unit1 maths
TRANSCRIPT
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UNIT-I
ORDINARY DIFFERENTIAL EQUATIONS
Higher order differential equations with constant coefficients Method of variation ofparameters Cauchys and Legendres 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, calledmodeling.
(ii) Solutions of this differential equation, evaluating the arbitrary constantsfrom 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
yd
nn
n
n
n
n
n
=++++
..............2
2
21
1
1
.. (1)
Where KnK
K ...............,.........21 , are constants.
The symbol D stands for the operation of differential
(i.e.,) Dy = ,dx
dysimilarly D ...,,
3
33
2
22 etc
dx
ydyD
dx
dy
y
==
The equation (1) above can be written in the symbolic form
(D XyKDK nnn
=+++ )..........11 i.e., f(D)y = X
Where f (D) = D nnn
KDK+++
...........
1
1
Note
1. = XdxXD1
2. dxXeeXaD
axax
=
1
3. dxXeeXaD
axax
=+
1
(i) The general form of the differential equation of second order is
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(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
dyD
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 formRQyPDyyD =++
2 (Or) ( RyQPDD =++ )2
(iv) Complete solution = Complementary function + Particular Integral
PROBLEMS
1. Solve (D 0)652 =+ yD Solution: Given (D 0)65
2=+ yD
The auxiliary equation is m `0652 =+ m
i.e., m = 2,3xx BeAeFC 32. +=
The general solution is given byxx 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
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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
3642
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 + + xe29
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.Ix
eDD 23
121
+=
= xe431
1
+
= xe0
1
=x xeD 32
1
= x xe32
1
=x
xe
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P.I ( )xeDD
2
22 23
1
+=
= - xe2
264
1
+
= - x xeD2
32
1
= - x xe2
34
1
= - xe x2
P.I. = P.I 21 .IP+
= xxe + (- xe x2 )
= -x( xe + ex2 )
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=
22
20164
C.F = e ( )xBxAx
sincos
2+
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 D 22 aby
7. Solve xy
dx
dy
dx
yd3sin23
2
2
=++
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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 xD
sin2
1
= xD
xsin
1
2
= xdxx
sin2
P.I =-
2
cosxx
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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.I1 =
+x
D3cos
2
1
1
12
= x3cos19
1
2
1
+
= x3cos16
1
P.I 2 =
+
xD
cos2
1
1
12
= xcos11
1
2
1
+
= xD
x cos2
1
2
1
= xdxx
cos4
= xx
sin
4
P.I = x3cos16
1+ xx
sin4
Problems based on R.H.S = e axeoraxaxax cos)(cos ++
10. Solve (D xeyDx 2cos)44 22 +=+
Solution: Given (D xeyD x 2cos)44 22 +=+
The Auxiliary equation is m 0442 =+ m
(m 2 ) 2 = 0
m = 2 ,2C.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
01
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= x xe22
2
1
P.I2
=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
= xD21
1
= [ ] xD 121 = ( ) xDD .........1 22 +++ = [ ]...........000 +++ x = - x
12. Solve: ( ) 3234 2 xyDDD =+
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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 = 1,1C.F = (A + Bx)e xx eDxC )(0 ++
P.I = 3234 2
1x
DDD +
=( )[ ]
3
22 21
1x
DDD +
= ( )[ ] 3122
211
xDDD
+
= ( ) ( ) ( )[ ]...............222113
22
222 ++ DDDDDD
D
= [ ] 34322
43211
xDDDDD
++++
= [ ]241861 232
+++ xxxD
=
+++ x
xxx
D24
2
18
3
6
4
1 334
=
2
24
6
18
12
6
20
2345xxxx
+++
= 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 ( xeyDDx 2cos)522 =+
Solution: Given ( xeyDD x 2cos)522 =+
P.I = xeDD
x 2cos52
12
+
=
( ) ( )( )1Re2cos
5121
1
2
=
+++
placeDbyDxDD
ex
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= xDDD
ex 2cos
52212
12
+++
= xD
ex 2cos4
12
+
= xex 2cos44
1+
= xD
xex 2cos2
1
= xdxxex
2cos2
P.I =4
2sin xxex
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 22
+
= e( )
xD
x sin22
12
2
+
= e xD
x sin12
2
= e xx sin
1
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 '
[ ]V
DfDfV
DfxxV
Df 2
'
)()(
)(1
)(1 =
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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
++=
=( ) ( )[ ]
xDD
sin3141
12
++
= e( )
xDD
x sin2
12+
= e( )
xD
Dx sin41
212
+
+
= [ ]xxe x
sincos25
+
P.I( )
xDD
e x
3)3(4)3
12
3
2++++
=
= e( )
xDD
x
2410
12
3
++
= xDDe x
123
24
101
24
++
= xDex
12
5124
3
=
12
5
24
3
xe x
General solution is IPFCy .. +=
y = Axx Bee 3 + [ ]xx
e xsincos2
5+
+
12
5
24
3
xe x
16 Solve xxydx
dy
dx
yd cos22
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
+
+
+
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=( ) ( )
( )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 = x
D
2
2sin
1
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 + x2cos61
21 +
18. Solve ( ) xexxxydx
yd++= 1sin
2
2
Solution: A.E : m 012 =
m = 1
C.F = A xx Bee +
P.I ( ) )(
)(
1
)(
)('
)(
11 V
DfDf
DfxxV
Df
==
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=( )
( )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
221
1
1+
=
=( )[ ]
)1(11
1 22
xD
ex ++
= e
( ))1(
2
1 22
xDD
x+
+
=( )
12
932 23 xxxex +
y = A xx Bee + +
( )12
932 23 xxxex +
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 ( ) ( )
+
+ xD
D
xDDx
x
sin12
22
sin2
122
sin ,sin xx = 1,122
=== putD
= e( )
( )
+
x
D
Dx
Dxx sin
12
22sin
12
12
= e( )
( )( )
( )
+
+
+
144
sin22sin
41
2122 DD
xDx
D
Dxx
Put D 12 =
= e( ) ( )( )
++
+ x
D
DDx
Dxx sin
169
4322sin
5
21
2
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= e [ ]
+++
x
D
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 orderequation.
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 =
dxffff
Xf
2
'
1
1
21
2
Q =
dxffff
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 222'
1
'
21 +=
= 2 xx 2sin2cos22
+
= 2 [ ]1 = 2
P =
dxffff
Xf
2
'
1
1
21
2
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= dxxx
2
2sec2sin
= dxx
x2cos
12sin
2
1
=
dxx
x
2cos
2sin2
4
1
= )2log(cos4
1x
Q =
dxffff
Xf
2
'
1
'
21
1
= dxxx
2
2sec2cos
= dxx
x2cos
12cos
2
1
= dx21
= x2
1
P.I = Pf 21 Qf+
= )2log(cos4
1x (cos2x) + x
2
1sin2x
2. Solve by the method of variation of parameters xxy
dx
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 222'
1
'
21 =+= xxffff
P =
dx
ffff
Xf
2
'
1
1
21
2
= dxxxx
1)sin(sin
= xdxx2sin
=( )
dxx
x2
2cos1
= ( ) dxxxx 2cos21
= + xdxxxdx 2cos2
1
2
1
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=
+
4
2cos)1(
2
sin
2
1
22
1 2 xxx
x
= xxxx
2cos8
12sin
44
2
++
Q =
dxffff
Xf
2
'
1
'
21
1
= dxxxx
1
)(sin)(cos
= xdxxx cossin
= dxx
x2
2sin
= xdxx 2sin21
=
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
( ) 02 2 =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
exeeexffff22222'
12
'
21 22 +=
= 2x ( ) ( ) ( )222222 2 xxx eexe
= ( ) [ ]12222 xxe x
= ( )22xe = xe4
P=
dxffff
Xf
2
'
1
1
21
2
=
dx
e
ee
x
xx
4
22
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= xdx =
Q =
dxffff
Xf
2
'
1
'
21
1
= dxeexex
xx
4
22
= xdx
=2
2x
P.I = xxx ex
ex
ex 22
22
22
22=
y = C.F + P.I
= (Ax +B) ex2
+x
ex 2
2
2
5. Use the method of variation of parameters to solve ( ) xyD sec12 =+ Solution: Given xyD sec1
2=+
The A.E is 012 =+m
m = i
C.F = xcxc sincos 21 +
= 2211 fcfc +
xfxf sin,cos 21 ==
xfxf cos,sin '2
'
1=
1sincos22
2
'
1
'
21 =+= xxffff
P=
dxffff
Xf
2
'
1
1
21
2
= dxxx
1secsin
= dxxx
cos
sin
=
xdxtan
= log (cos x)
Q =
dxffff
Xf
2
'
1
'
21
1
= dxxx
1seccos
= dx = x
P.I = 21 QfPf +
= xxxx sincos)log(cos + y = xcxc sincos 21 + + xxxx sincos)log(cos +
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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(cos22 axaxa +
= a
P.I = 21 QfPf +
P=
dxffff
Xf
2
'
1
1
21
2
= dxa
axax
tansin
= dxaxax
a cos
sin12
=
dxax
ax
a cos
cos112
= ( )
dxaxaxa
cossec1
= ( )
+
a
axaxax
aa
sintanseclog
11
= ( )[ ]axaxaxa
sintanseclog12
+
= ( )[ ]axaxaxa
tanseclogsin1
2+
Q = dxffffXf
2'
1'
21
1
= dxaaxax tancos
= axdxa sin1
= axa
cos1
2
21. QfPfIP +=
= ( )[ ] [ ]axaxa
axaxaxaxa
cossin1
tanseclogsincos1
22
+
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= ( )[ ]axaxaxa
tanseclogcos1
2+
IPFCy .. +=
= axcaxc sincos 21 + ( )[ ]axaxax
a
tanseclogcos1
2+
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=
dxffff
Xf
2
'
1
1
21
2
= dxxx
1
tansin
= dxxx
cos
sin 2
= dxx
x
cos
cos1 2
= ( )dxxx
cossec
= xxx sin)tanlog(sec ++
Q =
dxffff
Xf
2
'
1
'
21
1
= dxxx
1tancos
= dxsin= xcos
21. QfPfIP +=
= )tanlog(seccos xxx + IPFCy .. +=
= xcxc sincos 21 + )tanlog(seccos xxx +
8. Solve by method of variation of parameters 144 22
'''+=+ xy
xy
xy
Solution: Given 144 2
2
'''+=+ xy
xyy
i.e., 24'''2 44 xxyxyyx +=+
i.e.,2422
44 xxyxDDx +=+ .(1)Put x = ze
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Logx = logze
= z
So that XD = D '
1''2= DDDx
(1) ( )[ ] ( ) ( )24'''
441zz
eeyDDD +=+ z
eeyDD z22 4'' 45 +=+
A.E is 0452 =+ mm
( )( ) 014 = mm 4,1=m
zz ececFC 24
1. +=
Here zz efef == 24
1 ,zz efef == '2
4'
1 ,zzz eeeffff 555'12
'
21 34 ==
21. QfPfIP +=
P=
dxffff
Xf
2
'
1
1
21
2
=[ ]
+ dx
e
eeez
zzz
5
24
3
= +
dxe
eez
zz
5
35
3
= [ ]dze z
+2
1
3
1
=
+
23
1 2zez
=zez 2
6
1
3
1
Q =
dzffff
Zf
2
'
1
'
21
1
=( )
+dz
e
eeez
zzz
5
344
3
= + dzeee
z
zz
5
68
31
= ( ) + dzeezz3
3
1
=
+
zz
ee
33
1 3
= zz ee3
1
9
1 3
zzzzz eeeeezIP
+
=
3
1
9
1
6
1
3
1. 342
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= zzzz eeeze 2424
3
1
9
1
6
1
3
1
IPFCy .. +=
=zz
ecec 24
1 + +zzzz
eeeze2424
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
(CAUCHYS HOMOGENEOUS LINEAR EQUATION)
Consider homogeneous linear differential equation as:
)1(............................1
11
1 Xyadx
ydxa
dx
yda nx
nn
n
nn
=+++
(Here as are constants and X be a function of X) is called Cauchys
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 Eulers 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 ,
===dzdxD
dxdx
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
dz
dy
xdx
yd+=
dzdy
dzyd
dxydx =
2
2
2
2
2
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( )1222 == Dx Similarly,
( )( )2133 = Dx
( )( )( )32144 = Dx and soon.
=xD
( )122 = Dx
( )( )2133 = Dx
and so on.
1. Solve )sin(log42
22 xy
dx
dyx
dx
ydx =++
Solution: Consider the transformation
xzorex z log)(, ==
=xD
( )122 = Dx
( ) )sin(log4122 xyxDDx =++ ( ) ( )zy sin412 =+
R.H.S = 0 : ( ) 012 =+ y
A.E : imm ==+ ,012
C.F = A cosz + B sinz
C.F = A cos (log x) +B sin (log x)
P.I = ( )zsin41
12+
= 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)(, ==
=xD
( )122
= Dx (1): ( )( ) zey z=++ 241
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zzey =++ 232
A.E : 0232 =++ mm
M = - 2, -1
C.F =12 loglog2
+=+ xxzz BeAeBeAe
C.F =xBA +
2
P.I =( )
( )zez23
12
++
=e( ) ( )
zz
2131
12
++++
=( )
zez
65
12
++
= z
ez1
2
6
5
16
++
=
=
6
5
66
51
6z
ez
e zz
=
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: Given222 43 xyxDDx =+ (1)
Take xzorex z log)(, ==
=xD
( )122 = Dx
(1): ( ) zey 22 44 =+ A.E : 2,2,0442 ==+ mmm
C.F = ( ) ( ) 22 log xxBAeBzA z +=+
P.I =( )
( )( )
)1(22
1
2
12
22
2+
=
zz ee
= ( )11
2
2
ze
P.I =
2
22 ze z
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=( )
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
1
11 xfyA
dxydbxaA
dxydbxa nn
n
nn
n
n =+++++
..(
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,, ==
( ) ( ) ( )ybDybxadzdy
bdx
dy
bxa =+=+
Similarly ( ) ( )ybyDbxa 1222 =+ (2)
( ) ( )( )21333 =+ byDbxa and so onSubstitute (2) in (1) gives: the linear differential equation of
constant Coefficients.
Solve ( ) ( ) xyyxyx 612'32''32 2 =++ Solution: This is Legendres linear equation:
( ) ( ) xyyxyx 612'32''32 2 =++ .(1)
Put z = log (2x + 3) , 32 += xez
( ) 232 =+ Dx
( ) ( )dz
dDx ==+ ,432 22
2
Put in (1) : ( ) 931264 2 = zey R.H.S =0
A.E : 01264 2 = mm
4
573,
4
57321
=
+= mm
C.F = zmzm BeAe 21 + C.F = 21 )32()34(
mmxBxA +++
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P.I 1 =1264
32
ze= ( )32
14
3+ x
P.I 2 =1264
92
ze
=43
129 =
Solution is
y = ( )( ) ( )
+
+++ 45734/573
3232 xBxA ( )3214
3+ x
4
3
SIMULTANEOUS FIRST ORDER LINEAR EQUATIONS WITH
CONSTANT COEFFICIENTS
1. Solve the simultaneous equations, 023,232 2 =++=++ yx
dt
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 22
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 +=
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)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
dy
tydt
dx
cos2,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 = ( )
+ tttBtA
dt
dcossin2sin2cos
2
1
x = A cos2t +Bsin2t cost
Solution is :
x = A cos2t +Bsin2t cost
y = tBtA 2sin2cos + - sin t
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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 = i1
C.F = ( )tBtAe t sincos +
P.I1 =( )
D
t
DD
t
+=
+
1
2sin
22
2sin22
=( )
41
)2(sin2sin2sin
1
12
+
=
tDtt
D
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( tBtAet
5
2cos22sin tt ( )
10
2sin22cos tt+
y = ( )2
2sinsincos
ttBtAe t