unit-i ordinary differential equations - qualify gate exam...
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UNIT-I
ORDINARY DIFFERENTIAL EQUATIONS
Higher order differential equations with constant coefficients – Method of variation of
parameters – Cauchy’s and Legendre’s linear equations – Simultaneous first order
linear equations with constant coefficients.
The study of a differential equation in applied mathematics consists of three phases.
(i) Formation of differential equation from the given physical situation, called
modeling.
(ii) Solutions of this differential equation, evaluating the arbitrary constants
from the given conditions, and
(iii) Physical interpretation of the solution.
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH
CONSTANT COEFFICIENTS.
General form of a linear differential equation of the nth order with constant
coefficients is
XyKdx
ydK
dx
ydK
dx
ydnn
n
n
n
n
n
=++++ −
−
−
−
..............2
2
21
1
1 ………….. (1)
Where KnKK ...............,.........21 , are constants.
The symbol D stands for the operation of differential
(i.e.,) Dy = ,dx
dy similarly D ...,,
3
33
2
22 etc
dx
ydyD
dx
dy
y
==
The equation (1) above can be written in the symbolic form
(D XyKDK n
nn =+++ − )..........1
1 i.e., f(D)y = X
Where f (D) = D n
nn KDK +++ − ...........1
1
Note
1. ∫= XdxXD
1
2. dxXeeXaD
axax
∫−=
−1
3. dxXeeXaD
axax
∫−=
+1
(i) The general form of the differential equation of second order is
2
…………………(1)
Where P and Q are constants and R is a function of x or constant.
(ii)Differential operators:
The symbol D stands for the operation of differential
(i.e.,) Dy = ,dx
dy D
2
22
dx
ydy =
D
1Stands for the operation of integration
2
1
DStands for the operation of integration twice.
(1) can be written in the operator form
RQyPDyyD =++2 (Or) ( RyQPDD =++ )2
(iv) Complete solution = Complementary function + Particular Integral
PROBLEMS
1. Solve (D 0)652 =+− yD
Solution: Given (D 0)652 =+− yD
The auxiliary equation is m `0652 =+− m
i.e., m = 2,3
xx BeAeFC 32. +=∴
The general solution is given by
xx BeAey 32 +=
2. Solve 0362
2
=+− ydx
dy
dx
yd
Solution: Given ( ) 0362 =+− yDD
The auxiliary equation is 01362 =+− mm
i.e., 2
52366 −±=m
= i23±
Hence the solution is )2sin2cos(3 xBxAey x +=
3. Solve (D 0)12 =+ given y(0) =0, y’’(0) = 1
RQydx
dyP
dx
yd =++2
2
3
Solution: Given (D 0)12 =+
A.E is 012 =+m
M = i±
Y = A cosx + B sinx
Y(x) = A cosx + B sinx
Y(0) = A =0
Y’(0) =B =1
∴A = 0, B = 1
i.e., y = (0) cosx + sinx
y = sinx
3. Solve xeyDD 22 )134( =+−
Solution: Given xeyDD 22 )134( =+−
The auxiliary equation is 01342 =+− mm
im 322
364
2
52164 ±=−±=−±=
( )xBxAeFC x 3sin3cos. 2 +=∴
P.I. = xeDD
2
2 134
1
+−
= xe2
1384
1
+−
= xe2
9
1
∴y = C.F +P.I.
y = ( )xBxAe x 3sin3cos2 + + xe2
9
1
5. Find the Particular integral of y’’- 3y’ + 2y = exx e2−
Solution: Given y’’- 3y’ + 2y = e xx e2−
xx eeyDD 22 )23( −=+−
P.I xeDD 23
121 +−
=
= xe431
1
+−
= xe0
1
=x xeD 32
1
−
= x xe32
1
−
= xxe−
4
P.I ( )xeDD
2
2223
1 −+−
=
= - xe2
264
1
+−
= - x xeD
2
32
1
−
= - x xe2
34
1
−
= - xe x2
∴P.I. = P.I 21 .IP+
= xxe− + (- xe x2 )
= -x( xe + e x2 )
6. Solve xydx
dy
dx
ydcosh254
2
2
−=+−
Solution: Given xydx
dy
dx
ydcosh254
2
2
−=+−
The A.E is m 0542 =+− m
m = i±−=−±−2
2
20164
C.F = e ( )xBxAx sincos2 +−
P.I = ( )
+++
−=−++
−
254
12cosh2
54
122
xx ee
DDx
DD
= xx eDD
eDD
−
++−+
++−
54
1
54
122
=541541 +−
−++
− xx ee
=210
xx ee −
−−
IPFCy .. +=∴
= e ( )xBxAx sincos2 +− - 210
xx ee −
−
Problems based on P.I = ( ) axDf
oraxDf
cos)(
1sin)(
1
⇒Replace D22 aby −
7. Solve xydx
dy
dx
yd3sin23
2
2
=++
5
Solution: Given xydx
dy
dx
yd3sin23
2
2
=++
The A.E is m 0232 =++ m
(m+1)(m+2) = 0
M = -1, m = -2
C.F = Ae xx Be 2−− +
P.I = xDD
3sin23
12 ++
= xD
3sin233
12 ++−
(Replace D 22 aby − )
= xD
D
Dx
D3sin
)73(
)73(
73
13sin
73
1
++
−=
−
= xD
D3sin
)7()3(
7322 −
+
= xD
D3sin
499
732 −
+
= xD
3sin49)3(9
732 −−+
= xD
3sin130
73
−+
= ( )xxD 3sin7)3(sin3130
1 +−
= ( )xx 3sin73cos9130
1 +−
IPFCy .. +=∴
Y = Ae xx Be 2−− + ( )xx 3sin73cos9130
1 +−
8. Find the P.I of (D xsin)12 =+
Solution: Given (D xsin)12 =+
P.I. = xD
sin1
12 +
= xsin11
1
+−
= x xDsin
2
1
= xD
xsin
1
2
= ∫ xdxx
sin2
P.I =-2
cos xx
6
9. Find the particular integral of (D xxy sin2sin)12 =+
Solution: Given (D xxy sin2sin)12 =+
=- ( )xx cos3cos2
1 −
= - xx cos2
13cos
2
1 +
P.I 1 =
−+
xD
3cos2
1
1
12
= x3cos19
1
2
1
+−−
= x3cos16
1
P.I 2 =
+x
Dcos
2
1
1
12
= xcos11
1
2
1
+−
= xD
x cos2
1
2
1
= ∫ xdxx
cos4
= xxsin
4
∴P.I = x3cos16
1 + x
xsin
4
Problems based on R.H.S = e axeorax axax cos)(cos ++
10. Solve (D xeyD x 2cos)44 22 +=+−
Solution: Given (D xeyD x 2cos)44 22 +=+−
The Auxiliary equation is m 0442 =+− m
(m – 2 ) 2 = 0
m = 2 ,2
C.F = (Ax +B)e x2
P.I 1 = xe
DD
2
2 44
1
+−
= xe2
484
1
+−
= xe2
0
1
= x xeD
2
42
1
−
= x xe2
0
1
7
= x xe22
2
1
P.I 2 = x
DD2cos
44
12 +−
= xD
2cos442
12 +−−
= xD
2cos4
1
−
=
−x
D2cos
1
4
1
= 8
2sin
2
2sin
4
1 xx −=−
IPFCy .. +=∴
y = (Ax +B)e x2
8
2sin x−
Problems based on R.H.S = x
Note: The following are important
• .......1)1( 321 +−+−=+ − xxxx
• .................1)1( 321 ++++=− − xxxx
• ..............4321)1( 322 +−+−=+ − xxxx
• ..............4321)1( 322 ++++=− − xxxx
11. Find the Particular Integral of (D xy =+ )12
Solution: Given (D xy =+ )12
A.E is (m 0)12 =−
m = 1±
C.F = Ae xx Be+−
P.I = xD 1
12 −
= xD 21
1
−−
= [ ] xD121
−−−
= ( )[ ]xDD .........122 +++−
= [ ]...........000 +++− x
= - x
12. Solve: ( ) 3234 2 xyDDD =+−
8
Solution: Given ( ) 3234 2 xyDDD =+−
The A.E is m 02 234 =+− mm
m 0)12( 22 =+− mm
0)1( 22 =−mm
m = 0,0 , m = 1,1
C.F = (A + Bx)e xx eDxC )(0 ++
P.I = 3
234 2
1x
DDD +−
= ( )[ ]3
22 21
1x
DDD −+
= ( )[ ] 312
221
1xDD
D
−−+
= ( ) ( ) ( )[ ]...............22211 32222
2+−−−+−− DDDDDD
D
= [ ] 3432
24321
1xDDDD
D++++
= [ ]241861 23
2+++ xxx
D
=
+++ x
xxx
D24
2
18
3
6
4
1 334
= 2
24
6
18
12
6
20
2345 xxxx +++
= 2345
123220
xxxx +++
IPFCy .. +=∴
y = (A + Bx)e xx eDxC )(0 ++ + 2345
123220
xxxx +++
Problems based on R.H.S = xeax
P.I = xaDf
exeDf
axax
)(
1
)(
1
+=
13. Obtain the particular integral of ( xeyDD x 2cos)522 =+−
Solution: Given ( xeyDD x 2cos)522 =+−
P.I = xeDD
x 2cos52
12 +−
= ( ) ( )( )1Re2cos
5121
12
=
++−+placeDbyDx
DDe x
9
= xDDD
e x 2cos52212
12
+−−++
= xD
e x 2cos4
12
+
= xe x 2cos44
1
+−
= xD
xe x 2cos2
1
= ∫ xdxxe x
2cos2
P.I = 4
2sin xxe x
14. Solve ( xeyD x sin)2 22 −=+
Solution: Given ( xeyD x sin)2 22 −=+
A.E is (m 0)12 =+
m = -2, -2
C.F. = (Ax + B)ex2−
P.I = ( )
xeD
x sin2
1 2
2
−
+
= e ( ) xD
x sin22
12
2
+−−
= e xD
x sin12
2−
= e xx sin1
12
−−
= -e xx sin2−
IPFCy .. +=∴
y = (Ax + B)e x2−- e xx sin2−
Problems based on f(x) = x axxorax nn cos)(sin
To find P.I when f(x) = x axxorax nn cos)(sin
P.I = axxoraxxDf
nn cos)(sin)(
1
VDfdD
dV
DfxxV
Df
+=
)(
1
)(
1)(
)(
1
i.e., ( )
VDfDf
DfV
DfxxV
Df
−=
)(
1
)()(
1)(
)(
1 '
[ ] VDf
DfV
DfxxV
Df 2
'
)(
)(
)(
1
)(
1 −=
10
15. Solve (xx xexeyDD 32 sin)34 +=++ −
Solution: The auxiliary equation is m 0342 =++ m
m= -1,-3
C.F =A xx Bee 3−− +
P.I ( )xeDD
x sin)34(
121
−
++=
= ( ) ( )[ ] x
DDsin
3141
12 +−+−
= e ( ) xDD
x sin2
12 +
−
= e( )
xD
Dx sin41
212+−
+−
= [ ]xxe x
sincos25
+−
−
P.I ( ) xDDe x
3)3(4)3
12
3
2 ++++=
= e ( ) xDD
x
2410
12
3
++
= xDDe x
123
24
101
24
−
++
= xDe x
−12
51
24
3
=
−12
5
24
3
xe x
General solution is IPFCy .. +=
y = Axx Bee 3−− + [ ]xx
e x
sincos25
+−−
+
−12
5
24
3
xe x
16 Solve xxydx
dy
dx
ydcos2
2
2
=++
Solution: A.E : m 0122 =++ m
m = -1,-1
C.F = xeBxA −+ )(
P.I = )cos()1(
12
xxD +
= ( ) ( )
( )xDD
Dx cos
1
1
1
)1(222 +
++−
11
= ( ) ( )( )xDDD
x cos12
1
1
22 ++
+−
= ( ) ( )xDD
x cos121
1
1
2
++−
+−
= 2
sin
1
2 x
Dx
+−
= 2
sin xx
= 1
sin
2
sin
+−D
xxx
= ( )
1
sin1
2
sin2 −
−−D
xDxx
= 2
sincos
2
sin xxxx −+
The solution is y = xeBxA −+ )( +
2
sincos
2
sin xxxx −+
17. Solve ( ) xyD 22 sin1 =+
Solution: A.E : m 012 =+
m = i±
C.F =A cosx +B sinx
P.I = xD
2
2sin1
1
+
=
−+ 2
2cos1
1
12
x
D
=
+−
+x
De
D
x 2cos1
1
1
1
2
12
0
2
=
+ x2cos
3
11
2
1
= x2cos6
1
2
1 +
=∴ y A cosx +B sinx + x2cos6
1
2
1 +
18. Solve ( ) xexxxydx
yd ++=− 1sin2
2
Solution: A.E : m 012 =−
m = 1±
C.F = A xx Bee +−
P.I ( ) )()(
1
)(
)('
)(
11 V
DfDf
DfxxV
Df
−==
12
= ( )( )xDD
Dx sin
1
1
1
222 −
−−
= 2
sin
1
22 −
−− x
D
Dx
= ( )
−+−
12
cos2
2
sin2D
xxx
= 2
cos
2
sin xxx −−
P.I ( ) xexD
2
22 11
1 +−
=
= ( )[ ] )1(
11
1 2
2x
De x +
−+
= e ( ) )1(2
1 2
2x
DD
x ++
= ( )
12
932 23 xxxe x +−
∴y = A xx Bee +− +( )
12
932 23 xxxe x +−
19. Solve xxeydx
yd x sin2
2
=−
Solution: A.E : m 012 =−
m = 1±
C.F = A xx Bee +−
P.I = ( ) xxeD
x sin1
12 −
= e( )[ ]( )xxD
x sin11
12 −+
= e ( )( )xxDD
x sin2
12 +
= e ( ) ( )
−+−
+x
D
Dx
DDxx sin
12
22sin
2
122
sin ,sin xx =α 1,1 22 −=−== αα putD
= e( )( )
−+−
−x
D
Dx
Dxx sin
12
22sin12
12
= e( )( )
( )( )
+−+−
−+−
144
sin22sin
41
2122 DD
xDx
D
Dxx
Put D 12 −=
= e( ) ( )( )
−−+++− xD
DDx
Dxx sin
169
4322sin
5
212
13
= e [ ]
−+−−++−
xD
DDxx
xx sin169
628cos2sin
5 2
2
= e ( ) ( )
−++−x
Dxx
xx sin25
214cos2sin
5
P.I = e ( ) ( )
−++−25
cos2sin14cos2sin
5
xxxx
xx
Complete Solution is
y = A xx Bee +− +
e ( ) ( )
−++−25
cos2sin14cos2sin
5
xxxx
xx
METHOD OF VARIATION OF PARAMETERS
This method is very useful in finding the general solution of the second order
equation.
Xyadx
dya
dx
yd =++ 212
2
[Where X is a function of x] ……………….(1)
The complementary function of (1)
C.F = c 2211 fcf +
Where c 21 ,c are constants and f 21andf are functions of x
Then P.I = Pf 21 Qf+
P = ∫ −− dx
ffff
Xf
2
'
1
1
21
2
Q = ∫ −dx
ffff
Xf
2
'
1
'
21
1
IPfcfcy .2211 ++=∴
1. Solve ( ) xyD 2sec42 =+
Solution: The A.E is m 042 =+
m = i2±
C.F = C xCx 2sin2cos 21 +
f x2cos1= f x2sin2 =
f x2sin2'
1 −= f x2cos2'
2 =
f xxfff 2sin22cos2 22
2
'
1
'
21 +=−
= 2 [ ]xx 2sin2cos 22 +
= 2 [ ]1 = 2
P = ∫ −− dx
ffff
Xf
2
'
1
1
21
2
14
= ∫− dxxx
2
2sec2sin
= dxx
x2cos
12sin
2
1∫−
= ∫−
dxx
x
2cos
2sin2
4
1
= )2log(cos4
1x
Q = ∫ −dx
ffff
Xf
2
'
1
'
21
1
= ∫ dxxx
2
2sec2cos
= dxx
x2cos
12cos
2
1∫
= ∫ dx2
1
= x2
1
P.I = Pf 21 Qf+
= )2log(cos4
1x (cos2x) + x
2
1sin2x
2. Solve by the method of variation of parameters xxydx
ydsin
2
2
=+
Solution: The A.E is m 012 =+
m = i±
C.F = C xCx sincos 21 +
Here xf cos1 = xf sin2 =
xf sin'
1 −= xf cos'
2 =
1sincos 22
2
'
1
'
21 =+=− xxffff
P = ∫ −− dx
ffff
Xf
2
'
1
1
21
2
= dxxxx
∫−1
)sin(sin
= ∫− xdxx 2sin
= ( )
∫−− dx
xx
2
2cos1
= ( )∫ −− dxxxx 2cos2
1
= ∫∫ +− xdxxxdx 2cos2
1
2
1
15
=
−−
+
−
4
2cos)1(
2
sin
2
1
22
1 2 xxx
x
= xxxx
2cos8
12sin
44
2
++−
Q = ∫ −dx
ffff
Xf
2
'
1
'
21
1
= ∫ dxxxx
1
)(sin)(cos
= ∫ xdxxx cossin
= ∫ dxx
x2
2sin
= ∫ xdxx 2sin2
1
=
−−
−4
2sin)1(
2
2cos
2
1 xxx
= xxx
2sin8
12cos
4+−
P.I = Pf 21 Qf+
= xxxx
xxxxx
sin2sin8
12cos
4cos2cos
8
12sin
44
2
+−+
++−
4. Solve (D xeyD 22 )44 =+− by the method of variation of parameters.
Solution: The A.E is 0442 =+− mm
( ) 022 =−m
m = 2,2
C.F = ( ) xeBAx 2+
= xx BeAxe 22 +
xxef 2
1 = xef 2
2 =
xx exef 22'
1 2 += , xef 2'
2 2=
( ) ( )xxxx exeeexffff 22222'
12
'
21 22 +−=−
= 2x ( ) ( ) ( )222222 2 xxx eexe −−
= ( ) [ ]12222 −− xxe x
= ( )22xe−
= xe4−
P= ∫ −− dx
ffff
Xf
2
'
1
1
21
2
= ∫ −− dxe
eex
xx
4
22
16
= xdx =∫
Q = ∫ −dx
ffff
Xf
2
'
1
'
21
1
= ∫ −dx
e
exex
xx
4
22
= ∫− xdx
= 2
2x−
P.I = xxx ex
ex
ex 22
22
22
22=−
y = C.F + P.I
= (Ax +B) e x2 + xex 22
2
5. Use the method of variation of parameters to solve ( ) xyD sec12 =+
Solution: Given ( ) xyD sec12 =+
The A.E is 012 =+m
m = i±
C.F = xcxc sincos 21 +
= 2211 fcfc +
xfxf sin,cos 21 ==
xfxf cos,sin '
2
'
1 −=
1sincos 22
2
'
1
'
21 =+=− xxffff
P= ∫ −− dx
ffff
Xf
2
'
1
1
21
2
= dxxx
∫−1
secsin
= ∫− dxx
x
cos
sin
= ∫− xdxtan
= log (cos x)
Q = ∫ −dx
ffff
Xf
2
'
1
'
21
1
= dxxx
∫ 1
seccos
= ∫ dx = x
P.I = 21 QfPf +
= xxxx sincos)log(cos +
y = xcxc sincos 21 + + xxxx sincos)log(cos +
17
6. Solve ( ) axyaD tan22 =+ by the method of variation of parameters.
Solution: Given ( ) axyaD tan22 =+
The A.E is 022 =+ am
aim ±=
C.F = axcaxc sincos 21 +
axfaxf sin,cos 21 ==
axafxaf cos,sin '
2
'
1 =−=
)sin(sincoscos'
12
'
21 axaaxaxaxaffff −−=−
= axaaxa 22 sincos +
= )sin(cos 22 axaxa +
= a
P.I = 21 QfPf +
P= ∫ −− dx
ffff
Xf
2
'
1
1
21
2
= dxa
axax∫− tansin
= ∫− dxax
ax
a cos
sin1 2
= ∫−−
dxax
ax
a cos
cos11 2
= ( )∫ −−dxaxax
acossec
1
= ( )
−+−a
axaxax
aa
sintanseclog
11
= ( )[ ]axaxaxa
sintanseclog12
−+−
= ( )[ ]axaxaxa
tanseclogsin12
+−
Q = ∫ −dx
ffff
Xf
2
'
1
'
21
1
= ∫ dxa
axax tancos
= ∫ axdxa
sin1
= axa
cos12
−
21. QfPfIP +=∴
= ( )[ ] [ ]axaxa
axaxaxaxa
cossin1
tanseclogsincos1
22−+−
18
= ( )[ ]axaxaxa
tanseclogcos12
+−
IPFCy .. +=
= axcaxc sincos 21 + ( )[ ]axaxaxa
tanseclogcos12
+−
7. Solve xydx
ydtan
2
2
=+ by the method of variation of parameters.
Solution: The A.E is 012 =+m
im ±=
C.F = xcxc sincos 21 +
Here xfxf sin,cos 21 ==
xfxf cos,sin '
2
'
1 =−=
1sincos 22'
12
'
21 =+=− xxffff
P= ∫ −− dx
ffff
Xf
2
'
1
1
21
2
= ∫− dxxx
1
tansin
= ∫− dxx
x
cos
sin 2
= dxx
x∫
−−cos
cos1 2
= ( )dxxx∫ −− cossec
= xxx sin)tanlog(sec ++−
Q = ∫ −dx
ffff
Xf
2
'
1
'
21
1
= dxxx
∫ 1
tancos
= ∫ dxsin
= xcos−
21. QfPfIP +=∴
= )tanlog(seccos xxx +−
IPFCy .. +=
= xcxc sincos 21 + )tanlog(seccos xxx +−
8. Solve by method of variation of parameters 144 2
2
''' +=+− xyx
yx
y
Solution: Given 144 2
2
''' +=+− xyx
yx
y
i.e., 24'''2 44 xxyxyyx +=+−
i.e., [ ] 2422 44 xxyxDDx +=+− ………….(1)
Put x = ze
19
Logx = log ze
= z
So that XD = D '
( )1''2 −= DDDx
(1) ( )[ ] ( ) ( )24''' 441 zz eeyDDD +=+−−⇒
[ ] z
eeyDD z 22 4'' 45 +=+−
A.E is 0452 =+− mm
( )( ) 014 =−− mm
4,1=m
zz ececFC 2
4
1. +=∴
Here zz efef == 2
4
1 ,
zz efef == '
2
4'
1 ,
zzz eeeffff 555'
12
'
21 34 −=−=−
21. QfPfIP +=
P= ∫ −− dx
ffff
Xf
2
'
1
1
21
2
= [ ]
∫ −+− dxe
eeez
zzz
5
24
3
= ∫+
dxe
eez
zz
5
35
3
= [ ]dze z
∫−+ 21
3
1
=
−+
−
23
1 2 zez
= zez 2
6
1
3
1 −−
Q = ∫ −dz
ffff
Zf
2
'
1
'
21
1
= ( )
∫ −+
dze
eeez
zzz
5
344
3
= ∫+− dze
eez
zz
5
68
3
1
= ( )∫ +− dzee zz3
3
1
=
+− z
z
ee
33
1 3
= zz ee3
1
9
1 3 −−
zzzzz eeeeezIP
−−+
−=∴ −
3
1
9
1
6
1
3
1. 342
20
= zzzz eeeze 2424
3
1
9
1
6
1
3
1 −−−
IPFCy .. +=
= zz ecec 2
4
1 + + zzzz eeeze 2424
3
1
9
1
6
1
3
1 −−−
= ( ) ( ) ( )3
1
9
1
6
1
3
1)(
424
2
4
1 −−−++ zzzzz eeezecec ( )2ze
= ( ) ( )29
log3
244
2
4
1
xxx
xecec zz −−++
DIFFERENTIAL EQUATIONS FOR THE VARIABLE COEFFICIENTS
(CAUCHY’S HOMOGENEOUS LINEAR EQUATION)
Consider homogeneous linear differential equation as:
)1(............................1
11
1 Xyadx
ydxa
dx
yda nx
nn
n
nn =+++ −
−−
(Here a’s are constants and X be a function of X) is called Cauchy’s
homogeneous linear equation.
)(.............. 011
1
1 xfyadx
dyxa
dx
yda
dx
ydxa
n
n
nn
nn
n =++++ −
−
−
is the homogeneous linear equation with variable coefficients. It is also known
as Euler’s equation.
Equation (1) can be transformed into a linear equation of constant Coefficients
by the transformation.
xzorex z log)(, ==
Then
dz
dy
xdx
dz
dz
dy
dx
dyDy
1===
==dz
d
dx
dD θ,
θ===dz
dxD
dx
dx
Similarly
=
=dz
dy
xdx
d
dx
dy
dx
d
dx
yd 12
2
=dx
dz
dz
dy
dz
d
xdz
dy
xdz
dy
dx
d
xdz
dy
x
+−=
+− 111122
2
2
222
2 11
dz
yd
xdz
dy
xdx
yd +−=
dz
dy
dz
yd
dx
ydx −=∴
2
2
2
22
21
( ) ( )1222 −=−=∴ θθθθDx
Similarly,
( )( )2133 −−= θθθDx
( )( )( )32144 −−−= θθθθDx
and soon.
θ=∴ xD
( )122 −= θθDx
( )( )2133 −−= θθθDx
and so on.
1. Solve )sin(log42
22 xy
dx
dyx
dx
ydx =++
Solution: Consider the transformation
xzorex z log)(, ==
θ=∴ xD
( )122 −= θθDx
( ) )sin(log4122 xyxDDx =++
( ) ( )zy sin412 =+θ
R.H.S = 0 : ( ) 012 =+ yθ
A.E : imm ±==+ ,012
C.F = A cosz + B sinz
C.F = A cos (log x) +B sin (log x)
P.I = ( )zsin41
12 +θ
= zzzz
cos22
cos4 −=
−
P.I = )cos(loglog2 xx−
∴Complete solution is:
y = A cos (log x) +B sin (log x) )cos(loglog2 xx−
y = ( ) )cos(loglog2)cos(loglog2 xxxxA −−
2. Solve xxydx
dyx
dx
ydx log24
2
22 =++
Solution: Given ( ) xxyxDDx log2422 =++ ……………(1)
Consider: xzorex z log)(, ==
θ=∴ xD
( )122 −= θθDx
(1) : ( )( ) zey z=++− 241 θθθ
22
( ) zzey =++ 232 θθ
A.E : 0232 =++ mm
M = - 2, -1
C.F = 12 loglog2 −−
+=+ −− xxzz BeAeBeAe
C.F = x
B
x
A +2
P.I = ( )( )ze z
23
12 ++ θθ
=e( ) ( )
zz
2131
12 ++++ θθ
= ( ) ze z
65
12 ++ θθ
= ze z
12
6
51
6
−
++ θθ
=
−=
−6
5
66
51
6z
ez
e zz
θ
=
−6
5log
6
log
xe x
=
−6
5log
6x
x
Complete solution is y = C.F +P.I
= x
B
x
A +2
+
−6
5log
6x
x
3. Solve ( giventhatxyxDDx ,)43( 222 =+− y(1) = 1 and y’(1) = 0
Solution: Given ( ) 222 43 xyxDDx =+− …………………(1)
Take xzorex z log)(, ==
θ=∴ xD
( )122 −= θθDx
(1) : ( ) zey 22 44 =+− θθ
A.E : 2,2,0442 ==+− mmm
C.F = ( ) ( ) 22 log xxBAeBzA z +=+
P.I = ( )
( )( )
)1(22
1
2
12
22
2 −+=
− θθzz ee
= ( )112
2
θze
P.I = 2
22 z
e z
23
= ( )2
log22 xx …………………(2)
Complete solution is : y = ( ) 2log xxBA + +( )2
log22 xx
Apply conditions: y(1) =1,y’(1) = 0 in (2)
A = 1, B = -2
Complete solution is y = ( ) +− 2log21 xx ( )2
log22 xx
EQUATION REDUCIBLE TO THE HOMOGENEOUS LINEAR
FORM (LEGENDRE LINEAR EQUATION)
It is of the form:
( ) ( ) )(...........1
11
1 xfyAdx
ydbxaA
dx
ydbxa nn
nn
n
nn =+++++ −
−−
…………..(
1)
nAAA ....,..........,.........21 , are some constants
It can be reduced to linear differential equation with constant
Coefficients,
by taking: )log()( bxazorebxa z +==+
Consider givesdz
dD
dx
d,, θ==
( ) ( ) ( )ybDybxadz
dyb
dx
dybxa θ=+⇒=+
Similarly ( ) ( )ybyDbxa 1222 −=+ θθ …………(2)
( ) ( )( )21333 −−=+ θθθbyDbxa and so on
Substitute (2) in (1) gives: the linear differential equation of
constant Coefficients.
Solve ( ) ( ) xyyxyx 612'32''322 =−+−+
Solution: This is Legendre’s linear equation:
( ) ( ) xyyxyx 612'32''322 =−+−+ ……………….(1)
Put z = log (2x + 3) , 32 += xe z
( ) θ232 =+ Dx
( ) ( )dz
dDx =−=+ θθθ ,432 222
Put in (1) : ( ) 931264 2 −=−− zeyθθ
R.H.S =0
A.E : 01264 2 =−− mm
4
573,
4
57321
−=+= mm
C.F =zmzm
BeAe 21 +
C.F = 21 )32()34(mm
xBxA +++
24
P.I 1=1264
32 −− θθ
ze= ( )32
14
3 +− x
P.I 2 = 1264
92 −− θθ
θze
= 4
3
12
9 −=−
Solution is
y = ( )( ) ( )
−+ +++ 4
5734/5733232 xBxA ( )32
14
3 +− x4
3−
SIMULTANEOUS FIRST ORDER LINEAR EQUATIONS WITH
CONSTANT COEFFICIENTS
1. Solve the simultaneous equations, 023,232 2 =++=++ yxdt
dyeyx
dt
dx t
Solution: Given 023,232 2 =++=++ yxdt
dyeyx
dt
dx t
Using the operator D = dt
d
( ) teyxD 2232 =++ ………………(1)
( ) 023 =++ yDx ……………….(2)
Solving (1) and (2) eliminate (x) :
( ) )3....(..........654)2()2()1(3 22 teyDDD −=−+⇒+×−×
A.E : 0542 =−+ mm
m = 1,-5
C.F = tt BeAe 5−+
P.I = t
t
eDD
e 2
2
2
7
6
54
6 −=−+
−
y = tt BeAe 5−+ te2
7
6−
put in (1) : x = ( )[ ]yD 23
1 +−
x = ttt eBeAe 25
7
8++− −
solution∴ is :
x = ttt eBeAe 25
7
8++− −
y = tt BeAe 5−+ te2
7
6−
2. Solve giventhattxdt
dyty
dt
dx,cos,sin =+=+ t=0, x = 1, y =0
Solution: Dx + y = sin t ……………..(1)
x + Dy = cost ……………… (2)
Eliminate x : (1) – (2)D ttyDy sinsin2 +=−⇒
25
( ) )3.....(..........sin212 tyD −=−
1,012 ±==− mm
C.F = Aett Be−+
P.I = tt
D
tsin
11
sin)2(
1
sin2
2=
−−−=
−−
y = Aett Be−+ + sint
(2) : x = cost –D(y)
x = cost - ( )tBeAedt
d tt sin++ −
x = tBeAet tt coscos −+− −
x = tt BeAe −+−
Now using the conditions given, we can find A and B
BAxt +−=⇒== 11,0
BAyt +=⇒== 00,0
B = 2
1, A =
2
1−
Solution is
x = tee tt cosh2
1
2
1 =+ −
y = tttee tt sinhsinsin2
1
2
1 −=++− −
3. Solve txdt
dyty
dt
dxcos2,sin2 =−−=+
Solution: Dx + 2y = -sin t ……………..(1)
- 2x +Dy = cos t ……………...(2)
(1) )(cossin24)2(2 2 tDtyDyD +−=+⇒×+×
( ) tD sin342 −=+⇒
2,042 imm ±==+
C.F = tBtA 2sin2cos +
P.I = tt
D
tsin
41
sin3
4
sin32
−=+−
−=+
−
y = tBtA 2sin2cos + - sin t
(2) : x = [ ]tDy cos2
1 −
x = ( )
−−+ tttBtAdt
dcossin2sin2cos
2
1
x = A cos2t +Bsin2t – cost
Solution is :
x = A cos2t +Bsin2t – cost
y = tBtA 2sin2cos + - sin t
26
4. Solve txdt
dy
dt
dxty
dt
dy
dt
dx2sin2,2cos2 =−+=+−
Solution: Dx + (-D +2)y = cos 2t …………..(1)
(D – 2)x +Dy = sin 2t ……………(2)
Eliminating y from (1) and (2)
( ) ttxDD 2cos2sin2222 +−=+−
R.H.S = 0 0222 =+−⇒ mm
m = i±1
C.F = ( )tBtAe t sincos +
P.I 1=( )
D
t
DD
t
+=
+−−
1
2sin
22
2sin22
= ( )
41
)2(sin2sin2sin
1
12 +
−=−− tDt
tD
D
= 5
2cos22sin tt −
P.I 2 = ( )tDD
2cos22
12 +−
= ( )
10
2sin22cos tt +−
x = ++ )sincos( tBtAe t
5
2cos22sin tt − ( )10
2sin22cos tt +−
(1) +(2) ttxydt
dx2sin2cos222 +=−+⇒
2y = cos2t +sin 2t + 2x -dt
dx2
y =
−++dt
dxxtt 222sin2cos
2
1…………(3)
Substitute x in (3)
y = ( )2
2sinsincos
ttBtAe t −−
Solution is :
x = ++ )sincos( tBtAe t
5
2cos22sin tt − ( )10
2sin22cos tt +−
y = ( )2
2sinsincos
ttBtAe t −−