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8/9/2019 Differential Equations - Solved Assignments - Semester Spring 2007
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ASSIGNMENT 01
Question 1:
Marks=10
Solve (1 )x ydye x e
dx
The equation is not separable
( )
2
2
2
1
1 ...............................(1)
1
. (1)
int
2
0
2
ln2
x y
x y
z
z
z
x y
dy xe e
dx
dyxe
dx
put z x y then
dz dy
dx dx
Equation no becomes
dzxe
dx
e dz xdx
egrating we have
xe c
r
xe c where z x y
x x y c
Question 2:
Marks=10
Solve2 2 2( ) 0 x xy y dx x dy
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2 2 2
2 2
2
2 2 2 2 2 2 2 2
2 2 2
2
2 2
2
( )
........................................(1)
(1)
( ) ( ) (1 )
1
1 1
1
x dy x xy y dx
dy x xy y
dx x
put y vx
dy dvv xdx dx
equation becomes
dv x x vx vx x x v v x x v vv x
dx x x x
v v
dv x v v v v
dx
dv v
dx xd
2
2
1
1
1
int
1
tan ln
tan ln
v dx
v x
egrating we get
dv dx
v x
v x c
yx c
x
is the required general solution
Question 3:
Marks=10
Determine whether the given equation is Exact, if so please
2 2 2(2 tan ) ( tan ) 0 xy y y dx x x y Sec y dy
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2 2 2
2 2
2
2 2 2
2
2 tan
tan sec
2 1 sec 2 tan
2 tan
( , )
2 tan (1)
tan sec (2)
int (1) . .
( , ) tan
M xy y y
N x x y y
M x y x y
Y
N x yX
the eqis exact thereforethe function f x y is
f xy y y
x
f x x y y
y
egrating w r t x
f x y x y xy x y h
2 2 / 2 2 2
( ) (3)
( ) tan int. .
sec ( ) tan sec
y
whereh y isthecons t of egrationdiff the above equation w r t y
fx x x y h y x x y y
y
/ 2
2
( ) sec
( ) tan
(3)( , ) tan tan
h y y
h y y
therefore eq becomes f x y x y xy x y y c
Question 4:
Marks=10
Solve (by finding I.F)
1ydy
dx e x
the equation may be written as
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2
2
2
(1)
.
. (1)
2
2
y
y
dy y
y y
y y
yy
yy
dxe x
dy
or
dxx e
dyIt is a linear in x
I F e e multiply by eq
therefore
d xe e
dy
or
xe e dy c
e xe c
or
e x ce
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ASSIGNMENT 02
uestion 1: Marks=10
olve the differential equation and mention the name of type of this D.E
3dy y xy
dx
3
3 2
2
3
2
3
.
1 1
1
. . .
2
olution
dy y xy
dx
t is known asa
dyx
y dx y
dyy x
y d
bernoulli different
x
putting y z
Diff w r t x
dy dzy
dx dx
ial equation
3
3
2
1
2
2
2 2
.
2, 2
dy dz
y dx dx
dy dz
y dx dx
dzz xdx
dzz x
dx
is linera differential equation of the first order
p q x
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2
2
2 2
2
2 2
22 2
22 2
2
,
.
( . ) ( . )
( ) ( ) 2
. ( 2)
. 1.
( )2
1
2
pdx
dx
x
x x
x
x x
x
x x
x
x x
sweknowthat Integrating factoris e so
F e
I F I F q dx
z e e x dx
x e dx
x e e dx
ez e x e c
ee x e c
y
uestion 2: Marks=10
nd an equation of orthogonal trajectories of the curve2
x cy
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2
2
2
2 2
22
2 2
(1)
2
(1)
2 . .
2.
2
,
2
2
nt
2
22 2
.
22 .
x cy
dycy
dx
xFromequation wehave c
y x dy
y y dx
x dy
y dx
dy y
dx x
Bythe rule of orthogonal trajectories we get
dy x
dx yy dy x dx
egrating both sides
y dy x dx
y xc
yx c
y x c
Question 3: Marks=10
fossilized bone is found to contain1
1000of the original amount of C-14. Determine the age of the fissile.
et A(t)be the amount present at any time tand A0the original amount of C14.herefore, the process is governed by the initial value problem.
0
A, (0)
dtkA A A
We know that the solution of the problem is
0( )ktA t A e
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nce half life of the carbon isotope is 5600 years, Therefore
0A(5600)=2
A
o that
560000
5600 ln 22
kA A e or k
= -0.00012378
ence(0.00012378)
0( ) tA t A e
tdenotes the time when fossilized bone was found then
0( )1000
AA t
(0.00012378)0
0000
tAA e
0.00012378 ln1000t
herefore
ln100055,800 .
0.00012378 years
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ASSIGNMENT 03
Maximum Marks: 30
Upload Date: April 11, 2007
uestion 1: Marks=10olve the initial value problem
2
2
'
6 9 0
(0) 2 , (0) 3
d y dyy
dx dx
with y y
olution:
iven equation is
2
26 9 0
d y dyy
dx dx
xpected solution of this equation is y = emx
ubstituting it in the differential equation, we get
2
26 9 0
mx mx mxd de e edx dx
fter differentiation, we get2 6 9 0mx mx mxm e me e
We factorize it
2 6 9 0mx m m nce 0
mxe ,we conclude that
2 6 9 0m m 23) 0
3 , 3
m
m
neral solution is3 3
1 2
x xy C e C xe
iven the initial conditions y(0) = 2 and y(0) = -3
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3(0) 3(0)
1 2
0 3(0)
1 2
1
(0) 2 (0)
2 (0)
2
y C e C e
C e C e
C
3 3 3
1 2
3 3 3
1 2 2
3
1 2 2
'( ) ( 3) ( 3)
3 3
3 3
x x x
x x x
x
y x C e C xe e
C e C xe C e
e C C x C
3(0)
1 2 2
0
1 2
1 2
'(0) 3 3 3 (0)
3 3
3 3
y e C C C
e C C
C C
ut from first initial value of y(0), we got C1 = 2erefore,
2
2
2
3 3(2)
3 6
3
C
C
C
ubstituting the values of C1 and C2 in3 3
1 2x x
y C e C xe , we can get
3 32 3
x xy e xe
uestion 2: Marks=10
olve
' ' ' ' ' '6 3 10 0 y y y y
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olution:
' ' ' ' ' '6 3 10 0 ( )
2 3, ,
( )
3 26 3 10 0
3 2( 6 3 10) 0
0
3 26 3 10 0
by synth
y y y y i
mxPut y e
mx mx mx y me y m e y m e
Putting these values in the eq i
mx mx mx mxm e m e me e
mxm m m e
mxSince e
m m m
Which can be factorized
etic division we find
1 6 3 101
1 7 10
____________________
1 7 10 0
-1 is a root.he coefficients of the quotient are-7, 10
hus we can write the auxiliary equation as:21)( 7 10) 0m m m
1m or 2 7 10 0m m ( 5)( 2) 0m m
5m or 2m he roots are -1, 2 and 5.herefore solution of the differential equation is
2 5
1 2 2
x x xy C e C e C e
uestion 3: Marks=10
iven that2
1y x is a solution of2 ' ' '3 4 0 x y xy y
nd the general solution of the differential equation on the interval (0, ) .olution:he equation can be written as
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2
3 40
2
2
2
( )2 1 2
1
32
2 4
3ln2
2 4
122
2ln
2
(0, )
1 1 2 2
1
Dividing by x
y y yx x
he nd solution y is given by
Pdxe
y y x dxy
dxde
y x dxx
xey x dx
x
y x dxx
y x x
Hencethe gernal solutionof thedifferential equation
n is given by
c y c y
c
2 2 ln2
x c x x
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Assignment 4
Answer# (1):-
22
2
22
2
2 3 3
:
2 3 3 .................( )
Associated homogeneous differential equation is:
2 '' 3 ' 0.....................(i)
Now solution to the equation (i) is det
d y dy y x Sinxdx dx
solution
d y dy y x Sinx A
dx dx
y y y
2
2
1 2
ermined by the solution
of associated auxillary equation given by:2 3 1 0.....................(ii)
2 2 1 0
2 ( 1) 1( 1) 0
( 1)(2 1) 0
Therefore the roots of the given equation(ii) are
1,
m m
m m m
m m m
m m
m m
1
21 2
1
2
Hence, the complimentary solution to the
associated homogeneous equation (i) is given as
xx
cy c e c e
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2
Now we assume that the particular integral
cos sin
. . .' 2 sin cos
. . .
'' 2 cos sin
Substituting the values of , ', '' in the equation (A)we
will ha
p
p
p
p p p
y Ax Bx C D x E x
Diff w r t x
y Ax B D x E x
Again diff w r t x
y A D x E x
y y y
2 2
2 2
ve:
4 2 cos 2 sin 6 3 3 sin 3 cos + x+ + cosx+ sin +3sin
(6 ) (3 )cos (3 )sin (4 3 ) +3sin .................(iii)
Equating like coefficient on both sides of equation (
A D x E x Ax B D x E x Ax B C D E x x x
Ax A B x E D x D E x A B C x x
iii), we have:1
6 0 6
4 3 0 14
3 - 0................................( )
3 3 3 3 .............(v)
A
A B B
A B C C
E D iv
and
E D D E
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2
Solving equation (iv) and (v) simultaneously we have:
3E-D=0
3E+9D=-9
......................
-10D =99
D=-10
Now,
93E+ 0
10
93
10
9 1 3*
10 3 10Hence the particular integral :
9 36 14 cos sin
10 10
an
p
p
E
E
y is
y x x x x
1
221 2
d the general solution to the given non-homogeneous differential equation is:
9 36 14 cos sin .
10 10
c p
xx
y y y
y c e c e x x x x
Answer# (2):-
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' '
' '
' '
2
4 36 3
To find the complementary function we solve the associated homogeneous
differential equation:
4 36 0
4
9 0
The auxillary equation is
9 0 3
roots of the auxillary equ
y y Csc x
y y
Dividing by
y y
m m i
1 2
1 2
ation are complex. Therefore, the complementary
function is
cos3 sin 3
From the complementary function, we have
y cos3 , sin 3cos3 sin3
(cos3 ,sin 3 ) 3-3sin3 3cos3
Now by
c y c x c x
x y xx x
W x xx x
1 2
dividing with 4, we put the given equation in the following
standard form:
1'' 9 csc3
4
1So we identify the function ( ) csc3
4
We shall now construct the determinants and
y y x
f x x
W W
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1
2
1 2
1 21 2
0 sin31 1
csc3 .sin314 4csc3 3cos3
4
cos 3 01 cos3
1 4 sin33sin3 csc34
Therefore the derivatives ' and ' are given by
1 1 cos3' , '
12 12 sin3
Now i
x
W x xx x
xx
W xx x
u u
W W xu u
W W x
1 2
ntegrating the last two equations . . , we obtain
1 1' , ' ln sin3
12 36
The particular solution of the non-homogeneous equation is:
1 1cos3 (sin3 ) ln sin 312 36
Hence the general solution of the
p
w r t x
u ux x
y x x x x
1 2
given differential equation is:
1 1cos3 sin3 cos3 (sin 3 ) ln sin3
12 36c py y y c x c x x x x x
Answer #(3):-
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' ' ' '
' ' '
3 2
tan
To find the complementary function we solve the associated homogeneous
differential equation
' 0The auxillary equation is
0 ( 1) 0
0,
Therefore, the complementary funct
y y x
y y
m m m m
m m i
1 2 3
1 2 3
1 2 3
1 2 3
ion is
cos3 sin 3
,
1, cos , sin
Now we compute the wronskian of , ,
1 cos sin
( , , ) 0 - sin cos
0 -cos - sin
c y c c x c x
Therefore
y y x y x
y y y
x x
W y y y x x
x x
1 3
2 2
By the elementary row operation , we have
1 0 0
0 - sin cos
0 -cos - sin
(sin cos ) 1 0
The given differential equation is alrea
R R
x x
x x
x x
st nd rd
1 2 3
dy in the required standard form
''' 0. '' ' 0. tan
The determinants , and are found by replacing 1 ,2 ,3 column of by the column
y y y y x
W W W W
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2 2
1
2
0 cos sin
0 -sin cos tan (cos sin ) tan
tan - cos - sin
1 0 sin
0 0 cos 1(0 cos tan ) s
0 tan -sin
x x
W x x x x x x
x x x
x
W x x x
x x
3
1 2 3
31 21 2 3
in
1 cosx 0
0 sin 0 1( sin tan ) 0 sin tan
0 cos tan
Therefore the derivatives ' , ' , ' are given by
' tan , ' sin , ' sin tan
N
x
and
W x x x x x
x x
u u u
WW Wu x u x u x x
W W W
1 2 3ow integrating these three derivatives of the function ' , ' , 'u u u
11
22
33
2
2 2
sintan ln cos
cos
sin cos
sin tan
sinsin sin seccos
(cos 1)sec (cos sec sec )
(cos sec ) cos sec
sin ln sec tan
W xu dx xdx dx x
W x
Wu dx xdx x
W
Wu dx x xdx
W
x x dx x xdxx
x xdx x x x dx
x x dx xdx xdx
x x x
2 2
Thus, the particular solution of the non-homogeneous equation
ln cos cos cos (sin ln sec tan )(sin )
ln cos cos sin sin ln sec tan
ln cos 1 sin ln sec tan
Hence, the general equation of th
p y x x x x x x x
x x x x x x
x x x x
1 2 3
e given differential equation is:
cos sin - ln cos 1 sin ln sec tan y c c x c x x x x x
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ASSIGNMENT 05
Maximum Marks: 30
uestion 1: Marks=10
olve the initial value problem
2
2
/
2 2 4cos 2 int
(0) 0 , (0) 3
d x dx x t S
dt dt
x x
Solution:
2
2
mx 2
2
associated homogeneous equation
2 2 0
x=e , ,
the auxiliary equation is
m 2 2 0.
2 4 81
2
mx mx
t
c
First of all
d x dxx
dt dt
By putting the
x me x m e
Then
mfor finding roots
m i
Thus the complementry function is
x e
1 2cos sinc t c t
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p
2
2 2
2
2 2
the particular integer
x cos sin
sin cos
cos sin
2 2 cos sin 2 sin 2 cos 2 cos 2 sin
2 2 2 cos 2 sin
in the give
p
p
p p
p
p p
p
For
A t B t
x A t B t
x A t B t
d x dxx A t B t A t B t A t B t
dt dt
OR
d x dx x A B t A B t
dt dt
By substituting
n differencial equation,we have
2 cos 2 sin 4cos 2sin
coefficient,we obtain
A+2B=42A+B=2
A B t A B t t t
The equation
p
c
1 2
1 2 1
By solving these two equation,we get
A=0,B=2
Thus
2sin
of the differencial equation is as given below,
(t) = x
cos sin 2sin
. . .
cos sin sin
p
t
t t
t
Hence general solution
x
t e c t c t t
Diff w r t x
t e c t c t e c t
2
1 2
1 2
1
2
-t
cos 2cos
Now we apply the boundry conditions
0 0 , .1 .0 0 0 ,
0 3 , .1 .1 2 3 ,
solution of the initial value problem is
=e sin 2
0
sin
1
c t t
c c
c
c
cc
hus
t t
uestion 2: Marks=10
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olve by using (tx e )
22 2
23 5 sin(ln )
d y dy x x y x x
dx dx
Solution:
22 2
2
22
2
2 2
t
2 2
3 5 sin(ln )
First of all associated homegeneous differential equation is.
3 5 0
3 5) 0
With the substitution of x=e , We have:
xD= , 1
So, the homegeneous di
d y dyx x y x x
dx dx
d y dyx x y
dx dx
x D xD y
x D
fferential equation becomes,
2
2
2
2
2
2
1 3 5 0
3 5 0
4 5 0
4 5 0
4 5 0
y
y
y
OR
d dy
dt dt
d y dyy
dt dt
mt
2 mt mt mt
2 mt
mt 2
22
By putting the y=e , We get the auxiliary equation as ,
e 4 e 5e 0
4 5 e 0
0 , 4 5 0
4 4 4 54 4 16 20
2 2 2
m m
m m
m m
b b acm
a
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2
1 2
2
1 2
2 2
22
2
4 4 4 22
2 2
cos sin
ln
cos(ln ) sin(ln )
Now the non-homogeneous equation becomes,
So equation becomes
2 5 sin
By the method
t
c
t
t
t
ii
y e c t c t
where t x
yc x c x c x
according to condition
x e
so
x e
d y dy y e t dt dt
2
2
2
2
2
2
2
2
2
1
of undetermined coefficient we try a particular
olution of the form
1( sin )
4 5
1(sin )
2 4 2 5
1(sin )
1
(sin )2
cos2
, ,
( cos
t
p
t
p
t
p
t
p
t
p
c p
t
y e t
y e t
y e t
ty e t
ty e t
So the General solution
y y y
y e c t c
2
2
22
1 2
sin ) cos2
lncos ln sin ln cos ln , ln2
ttt e t
x xy x c x c x x Where t x
Question 3: Marks=10
nd the interval of convergence of the power series
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0 (2 )!
n
n
x
n
Solution:or finding interval of the series we use ratio test
1
1
1
1
1
2 !lim lim .
2 1 !
2 !.lim lim .
2 1 !
. 2 !
lim lim 2 2 !
2 !lim lim
2 2 !
2 !lim
2 2 (2 1) (2 )!
1lim 0 1
2 2 (2 1)
n
n
nn nn
n
n
nn nn
n
n nn
n
n nn
n
n
na x
a n x
na x x
a xn
x na
a n
nax
a n
nx
n n n
xn n
Thus the power series converge
.s x
hus the interval of the convergence of the given series is ,
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Assignment 6
uestion 1: Marks=10
ind solution of the differential equation
/ /4 0y y
in the form of a powers series in x.
Solution:
We assume that a solution of the given differential equation is
0
0 1
n n
n n
n n
y c x c c x
Then term by term differentiation
1 1
1
1 2
n n
n n
n n y nc x c nc x
22
1n
n
n
y n n c x
Substituting the expression for/ /
y and y .
we get
2
2 04 4 1
n n
n nn ny y n n c x c x
Notice that both series start with 0x . If we, respectively, substitute
k= n - 2, k= n, k= 0,1, 2,K
In the first series and second series on the right hand side of the last equation.
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Then we after using, in turn, n = k+ 2 and n k, we get
20 0
4 4 2 1k k
k k
k k
y y k k c x c x
20
4 4 2 1k
k k
k
y y k k c c x
Substituting in the given differential equation, we obtain
20
4 2 1 0k
k k
k
k k c c x
24 2 1 0
k kk k c c
2, k=0,1,2,k
4 2 1
kk
cc
k k
From iteration of this recurrence relation it follows that
0 02 24.2.1 2 .2!
c cc
1 13 24.3.2 2 .3!
c cc
024 4
3 15 4
4.4.3 2 .4!
4.5.4 2 .5!
cc
c c
046 6
5 17 6
4.6.5 2 .6!
4.7.6 2 .7!
cc
c c
o on
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his iteration leaves both0
c and1
c arbitrary.
We have
2 3 4 5 6 7
0 1 2 3 4 5 6 7 c c x c x c x c x c x c x c x L
2 3 4 5 6 70 0 01 1 10 1 2 2 4 4 6 62 .2! 2 .3! 2 .4! 2 .5! 2 .6! 2 .7!
c c cc c cy c c x x x x x x x L
r
2 4 6 3 5 710 12 4 6 2 4 6
1 1 1 1 11
2 .2! 2 .4! 2 .6! 2 .3! 2 .5! 2 .7!
cy c x x x L c x x x x L
a general solution.
When the series are written in summation notation,
2 2 1
1 0 2 1
0 0
1 1and 2
2 ! 2 2 1 ! 2
k kk k
k k
x xy x c y x c
k k
he ratio test can be applied to show that both series converges for allx.
Maclaurin series as 1 0 cos / 2 y x c x and 2 12 sin / 2 y x c x
uestion 2: Marks=10
Whether 3x are singular points of the equation
2 2 / / / 9) ( 3) 0x y x y y
Solution:
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2 2 / / / ( 9) ( 3) 0 x y x y y
Dividing the equation by 2 2( 9)x = 2 2( 3) ( 3)x x
/ / /
2 2 2 2
/ / /
2 2 2
2 2 2
( 3) 10
( 3) ( 3) ( 3) ( 3)
1 10
( 3)( 3) ( 3) ( 3)
1 1and q x
( 3)( 3) ( 3) ( 3)
x y y y
x x x x
y y y x x x x
Where
p x x x x x
X=3 is regular singular points because power of x-3 in P (x) is 1 and in Q (x) is 2
X= -3 is an irregular singular point because power of x+3 in P (x) is 2.
Question 3(a) : Marks=5
ind the general solution of the equation
2 / / / 2 1( ) 016
x y x y x y
Solution:
The Bessel differential equation is
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2 / / / 2 2
2 / / / 2
( ) 0 1
1( ) 0 2
16
x y x y x v y
x y x y x y
Comparing 1 and 2 we get
2 1 1, therefore v=16 4
v
So, the general solution of 1 is
1 1/ 4 2 1/ 4y C J x C J x
Question 3(b) : Marks=10
erive the expression of3
( )2
n J x for n
Solution:
Consider
1 12
v v vv
J x J x J xx
For1
2v
1 1 11 12 2 2
122
J x J x J xx
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3 1 12 22
3 1 12 22
1
1
J x J x J xx
J x J x J xx
we know that
1/ 2 1/ 22 2
( ) sin , ( ) cos J x x J x xx x
3
2
32
1 2 2sin cos
2 sincos
J x x x x x x
x J x x
x x
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Assignment 7
Q No.1Solve the system of equation by systematic elimination
1 1 2
3 2 1
D x D y
x D y
Solution:
1 1 2 (1)
3 2 1 (2)
D x D y
x D y
From equation (2)
1 2
3
D yx
Putting in equation (1), then we have
1 2
1 1 23
D y D D y
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2 5 7D y
The associated homogenous equation is
2 5 0D y
The auxiliary equation is
2 5 0m
Roots are 5m i
1 2cos 5 sin 5
c y c t c t
Now to determine the particular solutionp
y
'
''
0
0
p
p
p
y A
y
y
Thus substituting in the given D.E, we have
'' 5 7p py y
Equating the coefficients and constants terms gives.
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5 7
7
5
A
A
Hence the solution of variable y is given by
c p y y y
1 2
7cos 5 sin 5
5 y c t c t
From equation (1) we have
1 3
2
xy
D
Putting in equation (2)
2 1 1 1 3 2 2 D D x D x D
2 1 1 3 1 2 2 D D x D x D D
2 5 3D x
The auxiliary equation of the differential equation for x is
2 5 0m
Roots are 5m i
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The roots of the auxiliary equation are complex. Therefore
3 4cos 5 sin 5
c x c t c t
We assume that
px A
'
''
0
0
p
p
x
x
Substituting in the differential equation we have.
'' 5 3p p
x x
Equating the coefficients we have
5 3A 3
5A
so that
c p x x x
3 4
3
cos 5 sin 5 5 x c t c t
Thus
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3 43
( ) cos 5 sin 55
x t c t c t
1 2
7( ) cos 5 sin 5
5 y t c t c t
Q No. 2: If possible rewrite the following equations in the canonical form
'' '
'' '
2
3
x y
x y
SOLUTION:
Writing the system in the operator form
2
2
2
3
D x Dy
D x Dy
2
2
2
3
0 1
D x Dy
D x Dy
This is absurd. Thus the given system cannot be reduced to a canonical form.Hence the system is a degenerate system.
Q No. 3: Solve the following system of linear equations by Gaussian elimination method
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1 3
1 2 3
1 2 3
2 1
2 4 2
8 3 2
x x
x x x
x x x
Solution
2 0 1 1
2 4 1 2
1 8 3 2
Theaugmented matrixof the systemis
By interchanging 13R
1 8 3 2
2 4 1 2
2 0 1 1
1 2
1 8 3 2
0 20 5 6 2
2 0 1 1
R R
1 3
1 8 3 2
0 20 5 6 2
0 16 7 3
R R
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2
1 8 3 2
5 60 1
20 20 20
0 16 7 3
R
2 3
1 8 3 2
5 60 1 16
20 20
90 0 3
5
R R
3
1 8 3 2
5 60 1
20 20 3
30 0 1
5
R
1 8 3 2
5 60 1
20 20
30 0 1
5
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1 2 3
2 3
3
8 3 2
5 6
20 203
5
Thelast matrix is in row echelon form and representsthe system
x x x
x x
x
2
2
1
1
1 2 3
5 3 6
20 5 20
9
20
9 38 3 2
20 5
1
5
lg
1 9 3, ,
5 20 5
x
x
x
x
Solution set of given system of linear a ebraic equation
x x x
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Assignment 8
Question # 1:- Find the Eigen values and Eigen vectors of the following matrix
7 2 1
2 15 2
1 2 7
A
Solution:
For Eigenvalues
det 0
7 2 1 1 0 0
2 15 2 0 1 0 0
1 2 7 0 0 1
A I
7 2 1
det 2 15 2 0
1 2 7
A I
Expanding the determinant
det 1( 2( 2) 1 15 ) 2 2 7 2
7 15 7 2 2
A I
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We have
27 ( 22 96) 0
7 16 6 0
Thus the Eigen values of the matrix are
1 2 37, 16, 6
For Eigenvectors are:
For 1 7 we have
=>
0
0
0
=>By row reducing the augmented matrix
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1 3
2 1
2
2 3 1 2
2
&
0
0
0
2
0
0
0
2
0
0
0
&
0
0
0
2
0
0
0
ExchangingR R
R R
R
R R R R
R
Thus we have the following equations in k1, k2, and k3. The number k3 can be chosen
arbitrarily.
1 3, 2 3
1
2k k k k
Choosing 3K = 1, we get k1=1 and k2 = , hence the eigenvector corresponding 1 7 is
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1
1
4
1
K
1
1
1
2
1
K
For2
16 we have
0
16 0 0
0
A
]
By the row reducing the augmented matrix
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1 3
2 1 3 1
2
3 2
0
0
0
&
0
0
0
2 & 9
0
0
0
50
0
0
20
0
0
0
Exchanging R R
R R R R
R
R R
1 22
0
0
0
R R
Thus we have the following equations in k1, k2, and k3. The number k3 can be
chosen arbitrarily.
1 3, 2 34k k k k
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Choosing k3 = 1, we have 1K =1 and k2 = -4, hence the eigenvector corresponding
216 is
2
1
41
K
For2
6 we have
0
6 0 0
0
A
2 1 3 1
2
1 2
0
0
0
2 &
0
0
0
5
0
0
0
2
0
0
0
R R R R
R
R R
Now, we have the following equations in k1, k2, and k3. The number k3 can be chosen
arbitrarily.
=> 1 3, 2 0k k k
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26
Choosing k3 = 1, we get k1= -1 and k2 = 0,
Hence the eigenvector corresponding
2 6
3
1
0
1
K
Question # 2:
Solve the homogeneous system of differential equations
2 3
2
dxx y
dt
dyx y
dt
Solution:
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The given system can be written in the matrix form as
2 32 1
dx
xdtdy y
dt
Therefore, the coefficient matrix
2 3
2 1A
Now se find the eigenvalues and eigenvectors of the coefficient A. The characteristics equation
is
2
det 0
2 30
2 1
3 4 0
A I
Therefore, the characteristic equations
1 4 0 1,4
Therefore roots of the characteristic equation are real and distinct and so ate the eigenvalues,
For 1 , we have
1
2
1 2
1 2
1 2
1 2
2 1 3
2 1 1
3 3
2
0
3 3 0
2 0
k A I K
k
k k A I K
k k
A I K
k k
k k
These two equations are nod different and represent the equation.
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Thus we can choose value of the constant k2 arbitrarily, If we choose2
K = -1 then
1K = 1.
Thus the corresponding eigenvector is
1
1
1K
For =4 we have
1
2
1 2
1 2
1 2
1 2
2 4 3
2 1 4
2 3
2 3
0
2 3 0
2 3 0
k A I K
k
k k A I K
k k
A I K
k k
k k
Again the above two equations are not different and represent the equation
21 2
32 3 0
2
kk k
Again m the constant k2 can be chosen arbitrarily. Let us choose k2 = 2 then k1 = 3.
Thus the corresponding eigenvector is
4
1 2
1 3,
1 2
t t X e X e
Hence the general solution of the system is the following
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1 1 2 2
4
1 2
4
1 2
41 2
1 3,
1 2
( ) 3
( ) 2
t t
t t
t t
X c X c X
X c e c e
x t c e c e
y t c e c e
4
1 2
4
1 2
( ) 3
( ) 2
t t
t t
x t c e c e
y t c e c e
Question # 3: Write the given system in matrix form
2
2
1
2 3
2
dxx y z t
dt
dy x y z t
dt
dzx y z t t
dt
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Solution:-
In the matrix notation
' 2
1 1 1 0 1 1
2 1 1 3 0 0
1 1 1 1 1 2
X X t t
'
2
1 1 1
2 1 1
1 1 1
0 1 1
3 0 0
1 1 2
X X g t
Where g t
t t
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