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Differential Equations - Solved Assignments - Semester Spring 2005
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Assignment 1 (Spring 2005)
Maximum Marks 60
Due Date 07, April 2005
Assignment Weight age 2%
Question 1
(a) Define initial value and boundary value problem elaborate with the help of examples (at least one).
05
Solution
INITIAL VALUE PROBLEM
Differential equation of first order or greater in which the dependent variable y or its derivative are
specified at one points such as
0 0
,dy
f x ydx
y x y
If equation of the second order
2
2 1 02
'
0 0 0 1, '
d y dya x a x a x y g x
dx dx
y x y y x y
Where 0y and 1'y are arbitrary constants
Is called the initial value problem and the 0 0y x y '
0 1, 'y x y is called the initial conditions
BOUNDARY VALUE PROBLEM
Differential equation of order two or greater in which the dependent variable y or its derivative are
specified at different points such as
2
2 1 02
0 0 1 1,
d y dya x a x a x y g x
dx dx
y x y y x y
Where 0y and 1'y are arbitrary constants
Is called the boundary value problem and the 0 0 1 1,y x y y x y is called the boundary conditions
NOT THE PART OF THE SOLUTION (MOTIVATION OF THE CONDITIONS)
Basically when we solve the differential equations then we find some constant in the solution so we
need to elaborate these constants to solve this problem we formulate two kinds of conditions(given
above) which basically helps us to elaborate the constants finding in the solution of the differential
equation.
Still the question is how many constants are present in the differential equation. Is there any hard and
fast rule? yes! It depends that what is the order of the differential equation. If we have the first order
DE then its solution contains must one constant similarly if second order DE then its solution contains
must two constants……and so on and definitely to elaborate the constant we always need same
number of condition.
Question 1
(b) Solve the following differential equation.
2 20; (0) 1
xdy ydxxdx ydy y
x y
10
Solution
2 2
2 2 2 2
2 2
2 2
2 2 2 2
3 2 2 3
2 2 2 2
2 2 3
( ) 0 , 0 1
0
( , )
( , )=
,
2 1 2
xdy ydxxdx ydy y
x y
y xx dx y dy
x y x y
f yM x y x
x x y
f xN x y y
y x y
y xM x N y
x y x y
x xy y x y y xM N
x y x y
xy x y y xM
y
2
22 2
3 3 2 2 3 3 2
22 2
3 3 2 2 3 3 2
22 2
2 2
22 2
2 2 2 3
22 2
3 3 2 2 3 3 2
22 2
3 3 2
2 2 2 2 2
2 2 2 2 2
2 1 2
2 2 2 2 2
2 2
xy y
x y
x y xy x y x y xy yM
y x y
M x y xy x y x y xy y
y x y
M y xnow
y x y
xy x y x x y y xN
x x y
x y xy x y x y xy xN
x x y
N x y xy x
x
2 3 3 2
22 2
2 2
22 2
2 2 2y x y xy x
x y
N y x M
x yx y
2 2 ( , )
soour givenequationis exact
nowas we know
f yM x y x
x x y
Integrating both sides with respect to “x”
2 2 2 2
21
1
2 2
21
'
2 2
2
( , )
1( , ) ( )
1( , ) tan ( )
2
1sin tan
( , ) tan ( ) (1)2
1( )
1
f x y Mdx
yf x y x dx xdx y dx y
x y x y
x xf x y y y
y y
dx xce
x a a a
x xf x y y
y
f xy
xy y
y
f
2'
2 2 2
'
2 2
2 2
'
2 2 2 2
2
21
21
( )
( )
( , )
( )
'( )
( )2
1
2( , ) tan
2 2
tan2
y xy
y y x y
f xy
y y x
as we knowthat
f xN x y y
y x y
we will get
x xy y
x y y x
y y
yy
now putting this valuein equation no
so
x y yf x y
x
x yc
x
21
2
2
1
2 2
1 2tan
2 2 2 2
y
applying initial condition we get
c
x y y
x
Question 2
(a) Define the order and degree of DE also write down the order and degree of following DE.
2 " ' 2
22 "' ' 2
1 ( 1) 0
2 19 4 (4 25) 0
x y xy x y
x y xy x
04
Solution
Order of Differential Equation:
Order of the differential equation is the highest order derivative in a differential equation e.g.3y''+(y') =0 has order two.
Degree of the Differential Equation:
Degree of the differential equation is the power of the highest order derivative in the differential e.g.
y''+y=2 has degree one. Although, (y'') 3+y'+y=0 has degree three.
Order
(1) 2
(2) 3
Degree
(1) 1
(2) 2
Question 2
(b) Find I.F from 3rd
and 4th
method of exactness i.e. (xM+yN) & (xM-yN) respectively if possible. If
not then explain the reason.
3 3 5 44 6 0x y xy dx x x y dy 05
Solution
Integrating factor for 3rd
case can not be found due to that 0xM yN is not homogenous
So first case is not applicable as M= 3 34x y xy , N= 5 46x x y
3 3 5 4
4 2 3 5 4 2
4 +y 6
4 + 6
xM yN x x y xy x x y
xM yN x y x y x y x y
To check it is homogeneous
Say
4 2 3 5 4 2
5 4 5 2 3 6 5 6 4 2
5 4 2 3 5 4 2
5
4 + 6 ,
, , ,
, 4 + 6
, 4 +t 6
, ,
n
xM yN x y x y x y x y f x y
f tx ty t f x y for some real number n
but
f tx ty t x y t x y t x y t x y
f tx ty t x y x y x y tx y
f tx ty t f x y
For 4th case it is not possible because condition ( ) ( ) 0yf xy dx xg xy dy is not being fulfilled here.
As
3 3 5 4
3 2 4 3
4 6 0
4 6 0
x y xy dx x x y dy
y x xy dx x x x y dy
But 3 24x xy and 4 36x x y are not the function of xy
NOTE (RMEMBER)
f xy Means each term should have the product of xy e.g.
2 2f xy xy x y And 2 2 4 4f xy xy x y x y
Both are functions of xy
Question 2
(c) Define ordinary and partial differential equation as well as check whether the following are
ordinary differential equation or partial differential equation.
2
21 4 2
d x dy dz
du du du Where x, y and z are dependent variable and u is independent variable
2
22 3 4
d x dx
du du Where x is dependent variable and u is independent variable
2 2
2 23 3
x x
u v
Where x is dependent variable, u and v are independent variable
2 2
2 24 4
u v
x y
Where u, v are dependent variable, x and y are independent variable 06
Solution
ORDINARY DIFFERENTIAL EQUATION (ODE)
If an equation contains only ordinary derivatives of one or more dependent variables depending on
only one independent variable is known as ODE.
PARTIAL DIFFERENTIAL EQUATION (PDE)
If an equation contains partial derivatives of one or more dependent variables depending on two or
more independent variables is known as PDE
(1) ODE
(2) ODE
(3) PDE
(4) PDE
Question 3
(a) Define explicit and implicit solution. Check whether 1 1 1y x c i.e. solution of the DE
01 1
dx dy
x y
can be defined explicitly or not? If it can be define then write down the explicit form?
05
Solution
Explicit solution:
A solution of the differential equation of the form y= f (x) is called the explicit solution of the
differential equation or roughly speaking if we can separate the dependent an independent variable in
the solution then we say solution is explicit. For example dy / dt = y2-1/x has the solution y = 3+x
2/3-
x2.
It is explicit solution because dependent and independent variable are separate here of it is form of
y = f(x).
Implicit solution:
A solution of the differential equation of the form G (x ,y)=0 is called the implicit solution of the
differential equation or roughly speaking if you cannot separate the dependent an independent variable
then that solution is said to be implicit.
For example, dy/dt = 1+1/y2
has the solution y – tan-1
(y) = t+c. it is implicit solution because you
cannot separate the dependent and independent variable or it is of the form of G (x, y) =0.
Solution can be defined explicitly as follows.
1 ( 1)( 1)
11
( 1)
11
( 1)
1 1
1
y x c
yx c
yx c
x cy
x c
Question 3
(b) Whether following differential equation is separable or not if it is separable then solve it?
2 2
2 2 3
1 ( 5 5) 20 20 0
2 y
r r dr r r d
dyt t e
dt
10
Solution
2 2
2 2
2
2
2
1 ( 5 5) 20 20 0
( 5 5) 20 20
5 1 5 20 1 20
1 5 1 20
1 20
51
5 5 20
1 5
5 5 20
1 5 5
25
1 5
r r dr r r d
r r dr r r d
r dr r d
r dr r d
rdr d
r
drd
r
drd d
r
drd d
r
Integrate both sides
25
1 5
ln 1 25ln 5
drd d
r
r c
3
3
3
2 2 3
2 2 3
2 3
2
3
32
3 3
32
3
32
3
3 3
3 3
3 3
3 3
3
1 1
2
0
1
1
1
3
3 1
1 3
3 1
ln(1 )
3 3
ln(1 ) 3
(1 )
(1 ) .
(1 ) ,
y
y
y
y
y
y y
y
y
y
y
y
y
y t c
y t c
y t
dyt t e
dt
dy t t e dt
dy t e dt
dyt dt
e
e dyt dt
e e
e dyt dt
e
edy t dt
e
e tc
e t c
e e
e e e
e c e say c e
3c
Question 4
(a) Define a homogenous function. Check whether 2
3 3,
x yf x y
xy x y
is homogeneous or not explain.
Also check whether the DE 2
3 3( )
dy x y
dx xy x y
is solvable by method of homogenous equation or not if
not then give reason? 05
Solution
Homogeneous Function:
A function ,f x y is said to be homogeneous if , , ,nf tx ty t f x y for somereal number n
e.g. 2
2
2x, =
3xy+yf x y is homogeneous function
2
3 3
3 2
4 3 3
2
3 3
1
,
,
,
, ,
but
x yf x y
xy x y
t x yf tx ty
t xy x y
x yf tx ty
t xy x y
f tx ty t f x y
This is homogeneous function
Now to solve 2
3 3( )
dy x y
dx xy x y
we can’t apply here the homogeneous method
To solve the homogeneous differential equation
),( yxfdx
dy
We use the substitution
x
yv
If ),( yxf is homogeneous of degree zero, which we don’t have there.
Question 4
(b) Solve the following differential equation.
2 10
2 1
dy x y
dx y x
10
Solution
2 10
2 1
dy x y
dx y x
,
the given equation reduces to
2X Y+2h k+10
2Y X+2k h 1
We choose h and k such that
2h k+1=0;2k h 1=0
h+2k 1=0 multiply this with 2 we get
2h+4k-2=0
then adding both
2h k + 1=
now put thevalues
x X h y Y k
dy
dx
0
2h+4k 2=0
3 1 0
k=1 3 2h k+1=0
2h 1 3 1 0
2h+ 2 3 0
2h= 2 3
h= 1 3
h= 1 3,k=1 3
2X Y=
X 2Y
This is a homogenous equation. We substitute Y=VX to obtain
s in
k
soultimately we get
dy
dx
dY dVX V
dX dX
by replacing all value the g
2X
X 2VX
2
1 2V
2 1 2V
1 2V
2 1 2V
1 2V
iven DE
dV VXX V
dX
dV VX V
dX
V VdVX
dX
V VdVX
dX
2
2
2
2
2 2
1 2V
2 2 2
1 2V
1 2V 2
1
2V 1 20
1
dV V V VX
dX
dV V VX
dX
dV dXV V X
dV dXV V X
Integrating both sides
2
2
2 2
22
2
2 2
2
2V 1 20
1
ln | 1| 2 ln | | ln | |
( 1)
now replacing V=Y/X we will get the
X 1
( )
replace the values of X,Y
x=X 1 3, 1 3
X=x+1/3,Y=y-1/3
y 1/3 y 1/3 x+1/3 x+
dV dXV V X
V V X c
X V V c
Y Yc
X X
Y YX X c
now
y Y
2
2 2
2 2
2 2
2 2
2 2
1/3
3 1 3 1 3 1 3 1 9
3 6 1 3 3 3 1 3x 6 1 9
3 3x 2 3 3 6 3 6 1 9
3 3x 3 9 9 3 9
x 3 3 1 3
c
y x y x c
y y xy y x x c
y xy y y x x c
y xy y x c
y xy y x c
Assignment 2 (Spring 2005)
Maximum Marks 60
Due Date 20, April 2005
Assignment Weight age 2%
Question 1
(a) What is the difference between a linear equation (first order) and Bernoulli equation gives at
least two examples of each. 05
Solution
LINEAR DIFFERENTIAL EQUATION (FIRST ORDER)
A linear differential Equation is an Equation of the form ( )dy
a x b x y c xdx
it’s more frequently
used form ( ) ( )dy
f x y g xdx
where f(x) and g(x) are both continuous functions of x. e.g.
3 2dyx y x
dx
dyxy x
dx
BERNOULLI DIFFEREMNTIAL EQUATION
Now the differential equation of the form ndyf x y g x y
dx where n is any real no is called
Bernoulli equation. e.g.
2
2
dy 1
dx
1
y xyx
dyx y
dx y
DIFFERENCE The difference is that linear equation is a special type of Bernoulli’s equation when n is equal to 0 and
1 simply for n=0, 1 the Bernoulli’s equation becomes a linear equation.
Question 1
(b) Solve the following linear differential equation 10
32dy
x y ydx
Solution
3
3
2
2
dyx y y
dx
dy y
dx x y
It is clearly not a linear equation it can be changed to a linear as in first order linear equation
independent and dependent variables can be changed so writing equation as.
3
2
1
1ln
2
12
" "
1 , 2
1
dyy
y
dx x y
dy y
dxx y
dy y
nowit is linear in x
p y y Q y
IF e
IF ey
Multiplying by 1/y on both sides we get
2
2 2
1 12
2
( )
dxx y
y dy y
d xnowit is a differential of
dx y
d xy
dx y
xy c x y y c
y
Question 2
(a) Solve the following initial value problem and mention the type of the equation
3
, (1) 22
dy y xy
dx x y 10
Solution
3
43
4
3 3
2
" " 3
1 2 ,
2
14 4
4
dy y x
dx x y
It is abernaoulli equtionin y where n
p x x Q x x
dy yy x
dx x
put t y
dt dy dt dyy y
dx dx dx dx
Now the equation becomes
22
2ln ln 2
2
2 3
1 1
4 2
24
2 4
dxx xx
dtt x
dx x
dtt x
dx x
now IF will be
IF e e e x
Now
multiplying both sides x
dtx xt x
dx
now
It is the differential of
2dtx
dt
2 3
2 3
2 4
( ) 4
4
dtx x
dx
tx x dx
tx x c
4 2 4y x x c
By applying initial conditions
Now put y=2, x=1
4 2 4
2 1 1
15
c
c
So we get 4 2 4 15y x x
Question 2
(b)Show that 2 24 4y cx c is self orthogonal
Note: A family of curves is said to be self orthogonal if the orthogonal trajectories of the family is
same as the DE of that family. 10
Solution
2 2
2
2
2
2
4 4 (1)
2 42
(1)
4 42 2
2
y cx c
now
dy y dyy c c
dx dx
substituting this valueinequation
we get
y dy y dyy x
dx dx
dy dyy x y y
dx dx
so DE of given faimly is
2
2 2 2dy dy
y xy ydx dx
To get the family of orthogonal trajectories
Replace 1dy dy
bydxdx
2
2
2
2
2
2 2
2
2 2
2
2
2
2 1
2 1
2
2 3
dy dyy x y y
dx dx
by replacing we get
dy dyy x y y
dx dx
dy dyy xy y
dx dx
dy dyy xy y
dx dx
dy dyy xy y
dx dx
If we compare both differential equations (2) and (3) then came to know that both are the same
So the given family of equation is self orthogonal
Question 3
(a)Determine either the following curves are orthogonal to each other or not? If not then find the
condition of orthognality.
2 2 31 , 3y x x y
Note:
Two functions are said to be orthogonal if product of the slopes is -1. 07
Solution
TO check both curves orthogonal or not,
First of all differentiating both the equations w.r.t x 2 2
1 1
1
2 2 0
2
2
( , )
y x
differentiatin wrt x
dyy x
dx
dy x
dx y
Now wecheck at x y
1 1
11
|( , ) 1x y
xdym
dx y
Now considering the second equation
1 1
3
2
1 1
2
2 1|( , )
3
3
( , )
3x y
y x
dyx
dx
now at x y
dym x
dx
Now by applying condition of orthogonal
1 2 1m m
It is not satisfied
Now both the curves are not orthogonal
To find the condition we may proceed as follow and condition for orthognality is
1 2
211
1
3
1
1
3
1 1
3
1 1
1 0
( 3 ) 1 0
31 0
3 0
3 0
m m
putting thevalues
xx
y
x
y
x y
y x
This is the required condition.
Question 3
(b) Solve the following DE by the proper substitution
2 2y ydy sinx
xye +e =dx x
08
Solution
2 2
2
2
2
2
y y
y
y
y
2
2
2
2
2
2
2 y
dy sinxxye +e =
dx x
put e =u
dy du2ye =
dx dx
dy 1 duye =
dx 2 dx
x du sinx+u=
2 dx x
du 2 sinx+ u=2
dx x x
sinxHere p(x)=2/x And Q(x)=
x
2I.F=exp( dx)=x
x
dux +2xu=2sinx
dx
d(x u)=2sinx
dx
Integrate
x u= 2cosx+c
x e = 2cosx+c
Question 4
(a) The population of a town grows at rate proportional to the population at any time. Its initial
population of 50 decreased by 9% in 15 years what will be the population in 25 years? (Just make
the model of the population dynamics as well as just describe the given conditions do not solve
further) 03
Solution
Suppose that P0 is the initial population of the town, as given P0 is 50 and P (t) the population at any
time t. Then population growth can be represented by
dPP
dt
dPkP
dt
dPkdt
P
Integrate both sides
0 0
ln
kt c
kt c
kt c
P kt c
P e
P e e
P P e say P e
0P is the initial population of the town
0P =50 = P (0)
P (15) =50- 9
50100
P (25) =?
Question 4
(b) If a small metal bar whose initial temperature 30C is dropped in to a container of boiling water,
how long will it take for the bar to reach 90C if it is known that its temperature increased 2C in a
second? (Just make the model as well as just describe the given conditions do not solve further)
04
Solution
Small metal bar has the initial temperature T (0) =30 0C=T when dropped into a container of boiling
water (temperature of surrounding matters a lot on the small metal bar) so suppose mT is the
temperature of the boiling water. As we know the boiling temperature of the water is 100c so
100mT
Change in the temperature can be represented as dT
dtand it is proportional to mT T i.e.
propotional to 100dT
Tdt
With conditions T (1) =32as 30+2and T (0) =30
We have too find the time when T=90
( 100)
100
dTk T
dt
dTkT k
dt
Question 4
(c)The initially there were 50 milligrams of the radioactive substance present after 5 hours the mass
increased by 4%. If the rate of decay is proportional to the amount of the substance present at any
time, determine the half- life of the radioactive substance. (Just make the model as well as just
describe the given conditions do not solve further) 03
Solution
Suppose, radioactive substances = R
Then its model would be
proportional to R dR
dt
=kR dR
dt
Where k is the constant of proportionality
With the initial conditions R (0) = 50, R (5) = 52
Since it is given after 5 hours radioactive substances
Increased by 4%, thus 4% of 50 is 2 milligram
That means 52, as we subtract 50+2=52
Then =k R
dRdt
After integration, we get
kt + c
kt c
0 0
ln = kt + c
=e
= e e
R
R
R R where R
Now we have to find half-life of radioactive
Substances, that means t is required if R = R/2 i.e. R=25
Assignment 3 (Spring 2005)
Maximum Marks 60
Due Date 02, May 2005
Assignment Weight age 2%
Question 1
(a) Define uniqueness of solution and condition of its existence? If 1 2, , , ny y y are n solutions, on
an interval I , of the homogeneous linear nth-order differential equation
0011
1
1
yxadx
dyxa
dx
ydxa
dx
ydxa
n
n
nn
n
n on interval I are linearly independent then
is it necessary 1 2, , , nw y y y must be non-zero? 05
Solution
Let )(),(),...,(),( 011 xaxaxaxa nn and )(xg be continuous on an interval I and let Ixxan ,0)( .
If Ixx 0 , then a solution )(xy of the initial-value problem exist on I and is unique
NOT PART OF THE SOLUTION:
Always remember that solution of the differential equation differ by a constant as I told in the
first assignment’s solution we have initial or boundary conditions to determine these unknown
constants in the differential equations
Secondly, If 1 2, , , ny y y are n solutions, on an interval I , of the homogeneous linear nth-order
differential equation 0011
1
1
yxadx
dyxa
dx
ydxa
dx
ydxa
n
n
nn
n
n on interval I are linearly
independent then it is necessary that W ( 1 2, , , ny y y ) must be non-zero since we know
Linear Independence of Solutions:
The solutions
nyyy ,,, 21
are linearly dependent if and only if
IxyyyW n ,0,,2,1
Question 1
(b) Verify that the given two-parameter family of functions is the complimentary solution of the non-
homogeneous differential equation
2 3
1 2 3"' 5 " 6 ' 2sin 8; x xy y y x y c c e c e 05
Solution
Consider the associated homogenous differential equation
"' 5 " 6 ' 0y y y ----------------------- (1)
Given function is
2 3
1 2 3
x xy c c e c e
Taking derivatives of the function 2 3
2 3
2 3
2 3
2 3
2 3
' 2 3
" 4 9
"' 8 27
x x
x x
x x
y c e c e
y c e c e
y c e c e
Putting all these in the given associated homogenous differential equation
2 3 2 3 2 3
2 3 2 3 2 3
2 3 2 3 2 3
2 3 2 3 2 3
"' 5 " 6 '
8 27 5(4 9 ) 6(2 3 )
8 27 20 45 12 18
0
x x x x x x
x x x x x x
LHS y y y
c e c e c e c e c e c e
c e c e c e c e c e c e
Hence "' 5 " 6 ' 0y y y is the complementary solution of the given non-homogeneous
differential equation.
Question 2
(a) Define linearly independence and linearly dependence of the functions?
Comment either the functions given below are linearly independent or dependent
2 3
1 2 3( ) , ( ) , ( )x x xf x xe f x x e f x x e 10
Solution
Linear Dependence A set of functions
)(,),(),( 21 xfxfxf n
is said to be linearly dependent on an interval I if constants nccc ,,, 21 not all zero, such that
Ixxfcxfcxfc nn ,0)(.)()( 2211
Linear Independence
A set of functions
)(,),(),( 21 xfxfxf n
is said to be linearly independent on an interval I if
Ixxfcxfcxfc nn ,0)()()( 2211 ,
only when
.021 nccc
2 3
1 2 3( ) , ( ) , ( )x x xf x xe f x x e f x x e , First of all finding the derivatives of the function
1
1
1
2
2
2
2
2
2
3
3
2 3 2
3
2
3
( )
' ( ) 1
'' ( ) 1 2
( )
' ( ) 2 2
'' ( ) 2 2 4 2
( )
' ( ) 3 3
'' ( ) 3 2 3
x
x x x
x x x
x
x x x
x x x x
x
x x x
x x
f x xe
f x e xe x e
f x e x e x e
f x x e
f x xe x e x xe
f x xe x e x xe x x e
f x x e
f x x e x e x x e
f x x e x xe
2 2
2 3
2 3
2
2 2
2
2
2 2
2
3 2
2 2
3
6 6
, ,
1 2 3
( 2) 4 2 6 6
1
1 2 3
( 2) 4 2 6 6
1
1 2 3
( 2) 4 2 6 6
1
x x
x x x
x x x
x x x
x x x
x x x
x
x
x x e xe x x
Now
w xe x e x e
xe x e x e
x e x xe x x e
x e x x e xe x x
x x
xe e e x x x x x
x x x x x x
x x
xe x x x x x
x x x x x x
x
xe
2
2
2 2
2 2
2 2 2
3
2
2
3 2 2 2 2 2 2
3 3 2
1 2 3
( 2) 4 2 6 6
2 3 1 3
4 2 6 6 ( 2) 6 6
1 2
( 2) 4 2
4 6 2 6 6 9 10 2
8 8 2 0
x
x
x
x
x x x x x
x x x x x x
x x x x x x xx
x x x x x x x x xxe
x x xx
x x x
xe x x x x x x x x x
x e x x
As Wronskian is not equal to zero so all function are linearly independent
Question 2
(b) If functions )(,),( ),( 21 xfxfxf n possess at least n-1 derivatives on an interval I are linearly
independent then is it necessary functions having Wronskian non-zero? If not then give the
example (search a set of linearly independent functions such as their Wronskian is zero). 05
Solution
If functions )(,),( ),( 21 xfxfxf n possess at least n-1 derivatives on an interval I are linearly
independent then it is not necessary that functions having Wronskian non-zero for example
1 2 3( ) 0, ( ) , ( ) xf x f x x f x e at least posses 2 derivatives on interval I
1 2 3
1 2 3
1 2 3
( ) 0, ( ) , ( )
'' ( ) 0, ' ( ) 1, ' ( )
'' ( ) 0, '' ( ) 0, '' ( )
x
x
x
f x f x x f x e
f x f x f x e
f x f x f x e
Then it can easily notice that all above functions are linearly independent since
Firstly, no function is the constant multiple of the other function
Secondly, no one can be written linear combination of remaining two.
Thus they all are linearly independent.
But 0
0, , 0 1 0
0 0
x
x x
x
x e
w x e e
e
This assures us if Wronskian of function is nonzero then there is no doubt set of functions will always
be linearly independent but when Wronskian of functions is zero then set of function may be linearly
dependent or may be linear independent .
Question 2
(c) Find the 2nd
solution of each of Differential equations by reducing order or by using the formula.
2 4 3
1'' 6 ' 4 0 ;x y xy y y x x 05
Solution
For solving this we use the formula
2
( )
1 2
1 ( )
p x dx
ey x y x dx
y x
2
2
'
2
'' 6 ' 4 0
6 4'' 0
x y xy y
nowdividing by x on both sides
we get
y yy
x x
1
4 3
1
6
4 3
2 4 3 2
6ln| |4 3
4 3 2
sec
( )( )
( )( )
dxx
x
nowas y is given and it is given as
y x x
now for ond solution we proceed as follows
ey x x dx
x x
ex x dx
x x
64 3
6 2
4 3
2
( )( 1)
1( )
( 1)
xx x dx
x x
x x dxx
14 3
4 3
3
3
( 1)( )
1
( )
( 1)
( 1)
( 1)
xx x
x x
x
x x
x
x
So this is the required second solution. Question 3
(a) Define fundamental solution set of a differential equation and what is criterion for existence of
fundamental solution set of equation 05
Solution
Any set 1, 2, 3,....... ny y y y of n linearly independent solutions of the homogenous linear differential
equation on an interval are said to be a fundamental set of solutions on the interval.
Criteria for the existence of the fundamental solution set of for a linear nth order homogenous
differential equation is,
There always exists a fundamental set of solution for a linear nth order homogenous differential
equation.1
1 11( ) ( ) ... ( ) ( ) 0
n n
n n on n
d y d y dya x a x a x a x y
dx dx dx
on the interval I Question 3
(b) Check weather 2 2, , lnx x x x in the interval (0, ) form the fundamental set of solution the
differential equation 2"' 6 4 ' 4 0y x y xy y if it is fundamental solution what will be general
solution. 10
Solution
Now for fundamental solution set of 2"' 6 4 ' 4 0y x y xy y differential equation we have to
show that all the three functions are linearly independent.
1 2 3
' "
1 1 1
2 ' 3 " 4
2 2 2
2 ' 3 3 " 4 4 4
3 3 3
2 ' 3 " 4
3 3 3
2 2
2 2
3 3
4 4
, ,
1 0
2 6
ln 2 ln 6 ln 2 3
ln 1 2ln 6ln 5
, , ln
ln
1 2 1 2ln
0 6 6
nowlet y y y are thethree functions
y x y y
y x y x y x
y x x y x x x y x x x x
y x x y x x y x x
w x x x x
x x x x
x x x
x x
2 2 1 1 2 2
2 3
2 2
1 1 1 1
4
2 2 2 2
4 1 2 1
ln 5
1 ln
1 2 1 2ln ,
0 6 6ln 5
2 1 2ln 1 1 2ln 1 21 ln
6 6ln 5 0 6ln 5 0 6
2 1 2ln 1 1 2l
6 6ln 5
x
x x
x x x x x by taking common x and x fromc c respectively
x x x
x x x x x xx x
x x x x x x
xx x x x
x
1
1 1
4 3 2 2
4 3 2
6 1
n 1 12 ln
0 6ln 5 0 3
12ln 10 6 12ln 6ln 5 6 ln
4 5
4 5 0
xx x
x x x
x x x x x x x x
x x x
x x
As not equal to zero so these are independent and so forms a fundamental solution set.
So general solution may be written as
2 2
1 2 3 lnY c x c x c x x
Question 4
(a) Find the solution of the following DE.
''' 9 '' 26 ' 24 0; 0 1, ' 0 2, '' 0 3y y y y y y 10
Solution
To solve this equation
''' 9 '' 26 ' 24 0y y y
Suppose
2
3
'
''
'''
mx
mx
mx
mx
y e
y me
y m e
y m e
Substituting in equation 3 2
3 2
3 2
9 26 24 0
( 9 26 24) 0
0 ( 9 26 24) 0
2
mx mx mx
mx
mx
m e m e me
e m m m
e or m m m
m is the root of equationthenby useof synthatic division we get the roots as
2 1 9 26 24
2 14 24
1 7 12 0
2
2
2 3 4
1 2 3
1 2 3
2 3 4
1 2 3
2 7 12 0
2 3 4 12 0
2 3 4 3 0
2 3 4 0
0 1
1 1
' 2 3 4
x x x
x x x
m m m
m m m m
m m m m
m m m
general solution may be written as
y c e c e c e
applying y
c c c
y c e c e c e
1 2 3
2 3 4
1 2 3
1 2 3
1 2 3
1 2 3
2 3
' 0 2
2 2 3 4 2
'' 4 9 16
' 0 3
3 4 9 16 3
2 2 1
2 2 3 4
2 2 2 2
0 2 4
x x x
applying y
c c c
y c e c e c e
applying y
c c c
EQ EQ
c c c
c c c
c c
3 4 1EQ EQ
1 2 3
1 2 3
2 3
3 4 9 16
4 4 4 4
1 5 12
c c c
c c c
c c
2 3
2 3
3 3
2 2 2
1 2 3 1 1
2 3 4
5 5 4
1 5 12
0 5 10
1 2 1 2
4 0 2 2 2 2 1
1
1 1 1 1/ 2 1/ 2
1 1
2 2
x x x
EQ EQ
c c
c c
c c
then EQ implies c c c
then EQ implies
c c c c c
y e e e
Question 4
(b) The roots of an auxiliary equation are 1 2 3
1, 2 , 2
3m m i m i . What is the corresponding
differential equation? 05
Solution
Given that 1 2 3
12 2
3
10 ( 2) 0 ( 2) 0
3
m m i m i
m m i m i
The corresponding auxiliary equation will be,
2
2
2
23 2
3 2 2
3 2
1( ( 2) )( ( 2) ) 0
3
1( 2)( 2) ( 2) ( 2) 0
3
12 2 4 2 2 1 0
3
14 5 0
3
4 54 5 0
3 3 3
3 12 15 4 5 0
3 11 11 5 0
m m i m i
m m m i m i m i
m m m m im i im i
m m m
mm m m m
m m m m m
m m m
Finally the corresponding differential equation will be,
3 23 2
3 23 11 11 5 0 3 11 11 5 0
d y d y dyD D D
dx dx dx
Assignment 4 (Spring 2005)
Maximum Marks 40
Due Date 12, May 2005
Assignment Weight age 2%
Question 1
Solve the differential equation /// //6 1 cosr r .find complimentary cr and particular solution (
pr )
by the undetermined coefficient (superposition approach). 10
Solution /// //6 1 cosr r
Complementary function
To find cr , we solve the associated homogeneous differential equation
/// //6 0r r
Put nr e , 2 3' , r'' , '''n n nr ne n e r n e
Substitute in the given differential equation to obtain the auxiliary equation
/// //
3 2 2
6 0
6 0 6 0
0,0,6
r r
n n n n
n
Hence, the auxiliary equation has complex roots. Hence the complementary function is
1 2 3
6cr c c c e
Particular Integral Corresponding to cos( )g :
1cos sinpr A B
Corresponding to 1f : 2pr C
Therefore, the normal assumption for the particular solution is
1 2p p pr r r
1
cos sinpr A B +C
Clearly there is duplication of
(i) The constant function between cr and2pr .
To remove this duplication, we multiply 2pr with 2 . This duplication can’t be removed by multiplying
with . Hence, the correct assumption for the particular solution 2pr is
2cos sinpr A B C
Then ' sin cos 2pr A B C
'' cos sin 2
''' sin cos
p
p
r A B C
r A B
Therefore
1 1
1 1
''' 6 '' sin cos 6 cos sin 2
''' 6 '' sin 6 cos 6 12
p p
p p
r r A B A B C
r r A B A B C
Substituting the derivatives of pr in the given differential equation and grouping the like terms, we
have
sin 6 cos 6 12 1 cosA B A B C
6 0, 6 1, 12 1 21 1A B A B C C
Solving these equations, we obtain
6 6 6 1 37 1 1 37
6 37
BB
A
A B B B
A particular solution of the equation is
21 37 6cos sin 1 12pr
So
2
1 2 3
6 1 37 6cos sin 12
c pr r r
r c c c e
Question 2
Factor the given differential operator also find the function annihilated by this 4 64D D 10
Solution
4 3
34 3
4 2
2 1
64 64
64 4
64 4 4 16
Since 0
Annihilates each of the functions
1, , , ,
D Annihilates 1
n
n
D D D D
D D D D
D D D D D D
D y
x x x
Since nD )(
Annihilates each of the functions
xnxxx exexxee 1 2 , , , ,
( 4)D Annihilates -4 xe
Since differential operator
nDD 2 222
is the annihilator operator of the functions
2 1cos , cos , cos , , cosx x x n xe x xe x x e x x e x
2 1sin , sin , sin , , sinx x x n xe x xe x x e x x e x
22 2 2 2 2
2 4 2
16 2 16 16 4 12 2 3
2, 2 3
2 4 16D D Annihilates 2 2cos2 3, sin 2 3x xe e
Therefore, the operator 4 264 3 4 16D D D D D D annihilates linear combination 1,-4 xe ,
2 2cos2 3, sin 2 3x xe e
Question 3
Find the annihilator operator of the function 22e e e 10
Solution
Suppose that
2
21 2 3
2
,( ) , y ( ) 2 y ( )
e e e
y e e e
Then
1
2 2
2
3 3 2
2
1 1 0
1 1 2 0
1 1 0
D y D e
D y D e
D y D e
Therefore, the product of two operators
3
1D
Annihilates the given function 2( ) 2f e e e
Question 4
Solve the differential equation // 2 5xy y x e .find complimentary cy and particular solution ( py ) by
the undetermined coefficient (annihilator approach). 10
Solution
Step 1 The given differential equation can be written as
2 2( 1) 5xD y x e
Step 2 The associated homogeneous differential equation is
2( 1) 0
1 1 0
D y
D D
Roots of the auxiliary equation are complex
1, 1m
Therefore, the complementary function is
1 2
x x
cy c e c e
Step 3 Since 3 21 0, 5 0xD x e D
Therefore the operators 3
1D and D annihilate the functions 2 xx e and 5 . We apply 3
1D D to
the non-homogeneous differential equation
3 21 ( 1) 0D D D y .
This is a homogeneous differential equation of order 6.
Step 4 The auxiliary equation of this differential equation is
3 21 ( 1) 0 0
0, 1,1,1,1, 1
m m m y
m
Therefore, the general solution of this equation must be
2 31 2 53 4 6
x x x x xc c e xe e x ey c c x c c e
Step 5 Since the following terms are already present in cy
2 6
x xc e c e
Thus we remove these terms. The remaining ones are
2 31 53 4
x x xc xe e x ec c x c
Step 6 The basic form of the particular solution of the equation is
2 31 53 4
x x xp c xe e x ey c c x c
The constants 1c , 43 ,cc and 5c have been replaced with BA , and C .
Step 7 Since
2 3
2 3
x x x
p
x
p
y A Bxe C e Dx e
y A Bx C Dx e
x
x
Then taking derivative
2 3 2
2 3 2
2 3
'
'
' 2 3
2 3
2 3
x x
p
x
p
x
p
y Bx C Dx e B C Dx e
y Bx C Dx B C Dx e
y B C x C D Dx B e
x x
x x
x
Again taking derivative
2 2 3
2 2 3
2 3
'' 3 2 3
'' 3 2 3
'' 2 2 6 2 3
2 2 3
2 2 3
2 3
x x
p
x
p
x
p
y B C C D e B C x C D Dx B e
y B C C D B C x C D Dx B e
y B C C D B C C D Dx e
x Dx x
x Dx x
x D x
Therefore
2 3 2 3
2 3 2 3
2 3 3
2
2 2 6 2 3
2 2 6 2 3
2 2 6 2 3
2 4 6 6
2 3
2 3
2 3
2
x x
p p
x
p p
x
p p
p p
y y B C C D B C C D Dx e A Bx C Dx e
y y B C C D B C C D Dx Bx C Dx e A
y y B C C D B C B C D C Dx Dx e A
y y B C C D D
x D x x
x D x x
x D x
x x
xe A
Substituting in the given differential equation, we have
2
2 2
5
2 4 6 6 52
x
p p
x x
y y x e
B C C D D e A x ex x
Equating coefficients of 2,x xe x e and the constant terms, we have
2 0, 4 6 0, 6 1, 5
1, 2 3 , , 5
6
1 1 1, , , 5
4 4 6
2B C C D D A
B C C D D A
B C D A
Thus
2 3
2 3
1 1 15
4 4 6
5 3 3 212
x
p
x
p
y x x e
ey x x
x
x
Step 8 Hence, the general solution of the given differential equation is
pc yyy
Or 2 31 2 5 3 3 2
12
xx x e
c e c e x xy x
Assignment 5 (Spring 2005)
Maximum Marks 40
Due Date 21, June 2005
Assignment Weight age 2%
Question 1
State in words a possible physical interpretation of the given initial-value problems
(a)
2
2
15 0
32
(0) 2, '(0) 1
d xx
dt
x x
(b)
2
2
12.5 0
2
(0) 0.5, '(0) 0.25
d xx
dt
x x
(c)
2
2
13.5 0
8
(0) 1.5, '(0) 0
d xx
dt
x x
15
Solution
(a)Comparing 2
2
15 0
32
d xx
dt with
2
20
d xm kx
dt , we got
1m= , k=5
32
W=mg
W=32/32=1
Weight of 1 lb attached to a spring, is released from a point 2 units below the equilibrium position
with initial downward velocity and the spring constant is 5lb/ft.
(b)Comparing 2
2
12.5 0
2
d xx
dt with
2
20
d xm kx
dt , we got
1m= , k=2.5
2
W=mg
W=32/2=16
Weight of 16lb attached to a spring, is released from a point 0.5 units above the equilibrium position
with initial upward velocity and the spring constant is 2.5lb/ft.
(c)Comparing 2
2
13.5 0
8
d xx
dt with
2
20
d xm kx
dt , we got
1m= , k=3.5
8
W=mg
W=32/8=4
Weight of 4lb attached to a spring, body starts from rest from point 1.5 units below the equilibrium,
and spring constant is 3.5 lb/ft.
Question 2
A 24lbweight, attached to a spring, stretches it 4 in. find the equation of motion if the weight is
released from the rest from a point 3 in above the equilibrium point also find the time period(period of
oscillation) and frequency as well as amplitude. 10
Solution
As we know general equation of simple harmonic motion (un-damped free motion) is
kxdt
xdm
2
2
----- (1)
Or 2
2
20
d xx
dt
So basically we needed m, k thus we can find
For consistency of units with the engineering system, we make the following conversions
12 inches 1 foot
11 inches foot
12
4 1
4 inches foot= foot12 3
.
So, 4 1
4 12 3
Stretch s inc ft ft
Further weight of the body is given to be
W 24 lb
But mgW
Therefore W 24 3
32 4m
g
or 3
slugs.4
m
Since 4 1
4 12 3
Stretch s inc ft ft
Therefore by Hook’s Law, we can write
1
243
k
72 lbs/ftk
Equation 1 is becomes
2
2
2
2
372
4
472
3
d xx
dt
d xx
dt
or 2
296 0
d xx
dt .
Since the initial displacement is 3 1
3 f12 4
inches ft t
1
0 , 0 04
x x
The negative sign indicates that body starts from rest above the equilibrium point at the distance of 3
inch from equilibrium point or ¼ foot from equilibrium point
2
296 0
d xx
dt
Subject to 1
0 , 0 04
x x
Putting mtmt emdt
xdex 2
2
2
,
We obtain the auxiliary equation
2 96 0m
or 4 6m i
The general solution of the equation is
1 2cos4 6 sin 4 6x t c t c t
Now, we apply the initial conditions.
1
04
x 1 2
1.1 .0
4c c
Thus 1
1
4c
So that 2
1cos4 6 sin 4 6
4x t t c t
Since
1 26 sin 4 6 4 6 cos4 6x t c t c t .
Therefore
0 0x 20 4 6 .1 0c
Thus
2 0c .
Hence, solution of the initial value problem is
1
cos 4 64
x t t
Time period
2T
Comparing 2
2
20
d xx
dt with
2
296 0
d xx
dt we have
2 96 4 6
It implies
2
4 6 2 6T
Frequency
1 2 6
2f
T
Amplitude
22 2 2
1 2 1 4 0 1 4A c c Question 3
Physically interpret the following differential equations.
(a)
2
2
12 0
16
0 0, ' 0 1.5
d x dxx
dt dt
x x
(b)
2
24 4 0
0 1, ' 0 2.01
d x dxx
dt dt
x x
(c)
2
216 0
0 1, ' 0 0
d x dxx
dt dt
x x
15
Solution
(a)Comparing the given differential equation
2
2
12 0
16
d x dxx
dt dt
With the general equation of the free damped motion
2
2
2
2
22
2
0
0
2 0
d x dxm β kx
dt dt
d x β dx kx
dt m dt m
d x dxλ x
dt dt
So here
m =1/16, β =2, k =1
W=mg
W=32/16=2lb
we see that
22 12 16
1 16 1 16
16, 4
β kλ ,
m m
λ
So that 022 λ
System is Over-damped motion.
The problem represents
“A 2lb weight is attached to a spring whose spring constant is 1lb/ft i.e. k=1lb/ft. the system is Over-
damped with resisting force numerically equal to 2 times of the instantaneous velocity i.e.
2dx dx
Damping force -βdt dt
”
Inspection of the boundary conditions:
0 0, ' 0 1.5x x
Reveals that the weight starts from equilibrium position with upward velocity of 1.5 ft/sec
Solution
(b)Comparing the given differential equation
2
24 4 0
d x dxx
dt dt
With the general equation of the free damped motion
2
20
d x dxm β kx
dt dt
2
20
d x β dx kx
dt m dt m
2
2
22 0
d x dxλ x
dt dt
So here
m =1, β =4, k =4
W=mg
W=32=32lb
we see that
24 42 4
1 1
2, 2
β kλ ,
m m
λ
So that 2 2 0λ
System is critically -damped motion.
The problem represents
“A 32lb weight is attached to a spring whose spring constant is 4lb/ft i.e. k=4lb/ft. the system is
critically damped with resisting force numerically equal to 4 times of the instantaneous velocity i.e.
4dx dx
Damping force -βdt dt
”
Inspection of the boundary conditions:
0 1, ' 0 2.01x x
Reveals that the weight starts from1 unit above the equilibrium position with a downward velocity of
2.01 ft/sec
Solution
(c)Comparing the given differential equation
2
216 0
d x dxx
dt dt
With the general equation of the free damped motion
2
20
d x dxm β kx
dt dt
2
20
d x β dx kx
dt m dt m
2
2
22 0
d x dxλ x
dt dt
So here
m =1, β =1, k =16
W=mg
W=32=32lb
we see that
21 162 16
1 1
1, 4
2
β kλ ,
m m
λ
So that 2 2 0λ
System under -damped motion
The problem represents
“A 32lb weight is attached to a spring whose spring constant is 16 lb/ft i.e. k=16 lb/ft. the system is
under -damped with resisting force numerically equal to 1 times of the instantaneous velocity i.e.
1dx dx
Damping force -βdt dt
”
Inspection of the boundary conditions:
0 1, ' 0 0x x
Reveals that the weight starts from rest 1 unit below the equilibrium position
Question 4
(a) A 16-lb weight stretches a spring 8/3 ft. Initially the weight starts from rest 2-ft below the
equilibrium position and the subsequent motion takes place in a medium that offers a damping force
numerically equal to ½ the instantaneous velocity. Find the equation of motion, if the weight is driven
by an external force equal to .3cos10 ttf 10
(b)Solve the following differential equation.
2 32 2 lnx y xy y x x
10
Solution
(a) Given w=16 lb and s=8/3 ft , β = ½ , .3cos10 ttf Initially the weight starts from rest 2-ft
below the equilibrium position and the subsequent motion takes place in a medium i.e
0 2, ' 0 0x x then differential equation of forced motion
2
2
d x dxm kx f t
dt dt
16 1
32 2
WW mg m
g
166
8 3
FF ks k
s
2
2
1 16 10cos3
2 2
d x dxx t
dt dt
or 2
212 20cos3
d x dxx t
dt dt
First consider the associated homogeneous differential equation.
2
212 0
d x dxx
dt dt
Put mtmtmt emdt
xd, me
dt
dx, ex 2
2
2
Then the auxiliary equation is:
2
2
2
2
12 0
1 112 0
4 4
1 48 10
2 4
1 47
2 4
1 47
2 2
m m
m m
m
m
m i
1 47
2 2m i
Thus the auxiliary equation has complex roots
1 2
1 47 1 47,
2 2 2 2m m i m m i
So that the complementary function of the equation is
47 472
cos sins1 22 2
tx e c t c tc
To find a particular integral of non-homogeneous differential equation we use the undetermined
coefficients, we assume that
cos3 sin3px t A t B t
Then 3 sin 4 3 cos4px t A t B t
9 cos4 9 sin 4px t A t B t
So that
12 9 cos 4 9 sin 4 3 sin 4 3 cos 4 12 cos3 12 sin3
3 3 cos 4 3 3 sin 4
p p px x x A t B t A t B t A t B t
A B t A B t
Substituting in the given non-homogeneous differential equation, we obtain
3 3 cos4 3 3 sin 4 20cos3A B t A B t t
Equating coefficients, we have
3 3 20A B
3 3 0A B
Solving these equations, we obtain
10 10
3 3A , B
Thus 10
cos3 sin33
px t t t
Hence the general solution of the differential equation is:
47 47 102
cos sins cos3 sin31 22 2 3
tx t e c t c t t t
1 47 472cos sins
1 22 2 2
47 47 47 102sin cos 3 sin 3 cos3
1 22 2 2 3
tx t x t e c t c t
te c t c t t t
Now 0 2x gives
1 1
10 10 4.1 2 2
3 3 3c c
Also 00 x gives
2
2
2
47 10 1 40 3
2 3 2 3
47 20 10
2 3
64
3 47
c
c
c
or
Hence the solution of the initial value problem is:
4 47 64 47 102
cos sins cos3 sin33 2 2 33 47
tx t e t t t t
Solution
(b)
2 3
1
2
2
2
1 2
2 2 ln
,
1
2 2 0
2 1 0
1,2
m m
m
m
c
x y xy y x x
put
y x y mx
y m m x
x m m m
m m
m
y c x c x
2
1 2
1 1 2 2
2
2 2 2
2
5
1 3
4
2 3
,
. .
21 2
0ln
ln 2
0ln
1 ln
p
y x y x
By V D P y u y u y
x xW x x x
x
xW x x
x x x
xW x x
x x
311
4
1
ln
int
1ln
4 4
Wu x x
W
On egrating
xu x
222
3
2
ln
int
1ln
3 3
Wu x x
W
On egrating
xu x
2
1 2
5 5
5
5 2
1 2
1 1ln ln
4 4 3 3
ln 1 ln 1
4 16 3 9
ln 7
12 16 9
p
p
p
y u x u x
x xy x x
x xy x
xy x c x c x
Assignment 6 (Spring 2005)
Maximum Marks 60
Due Date 05, July 2005
Assignment Weight age 2%
Question 1
(a) Define sequence and series define both with at least two examples, can every sequence be
represented as a series or not? Consider the series 2 3
0
1 ...! 2! 3!
n
n
x x xx
n
find the derived
series by differentiating the series. Show that both have same radius of convergence. 10
NOTE:
A series is said to be derived series which is obtained by integrating or differentiating term by term a
given series.
Solution
Sequence:
The sequence is defined as the function having the natural numbers as its domain N= {1, 2, 3…}.
Series:
The series is defined sum of the terms of a sequence.
Examples
1
2
1
1 2
1
1 2 3{ } , , ,... (1)
2 3 4 5
1 4 9{ } , , ... (2)
2 1 3 5 7
,
n
n
n
n
n
n
n
n
now for series series are the sum of the terms of a sequence
ar a ar ar
3
1
...
1 1 1 1...
( 1) 1.2 2.3 3.4k
ar
k k
Every sequence definitely can be represented as a series by simply adding the terms of that sequence.
Both the sequence 1 and 2 can be represented as a series like
1
2
1
1 2 3...
2 3 4 5
1 4 9...
2 1 3 5 7
n
n
n
n
n
n
Now we are going to find the derived series
The given series is
2 3
0
2 3
2 3
0
1 ...! 2! 3!
term by term
0 1 ...2 6
1 ...! 2! 3!
n
n
n
n
x x xx
n
now to get derived series we differentiate
x xx
we will get this series which is same as the above series
x x xx
n
so both the se
1
n n n n
1 1 ! ! !lim lim lim lim 0
1 ! 1 ! ! 1
n
n
ries will have the same radius of convergence but for is series
radius of convergence is
nc n n
c n n n n
so radius of convergence of orignal and derived series
.are same
Question 1
(b)If 3 5 7
...3 5 7
x x xSinx x the find the few terms of the power series of Cscx 05
Show all the steps involved each step has its own importance.
Solution
3 5 7 9
3 5 7 9
2 4 6 8
1sec
x x x xx-
3 5 7 9
x x x xx-
3 5 7 9
x x x x1-(
x 3 5 7 9
Co xSinx
I Now as we have that 3 5 7
...3 5 7
x x xSinx x
2
pplying the boinomial theorem i.e. (1+x)n =1+nX+n(n-1)/2! X2+n(n-1)(n-2)/3! X3...
When n is a ve integer or it is in the form of fraction then we use the following formula .
x1+(-1)
1=
x
4 6 8 2 4 6 8
2 4 6 8
2 4 6 8 2 4 6 8
2 4 6 8
x x x x x x x
3 5 7 9 3 5 7 9
x x x x...
3 5 7 9
x x x x x x x x1+(-1)( (
1 3 5 7 9 3 5 7 9
x x x x(
3 5 7 9
x
2 4 6 8 4 8 6 8
6 8
2 4 6 8 4
...
x x x x x x x x1+ neglecting higher tems neglecting higher terms
3 5 7 9 4 25 15 211
x xneglecting higher terms ...
27 45
x x x x x1
1 3 5 7 9 4
x
x
8 6 8
6 8
2 4 4 6 6 6 8 8 8
6 8
2 4 6 8
x x x
25 15 21
x x...
27 45
x x x x x x x x x1-
1 3 5 4 7 15 27 9 25 21
x x...
27 45
1 x 9x 296x 206x1- ...
3 20 945 945
x
x
So it is the required power series for the given function
Question 1
(c) Find the radius of convergence of 1
1
( 1) ( 2)n n
n
n x
05
.
Solution
1
0
1 1
1 2
1
( 1) ( 2) ( )
( 1) ( 1) 1
n n n
n
n n
n n
n n
compareit n x with c x x
c n c n
1
2
1
lim
( 1) ( 1)lim
( 1)
1lim 1
nn
n
n
n n
n
cR
c
nR
n
nR
n
So the radius of convergence is 1
Question 2
(a) Find the function represented by the following power series 0
n
n
x
05
Solution
2 3 4
1
0,1,2,3...
1 ...
inf
S1
1 and r=x
1so
1-x
n
n
now for n
we can get the series
x x x x x
so it is an inite geometric series so finding it sum
a
r
here a
Question 2
(b) Show that the given DE has a regular singular point at x=0.Determine the indicial equation, the
recurrence relation and the roots of the indicial equation. 10
" 0xy y
Solution
2
" 0
tan
" 0
sin int x=0 is a singular point
and p(x)=0 for which x-0) is analytic
1and q(x)= for which x-0) is analytic
x x
xy y
in s dard form it can be written as
yy
x
now by defination of gular po
so the point x=0 is regular singular point.
0
1
0
2
0
0
" 0 (1)
sup
' ( )
'' ( )( 1)
1
( )(
r n
n
n
r n
n
n
r n
n
n
n
n
xy y
now we pose there is a solution of the form
y a x
y a r n x
y a r n r n x
now substitutiong all theses values in equation no
x a r n
2
0
1
0 0
1 1
0
1 0
1
1
0 1
1) 0
( )( 1) 0
( 1) ( )( 1) 0
putting this value in the equation
r n r n
n
n
r n r n
n n
n n
r r n r n
n n
n n
r n r n
n n
n n
r n x a x
a r n r n x a x
a r r x a r n r n x a x
now we can write a x a x
now
1 1 1
0 1
1 1
1 1 1
0 1
1 1
1 1
0 1
1
( 1) ( )( 1) 0
( 1) ( )( 1) 0
( 1) [ ( )( 1) ]
( 1)
r r n r n
n n
n n
r r n r n
n n
n n
r r n
n n
n
a r r x a r n r n x a x
a r r x a r n r n x a x
a r r x a r n r n a x
now we have got indicial equation as r r
0
1
1
1
1
0 1
comparing the cofficent of
( )( 1) 0
( )( 1)
r 1
(1 )( )
r n
n n
nn
nn
r r
now x
we get a r n r n a
aa
r n r n
so for the greater root
we get the relation
aa
n n
01
012
023
0
1,2,3...
11.2
n=22.3 2.3!
33.4 3.4!
( 1)
!( 1)!
n
n
now replacing values of n
afor n a
aafor a
aafor n a
so continuing in this manner we will get
aa
n n
so for the series sol
2
01
( 1)( ) [1 ... ...]
1!2! 2!3! !( 1)!
n
ution we will get the following series solution
ax xy x x
n n
Question 3
(a) Define the ordinary and singular point. Is it necessary that ordinary and singular points are always
real? Find the singular point for the following differential equations.
(1) 2( 4) 2 6 0x y xy y
(2) 1
3 01
y xyx
08
Solution
Ordinary point:
A point 0x is said to be a ordinary point of a differential equation 0)()()( 012 yxayxayxa if
both 1
2
( )P(x) =
( )
a x
a x and) 0
2
( )Q(x)=
( )
a x
a x are analytic at 0x .
Singular point:
A point that is not an ordinary point is said to be singular point of the equation
Or
A point 0x is said to be a singular point of a differential equation 0)()()( 012 yxayxayxa if
both 1
2
( )P(x) =
( )
a x
a x and) 0
2
( )Q(x)=
( )
a x
a x or either one of these are not analytic at 0x .
Both singular and ordinary points need not be real numbers
(1) 2( 4) 2 6 0x y xy y
The equation 2( 4) 2 6 0x y xy y has the singular points at the solutions of 2 4 0x , namely,
2x i .
All other finite values, real or complex, are ordinary points.
(2) 1
3 01
y xyx
The equation 1
3 01
y xyx
has no singular point. But having ordinary point at the solutions of
1 0x , namely, 1x .
Question 3
(b) If 0xx is an ordinary point of the differential equation 0)()( yxQyxPy , can we always
find two linearly independent solutions in the form of power series centered at 0x :
.)(0
0
n
n
n xxcy
Give the reason? 02
Solution
Yes! We can always find the two linearly independent solutions and THEOREM (Existence of Power
Series Solution) gives us authentication .i.e.
“If 0xx is an ordinary point of the differential equation 0)()( yxQyxPy , we can always
find two linearly independent solutions in the form of power series centered at 0x :
.)(0
0
n
n
n xxcy
A series solution converges at least for Rxx 0 , where R is the distance from 0x to the closest
singular point (real or complex)”
Question 4
(a) Define the regular singular and irregular singular point. Find the singular point for the following
differential equations.
(1) 2( 4) 2 30 0x y xy y
(2) 1 2 1 5 0x y x y y 06
Solution
Regular singular point
A Singular point x x 0 of the given equation 2 1 0( ) ( ) ( ) 0a x y a x y a x y is said to be a regular
singular point if both 10 0
2
( )( ) ( ) ( )
( )
a xx x P x x x
a x and 2 2 0
0 02
( )( ) ( ) ( )
( )
a xx x Q x x x
a x are analytic
at 0x .
Irregular singular point
A singular point that is not regular is said to be an irregular singular point of the equation
Or
A Singular point x x 0 of the given equation 2 1 0( ) ( ) ( ) 0a x y a x y a x y is said to be an irregular
singular point if both 10 0
2
( )( ) ( ) ( )
( )
a xx x P x x x
a x and 2 2 0
0 02
( )( ) ( ) ( )
( )
a xx x Q x x x
a x or either one
of them are not analytic at 0x .
(1) 2( 4) 2 30 0x y xy y
2x are singular points of the equation
Because 2 4 0x or 2x .
Now write the equation in the form
2 2
2 300
( 4) 4
xy y y
x x
or 2 30
0( 2)( 2) ( 2)( 2)
xy y y
x x x x
2( )
( 2)( 2)
xP x
x x
and
30( )
( 2)( 2)Q x
x x
Then
Clearly 2x are regular singular points.
The factor ( 2)x appears at most to the first powers in the denominator of )(xP i.e.1 and at most to
the second power in the denominator of ( ) . .1Q x i e then 2x is a regular singular point.
Similarly, for x=-2.
(2) 1 2 1 5 0x y x y y
1x are singular point of the equation
Because 1 0x or 1x .
Now write the equation in the form
1 52 0
1 1
xy y y
x x
or 0
2 50
11y y y
xx
0
2( )
1P x
x
and
5
( )1
Q xx
Then
Clearly 1x are regular singular points.
The factor ( 1)x appears at most to the first powers in the denominator of )(xP i.e.0 and at most to
the second power in the denominator of ( ) . .1Q x i e then 1x is a regular singular point.
Question 4
(b) If x x 0 is a regular singular point of equation 2 1 0( ) ( ) ( ) 0a x y a x y a x y , then is it
necessary there exists always two series solution of the form if not then Give the reason? 02
Solution
No it is not necessary there exist two solutions Frobenius’ Theorem gives us authentication that there
exist just one power series solution .i.e.
If x x 0 is a regular singular point of equation 2 1 0( ) ( ) ( ) 0a x y a x y a x y , then there exists at
least one series solution of the form
0 0 00 0
( ) ( ) ( )r n n rn n
n n
y x x c x x c x x
where the number r is a constant that must be determined. The series will converge at least on some
interval .0 0 Rxx
Question 4
(c) Define second order Bessel differential equation and Bessel function. Find 3/ 2J x 07
Solution
Bessel differential equation:
A second order linear differential equation of the form
022
2
22 yvx
dx
dyx
dx
ydx
is called Bessel’s differential equation.
Bessel function:
Solution of the 022
2
22 yvx
dx
dyx
dx
ydx differential equation is usually denoted by xJv and is
known as Bessel’s function.
And now to find 3/ 2J x
Consider
1 12
v v vv
J x J x J xx
For 1
2v
1 1 11 12 2 2
122J x J x J x
x
3 1 12 22
3 1 12 22
1
1
J x J x J xx
J x J x J xx
As given (we know) 1/ 2 1/ 22 2
( ) sin , ( ) cosJ x x J x xx x
32
32
1 2 2cos sin
2 cossin
J x x xx x x
xJ x x
x x
Assignment 7 (Spring 2005)
Maximum Marks 60
Due Date 13, July 2005
Assignment Weight age 2%
Question 1
(a) write the give DE as a system in normal form 4 3
4 33 3 6 10
d y d yy
dt dt 07
Solution
1 2 3 4
' ' ' '
1 1 2
'
1 2
'
2
'
2 3
3
'
3
'
3 4
'
4
'
4 4 1
'
1 2
'
2 3
3
, ' , " "'
"
"
"'
""
103 2
4
y x y x y x y x
now x y so x y x
x x
now y x
so x x
asy x now differentiating both sides
y x
sox x
x y
x x x
so the required normal form is
x x
x x
x
'
4
'
4 4 1
103 2
4
x
x x x
Question 1
(b)What is degenerate system Explain in your own words with example( no definition is required from
books)
Write the following system in the normal form if possible
3 32
3 3
3 2 3
3 2 3
2 10 4 6
2 5 3 4 4
d y d x dx dyt
dt dt dt dt
d x d y d y dyx
dt dt dt dt
08
Solution
The system of DE is said to be degenerate system If it cannot be reduced to a linear system in the form
of normal form .Some time the situation arises that when we write the highest derivative in the form of
remaining variables then we try to eliminate the both the highest derivatives eliminate at same time so
it is not possible to write it in the form of linear normal from this type of systems are known as
degenerate systems.
The solution of the given system is not possible it is a degenerate system because
3 32
3 3
3 2 3
3 2 3
3 32
3 3
3 3 2
3 3 2
2 10 4 6
2 5 3 4 4
2 10 4 6 (1)
4 2 5 3 4 (2)
(1)
d y d x dx dyt
dt dt dt dt
d x d y d y dyx
dt dt dt dt
so writinging again
d y d x dx dyt
dt dt dt dt
d y d x d y dyx
dt dt dt dt
multiplying
3 32
3 3
3 3 2
3 3 2
2 by1
4 2 20 8 12
4 2 5 3 4
by and
d y d x dx dyt
dt dt dt dt
d y d x d y dyx
dt dt dt dt
It is clear that it is a degenerate system
Question 2
(a)Explain the difference between the echelon form and reduced echelon form. 05
Solution
Explain the difference between the echelon form and reduced echelon form
The difference in the reduced echelon and echelon form is that for each pivot element
ija in the echelon form ija =1 and for each element in that column in which pivot exists for
Which i is less than the pivot i are non zero and reaming all element are zero. Now in the case of
reduced echelon form each pivot element ija =1 and all element in that column should be zero in
which pivot exists, the best example of reduced echelon form is identity matrix. Question 2
(b)Solve the following system of equations by using gauss elimination method
1 2 3
1 2 3
1 2 3
6 6 6 6
2 4 6 12
10 5 5 30
x x x
x x x
x x x
10
Solution
The augmented Matrix may be written as
1
2
3
2 1
6 6 6 6 1 1/ 6
2 4 6 12 2 3 1/ 2
10 5 5 30 1 R
1R
1 4R
1
R
R R
RR
3
1 2
2
2
2
0R
0 5R
1
0
0 R
1
0
0
1
R
RR
R
R
R
3 2
23
R
0
R
R R
now by backward substitution we will get the values of all variables
1 2 3 11/5x x x Question 3
(a) Define are Eigen values and vectors.
Solution
The equation Ax y can be viewed as a linear transformation that maps a given vector x into a new
vector y .to find such a vector we set y= x, where is a scalar proportionality factor, and seek
solutions of the equations
Ax= x
(A- I)x=0 ---------------(1)
Where satisfy the equation det (A- I)=0and are called eigenvalues of matrix A
And non zero solution of equation 1 obtained by using the Eigen values are known as Eigen vectors.
05 Question 3
(b) Find the Eigen values and vectors of the following
5 2 4
1 1 1
4 3 3
10
Solution
Let
5 2 4
1 1 1
4 3 3
A
then Eigen values can be obtained as
The characteristic equation of the matrix A is
5 2 4
det 1 1 1 0
4 3 3
A I
Expanding the determinant by the cofactors of the second row, we obtain
2
2
2 2 3
2 3
3 2
5 1 3 3 2 3 4 4 3 4 1 0
5 3 3 3 2 3 4 4 3 4 1 0
2 5 2 7 4 7 4 0
10 5 2 14 2 28 16 0
10 3 14 2 28 16 0
3 24 42 0
1
2
3
4.4704
5.0697
1.6006
haveeigenvalues
But its have complex Eigen vectors corresponding to Each Eigen value.
Question 4
Solve the following system of DE by the method of undetermined coefficients.
6 2 1
4 3 5 7
t
t
dxx y e
dt
dyx y e t
dt
Solution
2
6 1' ( )
4 3
2 0 1' and F(t)=
1 5 7
the homogenous system
6 1X'=
4 3
6 1det( ) 0
4 3
(6 )(3 ) 4 0
9
t
X X F t
dx
xdtX X e t
dy y
dt
now first solve the
X
now A I
1 2
1
1
2
1
2
1 2
1 2
1 2
1
1
2
14 0
2, 7
for 2
6 2 1 0
4 3 2 0
4 1 0
4 1 0
4 0
4 0
1, 4
1
4
7
1 1
4 4
now
k
k
k
k
k k
k k
so solving we get
k k
K
now for
k
k
1 2
2
2 7
1 2
0
0
0
sin values
1K
1
1 1
4 1
t t
c
k k
now cho g arditrary
now as X c e c e
1
3 2 1
3 2 1
2t
2 0 1( )
1 5 7
F(t) on e is an eigen value so correct form of parti
t
t
p
t
now F t e t
so the particular solution is
a a aX e t
b b b
as by replacing the e in the
34 2 12 2
34 2 1
cular solution is
t t
p
aa a aX te e t
bb b b
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