n ----------(1)
4 3 2 4 2 2 3 2 2 2 212 41 18 72 36 12 12 2 2x x x x x x x x x m x mnx n
4 3 2 4 3 2 2 2 212 41 18 72 12 (36 2 ) ( 12 2 )x x x x x x m x mn x n
Equating the coefficients we get 2 2 2(36 2 ) 41, ( 12 2 ) 18, 72m mn n
2 2 22 5, 9 6 , 72m mn n
222 5 72 9 6
3 2 22 144 5 360 81 36 108 0 3 22 41 252 441 0 2( 3)(2 35 147) 0
2123,7,
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By taking 3 1, 9m n equation (1) becomes
4 3 2 2 2 212 41 18 72 ( 6 3) ( 9) 0x x x x x x x
2 2( 6 3 9)( 6 3 9) 0x x x x x x
2 2( 5 6)( 7 12) 0x x x x ( 6)( 1)( 3)( 4) 0x x x x
1,3, 4,6x (Ans)
April -May-2012
49. Solve the equation 3 6x 20 0 x , one root being 1 3 .i Ans . As we know that from the general properties of roots of an equation that In an equation with real coefficients, if i is a root then i is also one of the root, therefore two
roots of the equation 3( ) 6x 20 f x x are 1 3 .i and 1 3 .i
( ) 1-3i 1+3if x x x x a
( ) 1 3 1 3f x x i x i x a
( ) 1 3 1 3f x x i x i x a
2( ) 1 9f x x x a
2( ) 2 1 9f x x x x a
2( ) 2 8f x x x x a
2
( )2 8
f x x ax x
3
2
6x 20
2 8
xx a
x x
2 2x x a a
Roots of the equation 3 6x 20x are 1 3 .,1 3 2i i and and
50. Solve the equation Solve the equation 3 2x 4x 20x 48 0 , given that the roots 훼 푎푛푑 훽 are connected by the relation 0 Ans . Let us assume , , are the roots of the equation 3 2x 4x 20x 48 0 and According to the given condition 0 --------------(1) According to the property of roots Now, 4 --------------(2) From 1 we will have 0 4 4
20 ------------(3)
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48 -------------(4)
From 2 and 4 we will have 4 48 12 -------------(5)
By using the equation 1 and 5 we 0 and 12
We can have quadratic equation from above i.e 2 -12 0y and 2 3y
2 3 2 3and and three roots of the equation 3 2x 4x 20x 48 0 are
2 3 , 2 3 4and
51. Solve the equation 3 26x 11x 3x 2 0 , given that its roots are in H.P. Ans. We know that H.P is reciprocal of A.P then by taking 1/y x the new equation will be
3 23 21 1 16 11 3 2 0 2 3y 11 6 0y y
y y y
now roots of the equation
3 22 3y 11 6 0y y are in A.P As , ,d d are the roots of the cubic equation 3 22 3y 11 6 0y y .
By properties of relation between roots and coefficient of the equation we will have
3 / 2d d
3 3 / 2 1/ 2
Again we will have 3d d
2 2 3d
221 1 3
2 2d
21 64
d
2 1 2564 4
d
52
d by putting the value of 5 / 2d in , ,d d we will have roots of the
equation 3 22 3y 11 6 0y y are 2,1/ 2,3 and roots of the equation 3 26x 11x 3x 2 0 are 1/ 2, 2 1/ 3and
52. Solve by Cardan’s methods: 3 x 18x 35 0.
Ans: 3 x 18x 35 0. Here the equation having terms involving 2x term missing.
Let x u v be the solution of equation (1).
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3 3 3 3x u v uv u v
3 3 3 3x u v uvx
3 3 33 0x uvx u v -------------(2)
By comparing equation(1) and (2) we get 3 36, ( ) 35uv u v ------------(3)3 3 3 3 3 36, ( ) 35 216, ( ) 35uv u v u v u v
So, 3u and 3v are the roots of the equation 2 35 216 0t t 2 35 216 ( 27)( 8) 27, 8t t t t t .
So, let 3u = -27 and 23, 3 , 3u w w therefore from 3 we will have 22, 2 , 2v w w
So roots of given cubic equation (1) are (iv) 3 2 5u v
(v) 2 1 3 1 3 5 33 2 3 22 2 2i i iu v w w
(vi) 2 1 3 1 3 5 33 2 3 22 2 2i i iu v w w
So, roots are 5 3 5 35, ,
2 2i i
(Ans).
Dec –Jan -2012
53. From the equation of the fourth degree whose roots are 3 + i and √7 . Ans . As we know that from the general properties of roots of an equation that In an equation with real coefficients, if i is a root then i is also one of the root, and for the
irrational roots if an equation has ba is one of the root, then ba is also another root.
Therefore two roots of the equation we will have 3 i and 3-i , 7 7and and the
equation will be ( ) 3+i 3-i 7 7 f x x x x x
( ) 3 3 7 7 f x x i x i x x
22 2 2( ) 3 7f x x i x
22 2 2( ) 3 7f x x i x
2 2( ) 9 6 1 7f x x x x
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4 2 2 3 2( ) 7 9 63 6 42 7f x x x x x x x 4 3 2( ) 6 3 42 70f x x x x x
54. Solve the equations: 5 4 3 2 6x x 43x – 43x 6 0 .x
Ans: 5 4 3 26 43 43 6 0x x x x x
The given equation is a reciprocal equation of odd degree having coefficient of terms equidistance from beginning and end.So, 1x is its root.
4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x
Dividing by 2x we get
22
5 66 5 38 0x xx x
22
1 16 5 38 0x xx x
21 16 2 5 38 0x xx x
26 2 5 38 0t t where 1x tx
26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2
t t t
1 10 1 53 2
x and xx x
2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x
10 100 36 5 25 166 4
x and x
10 8 5 36 4
x and x
1 13, 2,2 2
x and x
So, roots are 1 12, 1, , , 32 2
x (Ans)
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55. Solve by Cardan’s methods : 3 – 27x 54=0 x
Ans: 3 – 27x 54=0 x Here the equation having terms involving 2x term missing.
Let x u v be the solution of equation (1). 3 3 3 3x u v uv u v
3 3 3 3x u v uvx
3 3 33 0x uvx u v -------------(2)
By comparing equation(1) and (2) we get 3 39, ( ) 54uv u v ------------(3)
3 3 3 3 3 39, ( ) 54 729, ( ) 54uv u v u v u v
So, 3u and 3v are the roots of the equation 2 54 729 0t t 2 2 254 27 ( 27) 27t t t t .
So, let 3u = -27 and 23, 3 , 3u w w therefore from 3 we will have 23, 3 , 3v w w
So roots of given cubic equation (1) are (i) 3 3 6u v
(ii) 2 1 3 1 3 63 3 3 3 32 2 2i iu v w w
(iii) 2 1 3 1 3 63 3 3 3 32 2 2i iu v w w
So, roots are 6, 3, 3 (Ans).
56. Solve by Ferrari;s method : X4 – 4X3 –X2 + 16X – 12=0.
Ans: 4 3 2 – 4x – 16x – 12 0.x x
Let 4 3 2 2 2 2( ) – 4x – 16x – 12 ( 2 ) ( ) 0f x x x x x mx n ----------(1)
4 3 2 4 2 2 3 2 2 2 2– 4x – 16x – 12 4 4 4 2 2x x x x x x x m x mnx n 4 3 2 4 3 2 2 2 2– 4x – 16x – 12 4 (4 2 ) ( 4 2 )x x x x m x mn x n
Equating the coefficients we get 2 2 2(4 2 ) 1, ( 4 2 ) 16, 12m mn n
2 2 22 5, 8 2 , 12m mn n ……………………..(2)
222 5 12 8 2
3 2 22 24 5 60 64 4 32
CSVTU II SEMESTER SOLUTION ALL UNITS
Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)
3 22 8 4 0
By solving the above equation we will have 2, 2, 1 / 2
By taking 2, in the equation 2 we will have 2 29, 12, 16m mn n
23, 12, 4m mn n
3, 4 3, 4m n or m n
By putting the value of above in the equation 1 we will have 4 3 2 2 2 2( ) – 4x – 16x – 12 ( 2 2) (3 4) 0f x x x x x x 4 3 2 2 2( ) – 4x – 16x – 12 ( 2 2) (3 4) ( 2 2) (3 4)f x x x x x x x x x 4 3 2 2 2( ) – 4x – 16x – 12 2 2 3 4 2 2 3 4f x x x x x x x x x
4 3 2 2 2( ) – 4x – 16x – 12 5 6 2f x x x x x x x
2 25 6 3 2 , 2 2 1x x x x x x x x
4 3 2 ( ) – 4x – 16x – 12 3 2 2 1f x x x x x x x
Roots of the equation 4 3 2 – 4x – 16x – 12 0.x x 3,2,-2and 1