differential equations - solved assignments - semester spring 2010
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Differential Equations - Solved Assignments - Semester Spring 2010TRANSCRIPT
Differential Equations
Solved Assignments
Semester Spring 2010
Assignment 1
Question 1: Marks=10
Solve the given differential equation by Separation of variables.
2 2(4 ) (2 ) 0y yx dy x xy dx+ − + =
Solution:
2 2
2 2
2 2
2 2
2 2
2 2
2 2 2 2
2 2 2 2
2 21 1
2 21
1
(4 ) (2 ) 0
(4 ) (2 )
2 4
2 4
1 2 1 2
2 2 2 4
1 1ln | 2 | ln | 4 | ln
2 2ln | 2 | ln | 4 |
(2 ) (4 )
2 (4 )
(4 ) 2
( (4
y yx dy x xy dx
y x dy x y dx
y xdy dx
y x
y xdy dx
y x
y xdy dx
y x
y x c
y x c
y c x
y c x wherec c
y c x
y c
+ − + =+ = +
=+ +
=+ +
=+ +
+ = + +
+ = ++ = +
+ = + =
= + −
=
∫ ∫
∫ ∫
12 2) 2)x+ −
Question 2: Marks=10
Solve the following Differential Equation.
2 2( ) 0y yx dx x dy+ + =
Solution:
2 2
2
2
2 2 22
2
2
2
( ) 0
(1).
(1).
2
1 1
21 1
( 2)
1 1 1( )2 2( 2)
y yx dx x dy
dy y yx
dx xHomogenous differential equtaionoffirst order
put y Vx
dy dVV x
dx dxput in
dV V x VxV x V V
dx xdV
x V Vdx
dV dxV V x
dV dxV V x
dV dxV V x
+ + =
+= − − −− >
=
= +
++ = − = − −
= − −
= −+
=+
− = −+
∫ ∫
∫ ∫
1
2
1
2
1 1ln ln( 2) ln ln
2 21
ln( ) ln /2 2
1ln( ) ln /
2 2
1ln( ) ln /
2 2
ln( ) ln /2
( ) /2
V V x c
Vc x
Vy
put Vx
y
x c xy
xy
c xy x
yc x
y x
yc x
y x
− + = − +
=+
=
=+
=+
=+
=+
Question 3: Marks=10
Solve the following Differential Equation subject to the indicated initial condition.
(4 2 5) (6 4 1) 0, ( 1) 2y x dx y x dy y+ − + + − = − =
Solution:
(4 2 5) (6 4 1) 0 ( )
4 2 5 0
6 4 1 0
tan , ;
13 9
2 213 9
,2 2
( )
9 13 9 13(4( ) 2( ) 5) (6( ) 4( ) 1) 0
2 2 2 2(4 18 2 13 5) (6 27 4
y x dx y x dy i
y x
y x
Solving simul eously we have
x and y
x X y Y
usethesevalues in i
Y X dX Y X dY
Y X dX Y X
+ − + + − = − −− >+ − =+ − =
= − =
= − = +
+ + − − + + + − − =
+ + − − + + + − 26 1) 0
(2 4 ) (4 6 ) 0
dY
X Y dX X Y dY
− =+ + + =
2
2
,
.
2 4 2
4 6 2 32 1 2
.2 3 2 3
1 2 2 3 1 2.
2 3 2 3
3 4 1.
2 3
Homogenous differential equation soreplace
Y VX
dY dVV X
dX dXdY X Y X Y
dX X Y X YdV X VX V
V XdX X VX V
dV V V V VX V
dX V V
dV V VX
dX V
=
= +
+ += − = −+ +
+ ++ = − = −+ +
+ + + += − − = − + +
+ += − +
2
2
2
2
12 2
3 4 1.
2 3
3 2 1
3 4 1
1 6 4 1
2 3 4 11
ln | 3 4 1| ln ln ln2
ln | 3 4 1| ln
dV V VX
dX V
VdV dX
V V X
VdV dX
V V Xc
V V X cX
cV V
X
+ += − +
+ = − + +
+ = −+ +
+ + = − + =
+ + =
∫ ∫
12 2
12 2
212
2
2 212
2
12 2 2
ln | 3 4 1| ln
{3 4 1}
, ;
{3 4 1}
3 4( )
(3 4 )
cV V
Xc
V VX
YreplaceV we haveX
Y Y c
X X X
Y XY X c
X X
Y XY X c
+ + =
+ + =
=
+ + =
+ + =
+ + =
12 2 2
12 2 2
9 13sin
2 29 13 9 13
(3( ) 4( )( ) ( ) )2 2 2 2
,
( 3 4 5 14) ( )
u g Y y and X x
y x y x c
by opening square and solvin we have
x y xy x y c ii
= − = +
− + + − + + =
+ + − − − = − − −
12
1 12 2
1 12 2 2 2
( 1) 2 1 2 ( ).
(1 12 8 5 2 14)
(1 6 8 5 2 14) ( 6)
sin ( ),
( 3 4 5 14) ( 6)
now y x and y put in ii we have
c
c
u g in ii we have
x y xy x y
− = ⇒ = − =
+ − + − − =
= + − + − − = −
+ + − − − = −
Question 4: Marks=10
Solve the initial value problem
22 , (1) 5dx
y x y ydy
− = =
Solution:
2
1 1( ) lnln
2
2
2
, ;
1( ) 2 (1)
1. .
(1) . . ;
1 1( ) 2
1( . ) 2
1( . ) 2
1( . ) 2
1. 2
2
dyy y y
dxy x y
dy
divideby y we have
dxx y
dy y
I F e e ey
multiply by I F we have
dxx
y dy y
dx
dy y
d x dyy
d x dyy
x y cy
x y cy
− −
− =
+ − = − − >
∫⇒ = = =
+ − =
=
=
=
= +
= +
∫ ∫
2
2
1, 5
2
1 50 5
49
549
25
x y
x y cy
c
c
x y y
= == += +
= −
= −
Assignment 2
Question 1: Marks=10
Solve the differential equation.
2 2dyx y xy
dx+ =
Solution:
2 2
2( ) (1)
.
dyx y xy
dxdy y y
dx x xy
Tx
y Tx
dy dTT x
dx dx
+ =
+ = − −− >
=
=
= +
2
2
2
2
1
1 1
1 1
1ln ln ln
1ln
T
dTx T T T
dxdT
x Tdx
dT dxT x
dT dxT x
x c cxT
cxT
e cx
+ + =
= −
= −
= −
− = − − = −
=
=
∫ ∫
x
ye cx=
Question 2: Marks=10
Verify that the given two-parameter family of functions is the general solution of the non-homogeneous differential equation on the indicated interval.
// / 2 51 27 10 24 , 6 , ( , )x x x xy y y e y c e c e e− + = = + + −∞ ∞
Solution:
// /
2
2
5 21 2
2 2
7 10 24 ( )
:
( 7 10) 0
7 10 0
( 5)( 2) 0 5,2
:
( )
:
1 1.24 .24
7 10 (1) 7(1) 10
1.24
4
x
x xc
x xp
xp
y y y e A
characteristics equation
D D y
D D
D D D
Complementary solution
y c e c e i
For Particular Solution
y e eD D
y e
− + = − −− >
− + =− + =− − = ⇒ =
= + − −− >
= =− + − +
= =
5 21 2
6 ( ).
( )&( ), :
6
x
x x x
e ii
By collecting i ii we have general solution
y c e c e e
− −− >
= + +
To verify:
5 21 2
5 21 2
5 21 2
6
' 5 2 6
'' 25 4 6
x x x
x x x
x x x
y c e c e e
y c e c e e
y c e c e e
= + +
= + +
= + +
Using all these in (A), we have
5 2 5 2 5 21 2 1 2 1 2
5 2 5 2 5 21 2 1 2 1 2
25 4 6 7(5 2 6 ) 10( 6 ) 24
25 4 6 35 14 42 10 10 60 24
24 24
x x x x x x x x x x
x x x x x x x x x x
x x
c e c e e c e c e e c e c e e e
c e c e e c e c e e c e c e e e
e e
+ + − + + + + + =
+ + − − − + + + =
=
Hence it is verified that the given two-parameter family of functions is the general solution of the non-homogeneous differential equation on the indicated interval.
Question 3: Marks=10
Determine whether the functions 2
1 2 3( ) cos 2 , ( ) 1, ( ) cos ( )f x x f x f x x= = = are linearly independent or
linearly dependent on ( , )−∞ ∞ .
Solution:
21 2 3
1 2 3
1 1 2 2 3 3
1 2 3
1 1 2 2 3 3
21 2 3
( ) cos 2 , ( ) 1, ( ) cos ( )
( ), ( ) ( )
( ) ( ) ( ) 0
0, 0, 0
.
,
( ) ( ) ( ) 0
cos 2 .1 os ( ) 0
f x x f x f x x
f x f x and f x will belinearly independent if
a f x a f x a f x
a a a
otherwiselinearly dependent
Now
a f x a f x a f x
a x a a c x
= = =
+ + =⇒ = = =
+ + =
⇒ + + =2
1 2 3cos 2 .1 osa x a a c x⇒ + = −
This is only possible when
1 2 3
2
1 2 3
1 0, 1 0, 2 0
cos2 1 2 os ( )
( ), ( ), ( ) .
a a a
so that
x c x
which is true
So
f x f x f x arelinearly dependent
= ≠ = ≠ = − ≠
+ =
Assignment 3
Question 1: Marks=10
Solve the given differential equation subject to the initial conditions.
/// // / / //2 5 6 0 (0) (0) 0, (0) 1y y y y y y y+ − − = = = =
Solution:
The given differential equation is
/// // /2 5 6 0y y y y+ − − =
For the corresponding auxiliary equation
2
3
Then,
'
''
'''
mx
mx
mx
mx
Let
y e
y me
y m e
y m e
=
===
Substituting in the given differential equation, we have
( ) ( ) ( )3 22 5 6 0mx mx mx mxm e m e me e+ − − =
3 2
3 2
( 2 5 6) 0
0
,
2 5 6 0
mx
mx
e m m m
but e
so
m m m
+ − − =≠
+ − − =
By using synthetic division roots of the above differential equation can be found as follows
3)1 2 5 6
3 3 6
1 1 2 0
− − −
−
− −
i.e. the cubic auxiliary equation takes the form
( )( )( )( )( )
3 2
2
2
5 3 9 0
3 2 0
3 2 2 0
m m m
m m m
m m m m
− + + =
+ − − =
+ − + − =
( ) ( ) ( )( )( )( )( )( ) ( )
3 { 2 1 2 } 0
3 2 1 0
3 1 2 0
. . 3 0, 1 0, 2 0
3, 1,2
m m m m
m m m
m m m
i e m m m
m
+ − + − =
+ − + =
+ + − =+ = + = − =
= − −
So, the general solution is
3 21 2 3
x x xy c e c e c e− −= + +
Now applying the initial conditions, we have
3 21 2 3
0 0 01 2 3
1 2 3
(0) 0
0
0 (1)
x x x
y
y c e c e c e
c e c e c e
c c c
− −
== + +
= + += + + − − −
3 21 2 3
x x xy c e c e c e− −= + +
3 21 2 3
3 21 2 3
0 0 01 2 3
1 2 3
'(0) 0
' 3 2
0 3 2
0 3 2 (2)
x x x
x x x
y
y c e c e c e
y c e c e c e
c e c e c e
c c c
− −
− −
== + +
= − − +
= − − += − − + − − −
3 21 2 3
3 21 2 3
0 0 01 2 3
1 2 3
''(0) 1
'' 9 4
1 9 4
1 9 4 (3)
x x x
x x x
y
y c e c e c e
y c e c e c e
c e c e c e
c c c
− −
− −
== + +
= + +
= + += + + − − −
So, the three simultaneous equations are
1 2 3
1 2 3
1 2 3
0 (1)
3 2 0 (2)
9 4 1 (3)
c c c
c c c
c c c
+ + = − − −− − + = − − −
+ + = − − −
Solving the above system of equations, we have
1 2 3
1 1 1, ,
10 6 15c c c= = − =
Hence, the solution of the given differential equation is
3 21 1 1
10 6 15x x xy e e e− −= − +
Question 2: Marks=10
Solve the following differential equations using the undetermined coefficients.
// / 3 25 2 4 6y y x x x− = − − +
Solution:
The associated homogenous differential equation is
// /5 0y y− =
We put
mxy e=
Then the auxiliary equation is
2
2
2
5 0
( 5 ) 0
0
5 0
( -5) 0
0,5
mx mx
mx
mx
m e me
e m m
but e
so
m m
m m
m
− =− =≠
− ==
=
Thus the complementary function is
51 2
xcy c c e= +
Now, the input function
3 2( ) 2 4 6g x x x x= − − +
So, we assume
3 2py Ax Bx Cx D= + + +
As the constant term of py is already a part of the complementary function, so we remove this duplication by
multiplying the particular function by x i.e.
4 3 2py Ax Bx Cx Dx= + + +
/ 3 2
// 2
4 3 2
12 6 2
p
p
y Ax Bx cx D
y Ax Bx c
= + + +
= + +
Substituting in the given differential equation, we have
2 3 2 3 2
2 3 2 3 2
12 6 2 5(4 3 2 ) 2 4 6
12 6 2 20 15 10 5 2 4 6
Ax Bx c Ax Bx cx D x x x
Ax Bx c Ax Bx cx D x x x
+ + − + + + = − − ++ + − − − − = − − +
Comparing coefficients of 3x , 2x , x and constants, we have
-20 2 (1)
12 15 4 (2)
6 10 1 (3)
2 5 6 (4)
A
A B
B C
C D
= − − − − − − −− = − − − − − −
− = − − − − − − −− = − − − − − − −
Solving the above equations, we have
1
1014
7553
250697
625
A
B
C
D
= −
=
=
= −
So, the particular solution is
4 3 21 14 253 697
10 75 250 625py x x x x= − + + −
Hence, the general solution of the given differential equation is
5 4 3 21 2
1 14 253 697
10 75 250 625
c p
x
y y y
c c e x x x x
= +
= + − + + −
Question 3: Marks=10
Solve the differential equations by variations of parameters.
/// // 22 6y y x− =
Solution:
Step 1
The associated homogenous equation is
/// //2 6 0y y− =
By putting mxy e= , the auxiliary equation of the given differential equation is
3 2
3 2
2
2 6 0
3 0
( 3) 0
0,0,3
m m
m m
m m
m
− =− =
− ==
Therefore 31 2 3
xcy c c x c e= + +
Step 2
From cy , we find that three linearly independent solutions of the homogenous differential equation are
31 2 31, , xy y x y e= = =
Thus the Wronskian of the solutions 1y , 2y and 3y is given by
( )
3
3 31 2 3
3
1
, , 0 1 3 9
0 0 9
x
x x
x
x e
W y y y e e
e
= =
Step 3
The standard form of the given differential equation is
2/// //3
2
xy y− =
Step 4
Next we find the determinants 1W , 2W and 3W by respectively replacing the 1st , 2nd and 3rd column of W by the
column 2
0
0
2
x
Thus
( )3
32 23 3 3 3 3 2 3
1 32
3
0 3 1
0 1 3 32 2 2 21 3
0 92
x
xx x x x x
x
x
x ex ex x
W e xe e x e x ee
xe
= = = − = −
( )3 3
23 3 2 32
2 32
3
1 0 0 33
0 0 3 1 32 2 9
20 92
x x
x x x
x
x
e ex
W e e x exe
xe
= = = − = −
23
2
1 01
0 1 02
0 0 2
x
W x
x
= =
Step 5
Therefore, the derivatives of the unknown functions 321 and , uuu are given by.
3 3 2 3
11 3
2 3
22 3
2
33 3
3 12 2
9329
129
x x
x
x
x
x
x e x eWu
W e
x eWu
W e
xWu
W e
−′ = =
−′ = =
′ = =
Step 6
Integrating these derivatives to find 321 and , uuu , we get
3 3 2 3
11 3
3 3 2
3
3 2
3 2
4 3
3 12 2
93 12 2
=9
1 3 1 =
9 2 2
1 1 =
6 18
1 1 =
6 4 18 3
x x
x
x
x
x e x eWu dx dx
W e
e x xdx
e
x x dx
x dx x dx
x x
−= =
−
−
−
−
∫ ∫
∫
∫
∫ ∫
4 31 1 =
24 54
x x−
2 3
22 3
2
3
3
329
1 =-
6
1 =-
6 3
1 =-
18
x
x
x eWu dx dx
W e
x dx
x
x
−= =
∫ ∫
∫
( )
( )
2
33 3
2 3
2 3 3 2
3 32
129
1
18
1 = .
18
1 = 2
18 3 3
x
x
x x
x x
xWu dx dx
W e
x e dx
dx e dx e dx x dx
dx
e ex x dx
−
− −
− −
= =
=
−
− − −
∫ ∫
∫
∫ ∫ ∫
∫
( )
( )
2 3 3
2 3 3 3
3 32 3
1 1 2 =
18 3 3
1 1 2 = .
18 3 3
1 1 2 = 1
18 3 3 3 3
x x
x x x
x xx
x e xe dx
dx e x e dx e dx x dx
dx
e ex e x dx
− −
− − −
− −−
− +
− + −
− + − − −
∫
∫ ∫ ∫
∫
2 3 3 3
32 3 3
2 3 3 3
1 1 2 1 1 =
18 3 3 3 3
1 1 2 1 1 =
18 3 3 3 3 3
1 1 2 1 1 =
18 3 3 3 9
x x x
xx x
x x x
x e xe e dx
ex e xe
x e xe e
− − −
−− −
− − −
− + − +
− + − + −
− + − −
∫
2 3 3 3
2 3 3 3
1 1 2 2 =
18 3 9 27
1 1 1 =-
54 81 243
x x x
x x x
x e xe e
x e xe e
− − −
− − −
− − −
− −
Step 7
A particular solution of the non-homogenous equation is
( )
1 1 2 2 3 3
4 3 3 2 3 3 3 3
4 3 4 2
4 3 2
1 1 1 1 1 1-
24 54 18 54 81 243
1 1 1 1 1 1 =
24 54 18 54 81 2431 1 1 1 1
=-72 54 54 81 243
p
x x x xp
y u y u y u y
y x x x x x e xe e e
x x x x x
x x x x
− − −
= + +
= − + − + − −
− − − − −
− − − −
Step 8
The general solution of the given differential equation is
c py y y= +
i.e.
3 4 3 21 2 3
1 1 1 1 1-72 54 54 81 243
xy c c x c e x x x x= + + − − − −
Assignment 4
Question 1: Marks=10
Write the solution of the given initial-value problem in the form ( ) ( )sinx t A tϖ φ= +
( ) ( )/0.1 10 0 0 1 0 1x x , x , x′′ + = = =
Solution:
The given differential equation is
0.1 10 0x x′′ + =
100 0x x′′ + = ---------(1)
Put mtx e= , 2
22
mtd xm e
dt=
So, the differential equation (1) becomes
2
2
2
100 0
( 100) 0
0
,
100 0
10
mt mt
mt
mt
m e e
e m
but e
so
m
m i
+ =+ =≠
+ =⇒ = ±
i.e. 10ω =
Thus the general solution of the equation is
1 2( ) cos10 sin10x t c t c t= +
Now, we apply the initial conditions
( )0 1x =
1 2
1
.1 .0 1
1
c c
c
⇒ + =⇒ =
Now
1 2( ) cos10 sin10x t c t c t= +
/1 2( ) 10 sin10 10 cos10x t c t c t= − +
Applying second initial condition
( )/ 0 1x =
1 210 .0 10 .1 1c c− + =
2
1
10c =
So
1( ) cos10 sin10
10x t t t= +
The amplitude of the motion is given by
( )2
2 1 1011 ft.
10 10A
= + =
And the phase angle is defined by
1 11tan ( ) tan (10) 1.47
110
radiansφ − −= = =
Hence, the solution of the given differential equation is given by
101( ) sin(10 1.47)
10x t t= +
Question 2: Marks=10
Solve the given differential equation subject to the indicated initial conditions.
2 // / /0, (1) 1, (1) 2x y xy y y y+ + = = =
Solution:
The given differential equation is
2 // / 0x y xy y+ + =
Let us suppose that
mxy =
then 1−= mmxdx
dy
and 2
22
( 1) md ym m x
dx−= −
so, the given differential equation becomes
2 2 1
2
2
2
( 1) 0
( 1) 0
( 1) 0
( 1) 0
1 0
m m m
m m m
m
m
x m m x xmx x
m m x mx x
x m m m
x m
m
− −− + + =− + + =
− + + =+ =
+ =
2 1m
m i
= −= ±
As the roots, here, are conjugate complex
So, comparing the roots with the general complex conjugate roots
m iα β= ± , we have
0 1andα β= =
When the roots are conjugate complex,
Then the general solution of the given differential equation is given by
1 2[ cos( ln ) sin( ln )]y x c x c xα β β= +
Putting the values of α and β , we get
01 2
1 2
[ cos(1.ln ) sin(1.ln )]
[ cos(ln ) sin(ln )]
y x c x c x
y c x c x
= += +
By applying the condition
(1) 1y = , we get
1 2
1 2
1 2
1
cos(ln1) sin(ln1) 1
cos(0) sin(0) 1
(1) (0) 1
1
c c
c c
c c
c
+ =+ =
+ ==
/1 2
1 1[ sin(ln ). cos(ln ). ]y c x c x
x x= − +
Now, By applying the condition
/ (1) 2y = , we have
1 2
1 2
2
1 1sin(ln1). cos(ln1). 2
1 1sin(0).1 cos(0).1 2
2
c c
c c
c
− + =
− + ==
Hence, the solution of the given differential equation subject to the indicated initial conditions is
cos(ln ) 2sin(ln )y x x= +
Question 3: Marks=10
Solve the following differential equation using tx e=
22 /
23 0, (1) 0, (1) 4
d y dyx x y y
dx dx+ = = =
Solution:
The given differential equation can be written as
2 2( 3 ) 0x D xD y+ =
With the substitution tx e= or xt ln= , we obtain
∆=xD , )1(22 −∆∆=Dx
Therefore the given differential equation becomes
2
2
2
2
[ ( 1) 3 ] 0
[ 3 ] 0
[ 2 ] 0
2 0 (1)
y
y
y
d y dy
dt dt
∆ ∆ − + ∆ =∆ − ∆ + ∆ =∆ + ∆ =
+ = − − −
Now substitute: mty e= then mtdy medt
= , 2
22
mtd ym e
dt=
Then eq.(1) becomes
2
2
2 0
( 2 ) 0
0,
mt mt
mt
mt
m e me
e m m
but e so
+ =+ =≠
The auxiliary equation is
2 2 0
( 2) 0
0, 2
m m
m m
m
+ =+ =
= −
The roots of the auxiliary equation are distinct and real, so the solution is
21 2
ty c c e−= +
But tx e= , therefore the solution is
21 2y c c x−= +
Using the initial condition (1) 0y = , we have
21 2
1 2
(1) 0
0 (2)
c c
c c
−+ =+ = − − − −
21 2y c c x−= +
/ 322y c x−= −
Using the second initial condition / (1) 4y = , we have
32
2
2
2 (1) 4
2 4
2
c
c
c
−− =− =
= −
Using (2)
1 2
1
1
0
2 0
2
c c
c
c
+ =− ==
Therefore , the solution of the given differential equation subject to the indicated initial conditions is
22 2y x−= −
Assignment 5
Question 1: Marks=10
For the following differential equation, find two linearly independent power series solutions about the ordinary
point 0 1x = .
// / 0y xy y− − =
Solution:
The given differential equation is
// / 0y xy y− − =
As the ordinary point is 0 1x =
So, we assume
0
( 1)nny c x
∞
= −∑
Now
/ 1
0
( 1)nny c n x
∞−= −∑
And
// 2
0
( 1)( 1)nny c n n x
∞−= − −∑
Putting the expressions of y and its derivatives in the given differential equation, we have
2 1
0 0 0
2 1
0 0 0
2 1
0 0 0 0
2 1
0 0
( 1)( 1) ( 1) ( 1) 0
( 1)( 1) ( 1 1) ( 1) ( 1) 0
( 1)( 1) ( 1) ( 1) ( 1) 0
( 1)( 1) ( 1)
n n nn n n
n n nn n n
n n n nn n n n
n nn n
c n n x x c n x c x
c n n x x c n x c x
c n n x c n x c n x c x
c n n x c n x
∞ ∞ ∞− −
∞ ∞ ∞− −
∞ ∞ ∞ ∞− −
∞ ∞− −
− − − − − − =
− − − − + − − − =
− − − − − − − − =
− − − − −
∑ ∑ ∑
∑ ∑ ∑
∑ ∑ ∑ ∑
∑ ∑0
0 2 32 3 4 5
0 1 2 31 2 3 4
0 1 2 30 1 2 3
0 22 3 4 5
( 1)( 1) 0
2 ( 1) 6 ( 1) 12 ( 1) 20 ( 1) ....
( ( 1) 2 ( 1) 3 ( 1) 4 ( 1) ........
( ( 1) 2 ( 1) 3 ( 1) 4 ( 1) .....) 0
2 ( 1) 6 ( 1) 12 ( 1) 20 ( 1)
nnc n x
c x c x c x c x
c x c x c x c x
c x c x c x c x
c x c x c x c x
∞
+ − =
− + − + − + − +
− − + − + − + − +
− − + − + − + − + =
− + − + − + −
∑
3
0 1 2 31 2 3 4
0 1 2 30 1 2 3
....
( 1) 2 ( 1) 3 ( 1) 4 ( 1) ........
( 1) 2 ( 1) 3 ( 1) 4 ( 1) ..... 0
c x c x c x c x
c x c x c x c x
+
− − − − − − − − +
− − − − − − − − + =
Equating coefficients of 0( 1)x − , 1( 1)x − and 2( 1)x − , we have
2 1 02 0c c c− − = -------(1)
3 2 16 2 2 0c c c− − = -----(2)
4 3 212 3 3 0c c c− − = -----(3)
From eq.1, we have
1 02 2
c cc
+= ------(4)
From eq.2 and eq.4, we have
3 1 0 1
3 0 1
6 2 0
6 3 0
c c c c
c c c
− − − =− − =
1 03
3
6
c cc
+= ---(5)
From eq.2, 3 and 5, we have
4 3 2
1 04 2
1 04 2
1 0 1 04
0 01 14
0 01 14
4 1 0
12 3 3 0
312 3 3 0
6
312 3 0
2
312 3 0
2 2
33 312 0
2 2 2 233 3
122 2 2 2
12 3 2
c c c
c cc c
c cc c
c c c cc
c cc cc
c cc cc
c c c
− − =+ − − =
+ − − =
+ + − − =
− − − − =
= + + +
= +
4 1 0
1 1
4 6c c c= + ------(6)
The required solution is given by
0
0 1 2 3 40 1 2 3 4
1 2 3 41 0 1 00 1 1 0
2 3 40 0 0 0
1 2 31 1 1
( 1)
( 1) ( 1) ( 1) ( 1) ( 1) ......
3 1 1( 1) ( 1) ( 1) ( )( 1) .....
2 6 4 61 1 1
( 1) ( 1) ( 1) ...2 6 6
1 1 1( 1) ( 1) ( 1)
2 2
nny c x
c x c x c x c x c x
c c c cc c x x x c c x
c c x c x c x
c x c x c x
∞
= −
= − + − + − + − + −+ += + − + − + − + + − +
= + − + − + − +
+ − + − + − +
∑
41( 1) ...
4c x − +
2 3 40
2 3 41
1 1 1(1 ( 1) ( 1) ( 1) ...)
2 6 61 1 1
(( 1) ( 1) ( 1) ( 1) ...)2 2 4
c x x x
c x x x x
= + − + − + − +
+ − + − + − + − +
Thus the two linearly independent power series solutions about the ordinary point0 1x = are
2 3 41
2 3 42
1 1 1( ) 1 ( 1) ( 1) ( 1) ...
2 6 61 1 1
( ) ( 1) ( 1) ( 1) ( 1) ...2 2 4
y x x x x
y x x x x x
= + − + − + − +
= − + − + − + − +
Question 2: Marks=10
Express the Bessel function7
2
( )J x−
in terms ofsinx , cosx and powers ofx .
Solution:
Consider
1 1
2( ) ( ) ( )v v v
vJ x J x J x
x− + = ------(1)
Putting 5
2v = − , we get
5 5 51 1
2 2 2
5 / 2( ) ( ) 2J x J x J
x− − − + −
−+ =
7 3 5
2 2 2
5( ) ( )J x J x J
x− − −
−+ =
7 5 3
2 2 2
5( ) ( )J x J J x
x− − −
−= − ------(2)
Again Putting 3
2v = − in the equation(1), we get
3 3 31 1
2 2 2
5 1 3
2 2 2
3/ 2( ) ( ) 2 ( )
3( ) ( ) ( )
J x J x J xx
J x J x J xx
− − − + −
− − −
−+ =
+ = −
5 3 1
2 2 2
3( ) ( ) ( )J x J x J x
x− − −= − − ----(3)
And to find the value of 3
2
( )J x−
, putting 1
2v = − in equation(1), we get
1 1 11 1
2 2 2
3 1 1
2 2 2
1/ 2( ) ( ) 2 ( )
1( ) ( ) ( )
J x J x J xx
J x J x J xx
− − − + −
− −
−+ =
+ = −
3 1 1
2 2 2
1( ) ( ) ( )J x J x J x
x− −= − − ----(4)
Since
1 1
2 2
2 2( ) sin , ( ) cosJ x x J x x
x xπ π−= = ------------(5)
Therefore equation(4) becomes
3
2
1 2 2( ) cos sin
2 cos 2sin
J x x xx x x
xx
x x x
π π
π π
−
= − −
= − −
3
2
2 cos( ) sin
xJ x x
x xπ−
= − +
---------(6)
Using (5) and (6), equation (3) becomes
5
2
3 2 cos 2( ) sin cos
xJ x x x
x x x xπ π−
= − − + −
5
2
3 2 cos 2( ) sin cos
xJ x x x
x x x xπ π−
= + −
5 22
3 2 3cos 3sin( ) cos
x xJ x x
x x x xπ−
= + −
-------(7)
Using Equation(6) and (7), equation (2) becomes
7 22
2
3 2
3 2
5 2 3cos 3sin 2 cos( ) cos sin
5 2 3cos 3sin 2 cos 2cos sin
2 5 15 1 2 15cos sin 1
2 6 15 2 15cos sin 1
x x xJ x x x
x x x x x x
x x xx x
x x x x x x x
x xx x x x x x
x xx x x x x
π π
π π π
π π
π π
−
= − + − − − +
= − + − + +
= − + + −
= − + −
Question 3: Marks=10
Determine the singular points of the given differential equation. Classify each singular point as regular or irregular.
0)5(7)2(3)2)(25( 223 =++′−+′′−− yxyxxyxxx
Solution:
The given differential equation is
0)5(7)2(3)2)(25( 223 =++′−+′′−− yxyxxyxxx
Dividing both sides by 3 2( 25)( 2)x x x− − , we get
// /3 2 2 3 2 2
// /2 2 3 2
3 ( 2) 7( 5)0
( 25)( 2) ( 25)( 2)
3 70
( 25)( 2) ( 5)( 2)
x x xy y y
x x x x x x
y y yx x x x x x
− ++ + = − − − −
+ + = − − − −
So,
2 2
3 2
3( )
( 25)( 2)
7( )
( 5)( 2)
P xx x x
Q xx x x
=− −
=− −
So, the singular points are 0,±5,2
We check the regularity of these points
Regularity at 0x =
2
22
3( )
( 25)( 2)
7( )
( 5)( 2)
x P xx x x
x Q xx x x
=− −
=− −
( )x P x And 2 ( )x Q x are not analytical at 0x = .
So, 0x = is an irregular singular point
Regularity at 5x =
2
23 2
3( 5) ( )
( 5)( 2)
7( 5)( 5) ( )
( 2)
x P xx x x
xx Q x
x x
− =+ −−− =−
2( 5) ( ) ( 5) ( ) 5x P x and x Q x are analytic at x− − =
So, 5x = is a regular singular point
Regularity at 5x = −
2
22
3 2
3( 5) ( )
( 5)( 2)
7( 5)( 5) ( )
( 5)( 2)
x P xx x x
xx Q x
x x x
+ =− −
++ =− −
2( 5) ( ) ( 5) ( ) 5x P x and x Q x are analytic at x+ + = −
So, 5x = − is a regular singular point
Regularity at 2x =
2 2
3( 2) ( )
( 25)x P x
x x− =
−
23
7( 2) ( )
( 5)x Q x
x x− =
−
2( 2) ( ) ( 2) ( ) 2x P x and x Q x are analytic at x− − =
So, 2x = is a regular singular point
Assignment 6
Question 1: Marks=10
Solve the given system of differential equations by systematic elimination method.
dxy t
dtdy
x tdt
= − +
= −
Solution:
Step 1
Writing the given differential equations of the system in the differential operator form, we get
Dx y t
Dy x t
= − += −
Or
(1)
(2)
Dx y t
x Dy t
+ = − − − − −− + = − − − − − −
Step 2
We eliminate the variabley from the given system. For this purpose we Operate on the first equation with the
operator D , we get
2
2
( )
1 (5)
D Dx y t
D x Dy Dt
D x Dy
+ =+ =+ = − − −
Subtracting Eq.(5) from Eq.(2), we get
2
2
2
2
1
1
1
( 1) 1 (6)
x Dy D x Dy t
x D x t
D x x t
D x t
− + − − = − −− − = − −
+ = ++ = + − − − − −
Step 3
The auxiliary equation of the differential equation for x is
2
2
(m +1) = 0
=> m = -1
=> m i= ±
The roots of the auxiliary equation are complex. Therefore, the complementary function for x
1 2cos sincx c t c t= +
We now use the method of undetermined coefficients to find
the particular integrals px
= At+Bpx
/
//
=
=0
p
p
x A
x
Substituting in the differential equation forx , we obtain
0 1
1 , 1
1p
At B t
A B
x t
+ + = += == +
Hence, the solution for the variable x is given by
1 2cos sin 1c px x x
c t c t t
= +
= + + +
Step 4
Now, we eliminate one of the variable, say x.
Operating on Second equation by operation D, we get
2 1 (3)Dx D y− + = − − − − −
Adding Eq.(1) and Eq.(3), we get
2
2
2
1
1
( 1) 1 (4)
Dx y Dx D y t
D y y t
D y t
+ − + = −+ = −+ = − − − − −
Step 5
The auxiliary equation of the differential equation found in the previous step is
2
2
(m +1) = 0
=> m = -1
=> m i= ±
Therefore, roots of the auxiliary equation are
2 3, m i m i= = −
So that the complementary function for the retained variable y is
3 4cos siny c t c tc
= +
To determine the particular solution py we use undetermined coefficients. Therefore, we assume
py At B= +
/
// 0
p
p
y A
y
=
=
Substituting in eq.4, we get
1At B t+ = −
So,
1, 1A B= = −
So, 1py t= −
Hence, the solution for the variable y is given by
pc yyy +=
Or
3 4cos sin 1y c t c t t= + + −
Step 6
Now 3c and 4c can be expressed in terms of 1c and 2c by substituting these values of x and y into the second
equation of the given system and we find, after combining the terms,
3 4 1 2
3 4 1 2
3 2
4 1
( cos sin 1) cos sin 1
sin cos 1 cos sin 1
Dy x t
D c t c t t c t c t t t
c t c t c t c t
c c
c c
= −+ + − = + + + −
− + + = + +⇒ = −
=
Hence, the solution of the given system of differential equations is
1 2( ) cos sin 1x t c t c t t= + + +
1 2( ) sin cos 1y t c t c t t= − + −
Question 2: Marks=10
Solve the given system of differential equations by systematic elimination method.
2 5
5
t
t
dx dyx e
dt dtdx dy
x edt dt
− + =
− + =
Solution:
Step 1
Writing the given differential equations of the system in the differential operator form, we get
( )2 5 (1)
( 1) 5 (2)
t
t
D x Dy e
D x Dy e
− + = − − − −
− + = − − − − −
Step 2
We eliminate the variabley from the given system. Subtracting eq.2 from eq.1, we get
( )2 5 1 5 (1)
( 4) 4 (3)
t t
t
D D x Dy Dy e e
D x e
− − + + − = − − − − −
− = − − − − − −
Step 3
The auxiliary equation of the differential equation for x is
4 0m − =
4m =
The root of the auxiliary equation is real. Therefore, the complementary function for x
41
tcx c e=
We now use the method of undetermined coefficients to find
the particular integrals px
/
=
=
tp
tp
x Ae
x Ae
Substituting in the differential equation forx , we obtain
4 4
3 4
4
3
t t t
t t
Ae Ae e
Ae e
A
− = −
− = −
=
So, 4
=3
tpx e
Hence, the solution for the variable x is given by
41
4
3
c p
t t
x x x
c e e
= +
= +
i.e. 1
4 4( )
3t tx t c e e= +
Step 4
Now, we eliminate one of the variable, say x.
Operating 1D − on equation (1), we get
( )2 2
( 1) 2 5 ( 1) ( 1)
(2 7 5) ( ) 0 (4)
tD D x D Dy D e
D D x D D y
− − + − = −
− + + − = − − − −
Now operating 2 5D − on equation (2), we get
( )2 2
2 2
2 5 ( 1) (2 5) (2 5)5
(2 7 5) (2 5 ) 10 25
(2 7 5) (2 5 ) 15 (5)
t
t t
t
D D x D Dy D e
D D x D D y e e
D D x D D y e
− − + − = −
− + + − = −
− + + − = − − − − − −
Subtracting Equation (4) from Equation (5), we get
2( 4 ) 15 (6)tD D y e− = − − − − − −
Step 5
The auxiliary equation of the differential equation found in the previous step is
2 4 0
( 4) 0
0,4
m m
m m
m
− =− =
⇒ =
Therefore, roots of the auxiliary equation are
1 20, 4m m= =
So that the complementary function for the retained variable y is
42 3
tcy c c e= +
To determine the particular solution py we use undetermined coefficients. Therefore, we assume
/
//
tp
tp
tp
y Ae
y Ae
y Ae
=
=
=
Substituting in eq.(6), we get
4 15
3 15
5
t t t
t t
Ae Ae e
Ae e
A
− = −
− = −⇒ =
So, 5 tpy e=
Hence, the solution for the variable y is given by
pc yyy +=
Or
3 4cos sin 1y c t c t t= + + −
42 3
42 3
5
( ) 5
t t
t t
y c c e e
y t c c e e
= + +
= + +
Step 6
Now 3c can be expressed in terms of 1c by substituting these values of x and y into the first equation of the given
system and we find, after combining the terms,
( )
( )1
1 1
4 42 3
4 4 43
2 5
42 5 ( ) ( 5 )
38 20
8 5 4 53 3
t
t t t t t
t t t t t t t
D x Dy e
D c e e D c c e e e
c e e c e e c e e e
− + =
− + + + + =
+ − − + + =
Equating the like terms, we have
1 3
3 1
3 4 0
3
4
c c
c c
+ =
= −
So,
42 1
3( ) 5
4t ty t c c e e= − +
Hence, the solution of the given system of differential equations is
1
4 4( )
3t tx t c e e= +
42 1
3( ) 5
4t ty t c c e e= − +
Question 3: Marks=10
Find the Eigenvalues and Eigenvectors of the given matrix.
0 4 0
1 4 0
0 0 2
A
= − − −
Solution:
Eigenvalues
The characteristic equation of the matrix A is
( )4 0
det 1 4 0 0
0 0 2
A I
λλ λ
λ
−− = − − − =
− −
Expanding the determinant by the cofactors of the third Column, we obtain
2
2
3
( 2 )(( )( 4 ) ( 1)(4)) 0
( 2 )( 4 4) 0
( 2)( 2) 0
( 2) 0
2 0
2
λ λ λ
λ λ λ
λ λ
λλλ
− − − − − − − =
− − + + =
+ + =
+ =+ == −
As the roots are repeated so,
1 2 3 2λ λ λ= = = −
Eigenvectors For 1 2λ = − , we have
( )( )
| 0
2 | 0
A I
A I
λ−
= +
2 2 1
1 1
2 4 0 0
1 2 0 0
0 0 0 0
2 4 0 0
0 0 0 0 2
0 0 0 0
1 2 0 01
0 0 0 02
0 0 0 0
C C C
C C
= − −
= ⇒ +
= ⇒
Here from second and third row, we see that
3k and 2k are arbitrary variables,
From first row, we have
1 2 3
1 2
2 0 0
2
k k k
k k
+ + == −
Choosing 3 20, 1k k= = −
1 2k =
Hence, the eigenvector corresponding to 1 2λ = − is
1
2
1
0
K
= −
Choosing 3 21, 0k k= = , we get
1 0k =
So another eigenvector corresponding to 1 2λ = − is
2
0
0
1
K
=