ii sem (csvtu) mathematics unit 2 (linear differential equation )solustions
TRANSCRIPT
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
Unit I I Differential Equations of higher order
April -May, 2006
1. Solve: 2 2 32
2 cos2x xd y
y x e e xdx
Ans: 2 2 32 cos2x xD y x e e x ----------- (1)Here Auxiliary equation is 2 2 0 2D D i
So, cos 2 sin 2hy A x B x
2 321
. . cos22
x xpP I y x e e x
D
3 22 21 1
cos2( 3) 2 ( 1) 2
x xpy e x e x
D D
32
22
1 1
cos211 2 361
11
xx
p
e
y x e xD DD D
3
2
2 2
1 1cos2
611 4 2 31
11
xx
p
ey x e x
D D D
2 32 23 22
2
6 66 2 131 ... cos2
11 11 121 1331 4 169
xx
p
D D D De D D Dy x e x
D
3
2 2
2
6 1 36 2 131 ... cos2
11 11 11 121 4 169
xx
p
e D Dy D x e x
D
3
2 2
2
6 47 2 131 ... cos2
11 11 121 4 4 169
xx
p
e D Dy D x e x
3
2 6 47 2 132 2 0 0 ... cos211 11 121 233
xx
p
e Dy x x e x
3
2 12 94 4sin2 13cos211 11 121 233
x x
p
e x ey x x x
So, Solution h py y y
3
2 12 94cos 2 sin 2 4sin2 13cos211 11 121 233
x xe x ey A x B x x x x
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
2. Solve: 3 23 23 2
12 2 10( )
d y d yx x y x
dx dx x
Ans:3 2
3 2
3 2
12 2 10
d y d yx x y x
dx dx x
------------- (1)
Let logzx e z x
So,2 3
2 3
2 3', '( ' 1) , '( ' 1)( ' 2)
dy d y d yx D x D D y x D D D y
dx dx dx where
'd
Ddz
Putting all these values in (1) we get
'( ' 1)( ' 2) 2 '( ' 1) 2 10( )z zD D D D D y e e
3 2' ' 2 10( )z zD D y e e -------(2)
Its auxiliary equation is 3 2 2 0m m
2( 1)( 2 2) 0m m m
2 41,
2m
1, 1m i
So, . . cos sin )z zhC F y Ae e B z C z
. . cos(log ) sin(log )hA
C F y x B x C xx
-------------(3)
Now,3 2
1. . 10( )
' ' 2
z zpP I y e e
D D
3 2 3 2
1 1
10 10' ' 2 ' ' 2
z z
py e eD D D D
3 2 3 2
1 110 10 (1)
1 1 2 ( ' 1) ( ' 1) 2
z zpy e e
D D
2
15 10 (1)
'( ' 4 ' 5)
z zpy e e
D D D
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
1
21 4 15 10 1 ' ' (1)5 ' 5 5
z zpy e e D D
D
1 45 2 1 ' ..... (1)
' 5
z zpy e e D
D
15 2 (1)'
z zpy e e
D
5 2z zpy e ze
2log5p
xy x
x
So, Solution h py y y
2log
cos(log ) sin(log ) 5A x
y x B x C x xx x
(Ans)
3. Solve the following simultaneous equation: 2 , 2 , 2dx dy dzy z xdt dt dt
(Ans.) The given equation is:
2dx
ydt
..(1), 2 ..........(2)dy
zdt
,
2 ..........(3)dz
xdt
Differentiating (1)w . r. t., we get
2
22 2(2 )
d x dyz
dt dt using(2)
Differentiating again w. r. t. t,, we get
3
34 4(2 )
d x dzx
dt dt
3
3 2
21 2 3
2
1 2 3 2 3
2
1 2 3 3
2
1 2
( 8) 0,
8 0 ( 2)() 2 4) 0
2, 1 3
cos( 3 )
1
2
12 cos( 3 ) 3 sin( 3 )
2
2 2cos cos( 3 ) sin sin( 3 )
3 3
cos 3
t t
t t t
t t
t t
dD x whereDdt
D or D D D
D i
x ce c e t c
dxy
dt
ce c e t c c e t c
ce c e t c t c
ce c e t
32
3c
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
From(2)..2
1 2 3 2 3
21 2 3
1
2
1 2 22 cos( 3 ) 3 sin( 3 )
2 3 3
4cos( 3 )3
t t t
t t
dyz
dt
ce c e t c c e t c
ce c e t c
Nov-Dec 2006
4. Solve the following : sin , cosdx dxy t x tdt dt
, Given that , 2, 0, , 0x y when t
Ans: Given simultaneous differential equation are
sindx
y tdt
------------- (1)
cosdy
x tdt
------------(2)
From equation (2) we get cosdy
x tdt
---------(3)
Differentiating (3) with respect to t we get2
2sin
dx d yt
dt dt ---------(4)
From (1) and (4) we get2
2sin sin
d yt y t
dt
2
2 2sin
d y
y tdt 2( 1) 2sinD y t -----------(5)
Its Auxiliary equation is 2 1 0 1,1m m
So, . . t thC F y Ae Be
And2
1. . ( 2sin )
1pP I y t
D
2
1( 2sin ) sin
1 1py t t
So, h py y y
sint ty Ae Be t -------(6)
Putting the value of y in (2) we get
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
cos cost tAe Be t x t
t tx Ae Be ---------(7)
Given that 2, 0, 0x y when t
So, equation (6) and (7) becomes 0 A B and 2 A B
1, 1A B
So, solutiont tx e e and sint ty e e t (Ans)
5. Solve the differential equation: 3 23 23 2
12 2 10( )
d y d yx x y x
dx dx x
Ans:3 2
3 2
3 2
12 2 10
d y d yx x y x
dx dx x
------------- (1)
Let logzx e z x
So,2 3
2 3
2 3', '( ' 1) , '( ' 1)( ' 2)
dy d y d yx D x D D y x D D D y
dx dx dx where '
dD
dz
Putting all these values in (1) we get
'( ' 1)( ' 2) 2 '( ' 1) 2 10( )z zD D D D D y e e
3 2' ' 2 10( )z zD D y e e -------(2)
Its auxiliary equation is3 2 2 0m m
2( 1)( 2 2) 0m m m
2 41,
2m
1, 1m i
So, . . cos sin )z zhC F y Ae e B z C z
. . cos(log ) sin(log )hA
C F y x B x C xx
-------------(3)
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
Now,3 2
1. . 10( )
' ' 2
z zpP I y e e
D D
3 2 3 2
1 110 10
' ' 2 ' ' 2
z zpy e e
D D D D
3 2 3 2
1 110 10 (1)1 1 2 ( ' 1) ( ' 1) 2
z zpy e e D D
2
15 10 (1)
'( ' 4 ' 5)
z zpy e e
D D D
1
21 4 15 10 1 ' ' (1)5 ' 5 5
z zpy e e D D
D
1 45 2 1 ' ..... (1)
' 5
z zpy e e D
D
15 2 (1)
'
z zpy e e
D
5 2z zpy e ze
2log5p
xy x
x
So, Solution h py y y
2log
cos(log ) sin(log ) 5A x
y x B x C x xx x
(Ans)
6. Using method of variation of parameters: 22
4 tan2d y
y xdx
Ans: Homogeneous equation is 4 0y y
Its characteristics equation is2 4 0 2i
Hence, Homogeneous solution is cos2 sin2hy A x B x
1 2cos2 , sin2y x y x
1 1 2 2
2 2
cos2 sin22cos 2 2sin 2 2
2sin2 2cos2
y y x xW x x
x xy y
rdxy
y
W
rdxyyyp
.. 12
21
sin2 .tan2 cos2 tan2cos2 sin2
2 2p
x xdx x xdxy x x
cos2 sin2(sec2 cos2 ) sin2 .
2 2p
x xy x x dx xdx
cos2 log(sec2 tan2 ) sin2 sin2 cos2
2 2 2 2 2p
x x x x x xy
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
cos2 .log(sec2 tan2 )
4p
x x xy
So, Solutioncos2 .log(sec2 tan2 )
cos2 sin24
h p
x x xy y y A x B x
(Ans)
April -May, 20077. Solve the differential equation 2
22 sinx
d y dyy xe x
dx dx .
Ans:2
22 sinx
d y dyy xe x
dx dx
Its auxiliary equation is 2 22 1 0 ( 1) 0 1,1m m m m
So, . . x xhC F y Ae Bxe
Now,2
1. . sin
( 1)
xpP I y xe x
D
2 21 1sin sin
( 1 1)
x xpy e x x e x x
D D
1sinxpy e x xdx
D
1
cos sinxpy e x x xD
cos sinxpy e x x x dx
sin cos cosxpy e x x x x
sin 2cosx
py e x x x So, general solution h py y y
sin 2cosx x xy Ae Bxe e x x x (Ans)
8. Solve by method of variation of parameters of : - 22
sind y
y x xdx
.
Ans: Homogeneous equation is 0y y
Its characteristics equation is2 1 0
2i
Hence, Homogeneous solution is cos sinhy A x B x
1 2cos2 , sin2y x y x
1 1 2 2
2 2
cos sincos sin 1
sin cos
y y x xW x x
x xy y
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
Wrdxy
yW
rdxyyyp
.. 12
21
sin . sin cos . sincos sin
1 1p
xx xdx xx xdxy x x
2
cos sin . sin sin cospy x x xdx x x x xdx cos sin(1 cos2 ) sin2 .
2 2p
x xy x x dx x xdx
2cos sin2 cos2 sin cos2 sin2
2 2 2 4 2 2 4p
x x x x x x x x xy
2 cos cos sin2 cos cos2 sin cos2 sin sin2
4 4 8 4 8p
x x x x x x x x x x x xy
2 cos sin2 sin cos2cos cos cos2 sin sin24 4 8
p
x x x x xx x x x x xy
2 cos sin cos
4 4 8p
x x x x xy
So, Solution2 cos sin cos
cos sin4 4 8
h p
x x x x xy y y A x B x (Ans)
9. Solve the differential equation 22 22
3 4 (1 )d y dy
x x y xdx dx
.
Ans:2
2 2
23 4 (1 )
d y dyx x y x
dx dx --------- (1)
Let logzx e z x and
2
22' , '( ' 1)
dy d yx D y x D D ydx dx where ' dD dz
Putting all the values in (1) we get
2'( ' 1) 3 ' 4 (1 )zD D D y e 2 2' 4 ' 4 1 2 z zD D y e e ----------- (2)Its auxiliary equation is 2 24 4 0 ( 2) 0 2,2m m m m
So, 2 2 2 2. . logz zhC F y Ae BZe Ax Bx x
Now,2
2
1. . (1 2 )
( ' 2)
z zpP I y e e
D
0 2
2 2 2
1 1 1
2( ' 2) ( ' 2) ( ' 2)
z z z
py e e eD D D
2
2 2 2
1 1 1.1 2 1
(0 2) (1 2) ( ' 2 2)
z zpy e e
D
2
2
1 12 1
4 '
z zpy e e
D 2
12 1
4
z zpy e e dzdz
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
2 2 2 21 1 (log )2 2
4 2 4 2
zz
p
z e x xy e x
So, general solution is2 2
2 2 1 (log )log 24 2
h p
x xy y y Ax Bx x x (Ans).
Nov-Dec 2007
10.Solve: 22
3 4 0d y dy
ydx dx
Sol. Its symbolic form is:2
2
3 4 0
( 3 4) 0
D y Dy y
D D y
Its auxiliary equation is :2 3 4 0D D
4, 1D
Hence C.F.=4
1 2x xy ce c e
(Ans.)
11.Solve: 22
5 6 sin3d y dy
y xdx dx
Sol. It s symbolic form is 2 5 6 tan2D D y x
Its auxiliary equation is: 2 5 6 0D D ( 2)( 3) 0
2,3
D D
D
Hence C.F.= y=2 3
1 2x xce c e
And P.I.=
2 2
2 2
1 1sin3 sin3
( 5 6) 3 5 6
1 1 5 3sin3 [ ][ ]sin3
5 3 5 3 5 3
5 3 5 3sin3 sin3
25 9 25( 3 ) 9
1 15cos3 3sin3(5 3)sin3234 234
5cos3 sin3
78
x xD D D
Dx x
D D D
D Dx x
D
x xD x
x x
Hence complete solution is:y=C.F.+P.I.
2 31 2
1(5cos3 sin3 )
78
x xy ce c e x x
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
12.Solve by method variation of parameters:2
24 tan2
d yy x
dx
Ans: Homogeneous equation is 4 0y y
Its characteristics equation is2 4 0 2i
Hence, Homogeneous solution is cos2 sin2hy A x B x
1 2cos2 , sin2y x y x
1 1 2 2
2 2
cos2 sin22cos 2 2sin 2 2
2sin2 2cos2
y y x xW x x
x xy y
rdxy
yW
rdxyyyp
.. 12
21
sin2 .tan2 cos2 tan2cos2 sin22 2
p x xdx x xdxy x x
cos2 sin2(sec2 cos2 ) sin2 .
2 2p
x xy x x dx xdx
cos2 log(sec2 tan2 ) sin2 sin2 cos2
2 2 2 2 2p
x x x x x xy
cos2 .log(sec2 tan2 )
4p
x x xy
So, Solutioncos2 .log(sec2 tan2 )
cos2 sin2
4
h p
x x xy y y A x B x
(Ans)
13.Solve the simultaneously equation:2 5 , 4 3t
dx dxx y e x y t
dt dt
Sol. The given equation can be expressed as:
( 5) 3 ............(1)
2 ( 5) ......(2)tD x y t
x D y e
To eliminate y , operating equation (1) by (D+5) and equation (2) by 3 then subtracting , we get
2
( 5)( 4) 6 ( 5) 3
( 9 14) 1 5 3
t
t
D D x x D t e
D D x t e
..(3)
The root of auxiliary equation of the equation corresponding homogeneous equation
2( 9 14) 0D D x
of the equation (3) is given by
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
2( 9 14) 0
, 2, 7
D D
or D
Hence the complementary function of equation (3) is:
2 71 2. .
t tC F ce c e
The particular integral of equation (3) is
2
2 2
2
1. . (1 5 3 )
( 9 14)
1 1(1 5 ) 3
( 9 14) ( 9 14)
1 9 31 ............. (1 5 )
14 14 14 1 9 14
1 91 5 .5
14 14 8
1 315
14 14 8
t
t
t
t
t
P I t eD D
t eD D D D
D eD t
et
et
Hence the general solution of equation (3) is:
2 7
1 2
2 71 2
. . . .
1 315
14 14 8
5, 2 7
14 8
tt t
tt t
x C F P I
ex ce c e t
dx eNow ce c e
dt
.(4)
Substituting the above values of x and dx/dt in equation (1), we get;
2 7 2 71 2 1 2
2 71 2
5 20 1243 2 7 4 4
14 8 14 196 2
1 3 5 272 3 .......................(5)
3 7 8 98
t tt t t t
t t t
e ey ce c e ce c e t t
y ce c e t e
Since the degree of D in the determinant4 3
,22 5
Dis
D
it follows that the number of
independent constant in general solution must be two. Hence (4) and (5) together constitute the
general solution of the given system.
April -May, 2008
14.State Cauchys Linear equation.Ans: - A Differential equation of the form
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
Xyad
ydxa
d
ydxa
d
ydx nn
nn
n
nn
n
nn
........
2
22
21
11
1 is known as Cauchys
Linear differential equation.
15.Solve2
2
2 4 sin2xd y dy
y e xdx dx .
(ANS). Its symbolic form is :2 2( 4 1) sin2xD D y e x
Its auxiliary equation is:2( 4 1) 0D D
16.Solveby method of variation of parameters of : - 22
4 4tan2 .d y
y xdx
Ans: Homogeneous equation is 4 0y y
Its characteristics equation is2
4 0 2i Hence, Homogeneous solution is cos2 sin2hy A x B x
1 2cos2 , sin2y x y x
1 1 2 2
2 2
cos2 sin22cos 2 2sin 2 2
2sin2 2cos2
y y x xW x x
x xy y
rdxy
yW
rdxyyyp
.. 12
21
sin2 .tan2 cos2 tan2cos2 sin2
2 2p
x xdx x xdxy x x
cos2 sin2(sec2 cos2 ) sin2 .
2 2p
x xy x x dx xdx
cos2 log(sec2 tan2 ) sin2 sin2 cos2
2 2 2 2 2p
x x x x x xy
cos2 .log(sec2 tan2 )
4p
x x xy
So, Solutioncos2 .log(sec2 tan2 )
cos2 sin24
h p
x x xy y y A x B x
(Ans)
17.Solve 2 0dydx x ydt dt
and 5 3 0dydx
x ydt dt
(Ans). The given equation are:
2 0.....................(1)dydx
x ydt dt
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
5 3 0.....................(2)dydx
x ydt dt
Subtracting equation (2) from equation (1), we get
3 2 0...............(3)dx
x ydt
Writing equation (2)and (3) symbolically (i.e;D=d/dt),we have5 ( 3) 0..........(4)
( 3) 2 0..........(5)
x D y
D x y
Operating equation (5)by D+3 ,we get2( 9) 2( 3) 0.........(6)D x D y
Now, multiplying equation(4) by 2 and adding in equation (6),we get2( 1) 0D x .(7)
This is the linear equation in x with constant coefficient s for which the auxiliary equation is2( 1) 0D
D i
Hence the general solution of linear equation is:0
1 2
1 2
( cos sin )
( cos sin )..........................(8)
tx e c t c t
x c t c t
From equation (8)
1 2
1 2
sin cos
sin cos
dxc t c t
dt
Dx c t c t
Substituting these values in equation(5) we get
1 2 1 2
2 1 1 2
1 1( 3 ) ( sin cos ) 3( cos sin )
2 2
1 1( 3 )cos ( 3 )sin
2 2
y Dx x c t c t c t c t
y c c t c c t
Nov-Dec 200818.Define the linear differential equation.Ans. A differential equation in which the dependent variable & its derivatives occur only in the first
degree & are not multiplied together is known as Linear Differential equation.
General form of Linear Differential Equation of nth order is:
XyPd
ydP
d
ydP
d
ydnn
n
n
n
n
n
........2
2
21
1
1 , where XPPP n,,........,, 21 are functions of x.
19.Solve the differential equation:
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
2
23 2
xd y dy ey edx dx
Ans: - Its symbolic form is 2 3 2xeD D y e .
Its Auxiliary equation is 2 3 2 0D D 1, 2D
Hence, Complementary function is 2x xcy Ae Be
Now, 2 1 1 1 13 2 ( 2)( 1) ( 2) ( 1)x x xe e ePI e e e
D D D D D D
2 2 21 1
2 2
x x x x x x x xx x x xe e e ePI e e e dx e e e e e e dx e e e dxD D
2x xePI e e
So, Solution 2 2x x xc pxey y y Ae Be e e (Ans)
20.Solve by method variation of parameters:2 3
2 26 9
xd y dy ey
dx dx x
Ans: -2
3
2
2
96e
yd
dy
d
yd x
Its Homogeneous equation is 0962
2
yd
dy
d
yd
Its symbolic form is 0)96( 2 yDD
Its characteristics equation is 0)96( 2 DD
3,3 D
Hence, Homogeneous solution is 213)( ByAyeBxAy xh
xx xeyey 323
1 ,
xxxx
xxx
xx
exexeexeee
xee
yy
yyW 6666
333
33
21
2133
33''
Wrdxy
yW
rdxyyyp
.. 12
21
dxee
xedxexe
ey
x
x
xx
x
x
xx
p 2
3
6
33
2
3
6
33
..
dxxedxeyxx
p 2
33 11
xxexey xe
xp
1log 33 xe
xp exey
33 log
So, Solutionx
exxx
ph exeBxeAeyyy3333 log (Ans).
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
21.Solve the simultaneously equation:5 2 , 2 0
dx dyx y t x y
dt dt being x=y=0 when t=0.
Ans: - +5 2 = , +2 + = 0Its symbolic form is ( +5) 2 = ------------------(1)
2 +( +1) = 0-------------------(2)Now, eq(1) X (D+1) +eq(2) X 2 we get
[( +1)( +5) +4] = ( +1) +0 ( +6 +9) = 1+, which is linear differential equation.Its auxiliary equation is ( +6 +9) = 0 =3,3So,..= ( +)
Now, ..= () (1+ ) = 1+
(1+) = 1 2 +3
+ . (1+)
..= 1+ = + = + So, = ( +) + + ---------------------------------(3)Now, from (1) 2 = ( +5) = +5 2 = 5( +) + + 3( +) + + = (2 +) + + ---------------(4)Given that at = 0, = = 0, so equation (3) and (4) becomes
+ = 0 = and
(2 +) +
= 0 =
2 ==
+
=
=
So, Solution is
= ( ) + + and = + + +
= {(1+6) (3 +1)} and = {(4+6) +(6 4)}April -May, 2009
22.Explain briefly the method of variation of parameter.Ans: Let us solve
2
2
d y dyP Qy X
dx dx by variation of parameter method.
Let its complementary function is 1 1 2 2cy c y c y
Then find out wronskian1 2
1 2
1 2
( , )' '
y yW y y
y y
Now, particular integral 2 11 2
1 2 1 2
. .
( , ) ( , )p
y X y Xy y dx y dx
W y y W y y
Then general solution is c py y y (Ans).
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
23.Solve the differential equation 2 2 32 cos2x xD y x e e x .Ans: 2 2 32 cos2x xD y x e e x ----------- (1)
Here Auxiliary equation is 2 2 0 2D D i
So, cos 2 sin 2hy A x B x
2 321
. . cos22
x xpP I y x e e x
D
3 22 21 1
cos2( 3) 2 ( 1) 2
x xpy e x e x
D D
3
2
2 2
1 1cos2
611 2 31
11
xx
p
ey x e x
D D D D
3
2
2 2
1 1cos2
611 4 2 3111
xx
p
ey x e x
D D D
2 32 23 2
2
2
6 66 2 131 ... cos2
11 11 121 1331 4 169
xx
p
D D D De D D Dy x e x
D
3
2 2
2
6 1 36 2 131 ... cos2
11 11 11 121 4 169
xx
p
e D Dy D x e x
D
32 2
2
6 47 2 131 ... cos2
11 11 121 4 4 169
xx
p
e D Dy D x e x
3
2 6 47 2 132 2 0 0 ... cos211 11 121 233
xx
p
e Dy x x e x
3
2 12 94 4sin2 13cos211 11 121 233
x x
p
e x ey x x x
So, Solution h py y y
3
2 12 94cos 2 sin 2 4sin2 13cos211 11 121 233
x xe x ey A x B x x x x
24.Solve the equation : - 22 2(1 ) (1 ) sin 2log(1 )d y dyx x y xdx dx .Ans:
22
2(1 ) (1 ) sin 2log(1 )
d y dyx x y x
dx dx (1) is a Legendres equation.
Now let us put 1 tx e we get2
2
2(1 ) ( 1) , (1 )
d y dyx D D y x Dy
dx dx where
dD
dt
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
So, equation (1) becomes ( 1) sin2D D y Dy y t 2( 1) sin2D y t which is linear differential equation with constant coefficients.
Its auxiliary equation is 2( 1) 0D D i
So, 1 2. . cos sinC F c t c t ..(2)
Now,2 2
1 sin2 sin2. . sin2
1 2 1 3
t tP I t
D
So, solution 1 2sin2
. . . . cos sin3
ty C F P I c t c t (Ans)
25.Solve the simultaneous equation: -2 , 2t t
dx dyy e x e
dt dt
.
Ans: 2 , 2t tdx dy
y e x edt dt
2 ...............(1) 2 ..............(2)t tDx y e Dy x e whered
Ddt
(1) 2 (2) D we get22 4 2 2 t tDx y D y Dx e e
2( 4) 2 t tD y e e ..(3)
HereIts auxiliary equation is 2( 4) 0 2D D i
So, 1 2. . cos2 sin2C F c t c t ..(4)
Now, 2 2 21 1 1
. . 2 2
1 1 1
t t t tP I e e e e
D D D
2 2
1 1. . 2
1 1 ( 1) 1 2
tt t t eP I e e e
So, solution 1 2. . . . cos2 sin22
tt ey C F P I c t c t e
..(5)
Now putting the value of y in (2) we get
2 tDy x e
1 22 sin2 2 cos2 22
tt tec t c t e x e
1 22 2 sin2 2 cos22
tt ex c t c t e
1 2sin2 cos22 4
t te ex c t c t
(6)
So, solution 1 2 1 2sin2 cos2 , cos2 sin22 4 2
t t tte e ex c t c t y c t c t e
(Ans)
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
Nov-Dec 2009
26.Solve the equation: 2( 2) 0D D y .Ans: - 2( 2) 0D D y
Its auxiliary equation is 2( 2) 0 1, 2D D D
So, solution is 2x xy Ae Be (Ans).
27.Solve the following differential equation:2
2
2log
d y dyx x y x
dx dx .
Ans: - + = --------------------(1)
Given equation is Cauchys Homogeneous Linear differential equation.
Put
=
=
, if
=
,
=
,
=
(
1)
So, equation (1) is ( 1) + = ( 2 +1) = Which is linear differential equation with constant coefficients.
So, its Auxiliary equation is 2 + 1= 0 = 1,1So,..= + ------------------------(2)
Now, ..= () = (1 ) = (1+2 +3 + .)
..= +2So, solution is =..+..= + + +2 = + + +2 (Ans).
OR
28.Solve the simultaneous differential equation:
5 2 , 2 0dx dy
x y t x ydt dt
, given that 0x y when 0t .
Ans: - +5 2 = , +2 + = 0
Its symbolic form is ( +5) 2 = ------------------(1)2 +( +1) = 0-------------------(2)
Now, eq(1) X (D+1) +eq(2) X 2 we get
[( +1)( +5) +4] = ( +1) +0 ( +6 +9) = 1+, which is linear differential equation.Its auxiliary equation is
(
+6
+9) = 0
=
3,
3So, ..= ( +)
Now, ..= () (1+ ) = 1+
(1+) = 1 2 +3
+ . (1+)
..= 1+ = + = + So, = ( +) + + ---------------------------------(3)Now, from (1) 2 = ( +5) = +5
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
2 = 5( +) + + 3( +) + + = (2 +) + + ---------------(4)Given that at = 0, = = 0, so equation (3) and (4) becomes
+
= 0
=
and
(2 +) + = 0 = 2 == + = = So, Solution is
= ( ) + + and = + + +
= {(1+6) (3 +1)} and = {(4+6) +(6 4)}29.Solve the following differential equation: 2
24 sinh
d yy x x
dx .
Ans: - 4 =
4 = Its Symbolic form is ( 4) =
Its Auxiliary equation is ( 4) = 0 = 2So, ..= +Now, ..=
=
.
.=
()
()
=
..=
+
..=
1+ + +
1+
+
..=
+ + = (
) (
)
..= [3 2]So, solution is =..+..= + + [3 2] (Ans).
April -May, 201030.Define Cauchy and Legendre L inear differential equation.
Ans: - A Differential equation of the form
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
Xyad
ydxa
d
ydxa
d
ydx nn
nn
n
nn
n
nn
........
2
22
21
11
1 is known as Cauchys
Linear differential equation.
A Differential equation of the form
Xyad
ydbaxa
d
ydbaxa
d
ydbax nn
nn
n
nn
n
nn
........)()()(
2
22
21
11
1 is
known as Legendres Linear differential equation.
31.Solve 33
3 2xd y dy ey e
dx dx .
Ans: - Its symbolic form is 2 3 2xeD D y e .
Its Auxiliary equation is2 3 2 0D D 1, 2D
Hence, Complementary function is2x x
cy Ae Be
Now, 2 1 1 1 13 2 ( 2)( 1) ( 2) ( 1)x x xe e ePI e e e
D D D D D D
2 2 21 1
2 2
x x x xx x e x e x x x e x x ePI e e e dx e e e e e e dx e e e dxD D
2 xx ePI e e
So, Solution 2 2xx x x e
c py y y Ae Be e e (Ans)
32.Solve, by the method of variation of parameters: xeDD x log)12( 2 .Ans: - Its Auxiliary equation is
2 2 1 0D D 1, 1D
Hence, Complementary function is x
cy A Bx e
1 2,x xy e y xe
1 2 2 2 2 2
1 2
x xx x x x
x x x
y y e xeW e xe xe e
e e xey y
WXdxy
yW
Xdxyyyp
.. 12
21
2 2
. log . logx x x xx xp x x
xe e xdx e e xdxy e xe
e e
log logx xpy e x xdx xe xdx
2 2
log log2 4
x xp
x xy e x xe x x x
2 2 23
log 2log 32 4 4
x x x
p
x x xy e x e e x
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
So, Solution 2
( ) 2log 34
x x
c p
xy y y A Bx e e x (Ans)
33.Solve the simultaneous equation: -0 y
d
dxt , 0 x
dt
dyt , given that 0)1(,1)1( yx .
Ans: - 0 yd
dxt ------------------- (1), 0 x
d
dyt ------------- (2)
Differentiating (1) w.e.t. t we get2
20
dx d x dyt
dt dt dt , by multiplying t both side we get
22
20
dx d x dyt t tdt dt dt
-------------------- (3)
By putting (2) in (3) we get2 2
2 2
2 20
d x dx d xt t t x
dt dt dt -----(4) which is Cauchy
Linear Differential Equation.
Put logzt e z t and2
2
2( 1) ,
d x dxt D D x t Dx
dt dt where
dD
dz
So, eqn (4) becomes 2( 1) 0 ( 1) 0D D x Dx x D x
Its Auxiliary equation is 2( 1) 0 1,1D D
So, z zB
x Ae Be Att
-------(5)
Putting in (1) we get2
dx B By t t A At
dt t t
----------- (6)
So, Solution is Bx Att
and By Att
(Ans).
Given x(1) =1 and y(-1) =0, so2
1,
2
10,1 BAABBA
Hence solution is
t
tyBA
ttx
1
2
1,
1
2
1(Ans).
Nov-Dec 201034.Write the formula for P.I. for the method of variation of parameters.
Ans: - Let equation is rybaDD )( 2 , then P.I is given by
Wrdxy
yW
rdxyyyp
.. 12
21 where
'' 21
21
yy
yyW .
35.Solve )2sin(8)2( 222 xxeyD x .Ans: - )2sin(8)2( 222 xxeyD x --------------------- (1)
Its characteristics equation is 2,20)2( 2 DD
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
So, xxh BxeAeyFC22.. ----------------------- (2)
Now, )2sin(8)2(
1.. 22
2xxe
DyIP xp
2
22
2
2 )2(
1
2sin)2(
1
)2(
1
8 xDxDeDy
x
p
222
2
21
1
4
12sin
44
1
)2(2
18 x
Dx
DDe
Dxy xp
2
2
2
22 ..........4
32
214
12sin
442
1
2
18 x
DDx
Dexy xp
..........004
624
1
2
2cos
4
1
2822
2
xxx
ex
yx
p
3422cos4 222 xxxexy xp
So, Solution 3422cos4 22222 xxxexBxeAeyyy xxxph (Ans).
36.Solve, by the method of variation of parameters:2
3
2
2
96e
yd
dy
d
yd x .
Ans: -2
3
2
2
96e
yd
dy
d
yd x
Its Homogeneous equation is 0962
2
yd
dy
d
yd
Its symbolic form is 0)96( 2 yDD
Its characteristics equation is 0)96( 2 DD
3,3 D
Hence, Homogeneous solution is 213)( ByAyeBxAy xh
xx xeyey 323
1 ,
xxxx
xxx
xx
exexeexeee
xee
yy
yyW 6666
333
33
21
2133
33''
Wrdxy
yW
rdxyyyp
.. 12
21
dxee
xedxexe
eyx
x
xx
x
x
xx
p 2
3
6
33
2
3
6
33 ..
dxxedxeyxx
p 2
33 11
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
xxexey xe
xp
1log 33 xe
xp exey
33 log
So, Solutionx
exxx
ph exeBxeAeyyy3333 log (Ans).
37.Solve: xyd
dy
xd
yd
x elog2
22
.
Ans: - xyd
dyx
d
ydx elog2
22 --------- (1)
Let logzx e z x and2
2
2' , '( ' 1)
dy d yx D y x D D y
dx dx where '
dD
dz
Putting all the values in (1) we get
zyDDzyDDD ]12[1)1( 2 ----------- (2)Its auxiliary equation is 1,10122 mmm
So, xBxAxBzeAeyFC ezz
h log..
Now, zD
yIP p 2)1(
1..
zDDzDyp .........)321()1(22
xzzy ep log22........0002
So, general solution is xxBxAxyyy eeph log2log (Ans).
April -May, 2011
38.Find the particular integral of Solve
3
3. 4 sin2
d y dy
xdx dx
Ans.
3
3.4 sin2
d y dyx
dx dx
P.I will be3
1sin2
4PI x
D D
3 21 1
sin2 sin24 4
x xD D D D
3 2
1 1sin2 sin2
4 2 4
x x
D D D
31 1
sin2 sin24 0
x xD D D
By differentiating3 4D D as
1, 0in f D cannot bef D
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
3 21 1
sin2 sin24 3 4
x x xD D D
3 21 1
sin2 sin24 3( 2 ) 4
x x xD D
31 sin2 sin24 8
xx xD D
3
1 sin2sin2
4 8
x xx
D D
P.I of
3
3.4 sin2
d y dyx
dx dx will be
sin2
8
x x
39.Solve 22
3 2xd y dy ex y e
dx dx
Ans: - Its symbolic form is 2
3 2
xe
D D y e .Its Auxiliary equation is
2 3 2 0D D 1, 2D
Hence, Complementary function is2x x
cy Ae Be
Now, 21 1 1 1
3 2 ( 2)( 1) ( 2) ( 1)
x x xe e ePI e e eD D D D D D
2 2 21 1
2 2
x x x xx x e x e x x x e x x ePI e e e dx e e e e e e dx e e e dxD D
2 xx ePI e e
So, Solution2 2x x
c p
xx ey y y Ae Be e e
(Ans)
40.Solve the differential equation: 222
5 4 .log .d y dy
x x y x xdx dx
(Ans). Let
zx e andd
Ddz
. Then2
2
2, ( 1)
dy d yx Dy x D D y
dx dx
Thus the given differential equation reduces to the following fotm:
2
( 1) 5 4
( 4 4)
z
z
D D D y ze
D D y ze
Which is the linear differential equation with constant coefficient s, for which the auxiliaryequation is:
2 4 4 0D D 2, 2D
C.F= 2 21 2 1 2 logzc c z e c c x x
P.I Will be
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
1
( )
zzef D
i.e
2
1.
( 4 4)
zP I zeD D
2 2 21. ( 4 4) ( 1) 4( 1) 4 ( 6 9)
z z
z e eP I ze z zD D D D D D
26. (1 )
9 9 9
ze D DP I z
6.
9 9
zeP I z
2. log
9 3
xP I x
Complete Ans is C.F +P.I
21 2 logc c x x + 2log
9 3x x
41.Solve sin , cos ,dx dyy t x tdt dt
Given that x=2 and y=0 when t=0.
Ans: Given simultaneous differential equation are
sindx
y tdt
-------------(1)
cosdy
x tdt
------------(2)
From equation (2) we get cosdyx tdt
---------(3)
Differentiating (3) with respect to t we get
2
2sin
dx d yt
dt dt
---------(4)
From (1) and (4) we get
2
2sin sin
d yt y t
dt
2
2
2sind y
y tdt
2( 1) 2sinD y t -----------(5)
Its Auxiliary equation is2 1 0 1,1m m
So,. . t thC F y Ae Be
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
And2
1. . ( 2sin )
1pP I y t
D
2
1( 2sin ) sin
1 1py t t
So, h py y y
sint ty Ae Be t -------(6)
Putting the value of y in (2) we get
cos cost tAe Be t x t t tx Ae Be ---------(7)
Given that2, 0, 0x y whent
So, equation (6) and (7) becomes 0 A B and 2 A B 1, 1A B
So, solution t tx e e and sint ty e e t (Ans)
Nov-Dec 2011
42. Explain Cauchys homogeneous linear differential equation.Ans: - A Differential equation of the form
Xyad
ydxa
d
ydxa
d
ydx nn
nn
n
nn
n
nn
........
2
22
21
11
1 is known as Cauchys
Linear differential equation.
43.Solve the differential equation2
22 sinxd y dy y xe x
dx dx .
Ans:2
22 sinx
d y dyy xe x
dx dx
Its auxiliary equation is 2 22 1 0 ( 1) 0 1,1m m m m
So, . . x xhC F y Ae Bxe
Now,2
1. . sin
( 1)
xpP I y xe x
D
2 2
1 1sin sin
( 1 1)
x x
p
y e x x e x xD D
1sinxpy e x xdx
D
1
cos sinxpy e x x xD
cos sinxpy e x x x dx
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
sin cos cosxpy e x x x x
sin 2cosxpy e x x x
So, general solution h py y y
sin 2cosx x xy Ae Bxe e x x x (Ans)
44.Solve the differential equation 222
log .sin(log ) d y dy
x y x xdx dx
zx e andd
Ddz
. Then2
2
2, ( 1)
dy d yx Dy x D D y
dx dx
Thus the given differential equation reduces to the following fotm:
2
( 1) 1 (sin )
( 1) sin
D D D y z z
D y z z
Which is the linear differential equation with constant coefficient s, for which the auxiliary
equation is:2 1 0D
,D i i
C.F= 1 2 1 2cos sin cos log sin logc z c z c x c x P.I Will be
1sin
( )z z
f Di.e
2 2
1 1
. sin .( 1) ( 1)
iz
P I z z I P zeD D
22
1. sin .
( 1) 1)
izeP I z z I P z
D D i
21
. sin .( 1) 2
izeP I z z I P z
D D i
2
1 1 1. sin .
( 1) 2
1 2
izP I z z I Pe zDD i
i
1
2
1 1. sin . 1
( 1) 2 2
DizP I z z I Pe zD i i
2
1 1. sin . 1
( 1) 2 2
DizP I z z I Pe zD i i
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2
1 1 1. sin .
( 1) 2 2
izP I z z I Pe zD i i
2
1. sin .
( 1) 2 2
i iizP I z z I Pe zD
21 1. sin .
( 1) 2 4iizP I z z I Pe z
D
21 1
. sin . cos sin( 1) 2 4
iP I z z I P z i z z
D
2
1 cos sin sin. sin . cos
( 1) 2 4 2 4
i z z z i zP I z z I P z z
D
2
1 cos sin sin 1. sin . cos
( 1) 4 2 4 2
z z z zP I z z I P i z z
D
2
1 sin 1
. sin cos( 1) 4 2
z
P I z z z zD
2
1 sinlog 1. sin log coslog
( 1) 4 2
xP I z z x x
D
Ans.= C.F+P.I
1 2cos log sin logc x c x +sinlog 1
log coslog4 2
xx x
45.Solve the following simultaneous equation 2dx ydt
, 2dy
zdt ,
2 .dz
xdt
(Ans.) The given equation is:
2dx ydt
..(1), 2 ..........(2)dy zdt
,
2 ..........(3)dz xdt
Differentiating (1)w . r. t., we get
2
22 2(2 )
d x dyz
dt dt using(2)
Differentiating again w. r. t. t,, we get
3
34 4(2 )
d x dzx
dt dt
3
3 2
2
1 2 3
( 8) 0,
8 0 ( 2)() 2 4) 0
2, 1 3
cos( 3 )
1
2
t t
dD x whereD
dt
D or D D D
D i
x ce c e t c
dxy
dt
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7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions
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epared by Mrityunjoy Dutta
2
1 2 3 2 3
2
1 2 3 3
21 2 3
12 cos( 3 ) 3 sin( 3 )
2
2 2cos cos( 3 ) sin sin( 3 )
3 3
2cos 33
t t t
t t
t t
ce c e t c c e t c
ce c e t c t c
ce c e t c
From(2)..2
1 2 3 2 3
21 2 3
1
2
1 2 22 cos( 3 ) 3 sin( 3 )
2 3 3
4cos( 3 )
3
t t t
t t
dyz
dt
ce c e t c c e t c
ce c e t c
April -May, 2012
46.Solve3
3 0d y ydx
Ans: - Its symbolic form is Solve 013 yD .Its Auxiliary equation is 0)1)(1(01 23 DDDD
2
31,
2
31,1
iiD
So, solution 23
23221 sincos..x
xxx eCCeCyFC (Ans).
47.Solve 222
5 4 logd y dy
x x y x xdx dx
(Ans). Let
zx e andd
Ddz
. Then2
2
2, ( 1)
dy d yx Dy x D D y
dx dx
Thus the given differential equation reduces to the following fotm:
2
( 1) 5 4
( 4 4)
z
z
D D D y ze
D D y ze
Which is the linear differential equation with constant coefficient s, for which the auxiliary
equation is:2 4 4 0D D
2, 2D
C.F= 2 21 2 1 2 logzc c z e c c x x
P.I Will be
1
( )
zzef D
i.e
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2
1.
( 4 4)
zP I zeD D
2 2 2
1.
( 4 4) ( 1) 4( 1) 4 ( 6 9)
z zz e eP I ze z z
D D D D D D
26. (1 )9 9 9
ze D DP I z
6.
9 9
zeP I z
2. log
9 3
xP I x
Complete Ans is C.F +P.I
21 2 logc c x x +
2log
9 3
xx
48.Solve :
3 2
3. 22 sin2x
d y d y dye x
dx dx dx
Sol. Its symbolic form is xeyDDD x 2sin2 23 Its Auxiliary equation is 0)12(02 223 DDDDDD
1,1,0 D
So, xxc exCCeCyFC 32
01..
xc exCCCy 321 ---------------------(1)
Now, xeDDD
yIP xp 2sin2
1..
23
xDDD
eDDD
y xp 2sin2
1
2
12323
xDDDD
eDD
xy xp 2sin2.
1
13
1222
xDD
eD
xy xp 2sin)4(2)4.(
1
46
12
xD
exy
x
p 2sin83
1
4)1(62
xD
Dexy
x
p 2sin69
83
2 2
2
xDex
yx
p 2sin64)4(9
83
2
2
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10
2sin82cos6
2
2 xxexy
x
p
-------------------(2)
So, solution 10
2sin82cos6
2
2
321
xxexexCCCyyy
xx
pc
(Ans).
49.Solve the equations simultaneously5 2 , 2 0, 0
dx dyx t t x y x y
dt dt when t=0
Ans: - +5 2 = , +2 + = 0
Its symbolic form is ( +5) 2 = ------------------(1)2 +( +1) = 0-------------------(2)
Now, eq(1) X (D+1) +eq(2) X 2 we get
[( +1)( +5) +4] = ( +1) +0 ( +6 +9) = 1+, which is linear differential equation.Its auxiliary equation is ( +6 +9) = 0 =3,3So, ..= ( +)
Now, ..= () (1+ ) = 1+
(1+) = 1 2 +3
+ . (1+)
..= 1+ = + = + So, = ( +) + + ---------------------------------(3)Now, from (1) 2 = ( +5) = +5 2 = 5( +) + + 3( +) + + = (2 +) + + ---------------(4)Given that at = 0, = = 0, so equation (3) and (4) becomes + = 0 = and
(2 +) + = 0 = 2 == + = = So, Solution is
= ( ) + + and = + + +
= {(1+6) (3 +1)} and = {(4+6) +(6 4)}