ordinary differential equations mathematics ii(ma6251) i yr/unit 2(part i).pdf · 2 2 0 d y dy a b...
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UNIT 2
ORDINARY DIFFERENTIALEQUATIONS
MATHEMATICS II(MA6251)UNIT 2
-P.VEERAIAHDEPARTMENT OF APPLIED
MATHEMATICS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
1
UNIT 2 SYLLABUS Higher order linear differential equations with
constant coefficients
Method of variation of parameters
Cauchy’s and Legendre’s linear equations
Simultaneous first order linear equations with
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 2
Simultaneous first order linear equations with constant coefficients
Second-order linear differentialequations
2
2Differential equations of the form ( )
are called second order linear differential equations.
d y dya b cy Q x
dx dx
When ( ) 0 then the equations are referred to as homogeneous,Q x When ( ) 0 then the equations are non-homogeneous.Q x
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When ( ) 0 then the equations are non-homogeneous.Q x
Note that the general solution to such an equationmust include two arbitrary constants to becompletely general.
Second-order linear differentialequations
Theorem
If ( ) and ( ) are two solutions then so is ( ) ( )y f x y g x y f x g x 2
2we have 0d f df
a b cfdx dx
2
2and 0d g dg
a b cgdx dx
Adding 2 2
0d f df d g dg
a b cf a b cg
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Adding2 2 0
d f df d g dga b cf a b cg
dx dx dx dx
2 2
2 2 0d f d g df dg
a b c f gdx dx dx dx
And so ( ) ( ) is a solution to the differential equation. y f x g x
Second-order linear differentialequations
, for and , is a solution to the equation 0mx dyy Ae A m b cy
dx
It is reasonable to consider it as a possible solution for
2
2 0d y dy
a b cydx dx
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2dx dxmxy Ae mxdy
Amedx
2
22
mxd yAm e
dx
If is a solution it must satisfymxy Ae 2 0 mx mx mxaAm e bAme cAe assuming 0, then by division we getmxAe 2 0am bm c
The solutions to this quadratic will provide two values ofm which will make y = Aemx a solution.
Second-order linear differentialequations When the roots of the auxiliary equation are both
real and equal to m, then the solution would appear to be
y = Aemx + Bemx = (A+B)emx
A + B however is equivalent to a single constant
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A + B however is equivalent to a single constant and second order equations need two
With a little further searching we find that y = Bxemx is a solution. So a general solution is
mx mxy Ae Bxe
Roots are complex conjugates
( ) ( )p iq x p iq xy Ae Be px iqx px iqxAe e Be e
px iqx iqxe Ae Be We know that cos sinie i
When the roots of the auxiliary equation are complex, they will be of the form m1 = p + iq and m2 = p – iq.
Hence the general equation will be
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cos sin cos( ) sin( )pxe A qx i qx B qx i qx
cos sin cos sinpxe A qx i qx B qx i qx
cos sinpxe A B qx A B i qx
cos sinpxe C qx D qx
Where and ( )C A B D A B i
2
2 ( )d y dy
a b cy Q xdx dx
2
2 ( )d g dg
a b cg Q xdx dx
Non-homogeneousSecond-order lineardifferential equations
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Non homogeneous equations take the form
Suppose g(x) is a particular solution to this equation. Then
2
2
( ) ( )( ) ( )
d g k d g ka b c g k Q x
dx dx
Now suppose that g(x) + k(x) is another solution. Then
Non homogeneoussecond order differential equations
Giving
2 2
2 2 ( )d g d k dg dk
a a b b cg ck Q xdx dx dx dx
2 2
2 2 ( )d g dg d k dk
a b cg a b ck Q xdx dx dx dx
2
2( ) ( )d k dk
Q x a b ck Q xdx dx
2d k dk
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2
2 0d k dk
a b ckdx dx
From the work in previous exercises we know how to find k(x).
This function is referred to as the Complementary Function. (CF)
The function g(x) is referred to as the Particular Integral. (PI)
General Solution = CF + PI
22ndnd Order DEOrder DE –– Homogeneous LE withHomogeneous LE withConstant CoefficientsConstant Coefficients(2) If 1 and 2 are distinct real numbers (if b2 - 4ac > 0), then the general solution is:
xx ececy 2121
(3) If 1 = 2 (if b2 - 4ac = 0), then the general solution is:
xx xececy 1121
xececy 21
(4) If 1 and 2 are complex numbers (if b2 - 4ac < 0), then the general solution is:
xecxecy xx sincos 21
Where:
a
bac
a
b
2
4 and
2
2
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22ndnd Order DEOrder DE –– Homogeneous LE withHomogeneous LE withConstant CoefficientsConstant CoefficientsHomogeneous Linear Equations with Constant Coefficients
A second order homogeneous equation with constant coefficients is written as: 0 0 acyybyawhere a, b and c are constant
The steps to follow in order to find the general solution is as follows: The steps to follow in order to find the general solution is as follows:
(1) Write down the characteristic equation
0 02 acba This is a quadratic. Let 1 and 2 be its roots we have
a
acbb
2
42
2,1
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TYPE-1 PARTICULAR INTEGRALS f(x) = eax+b
Solve the equation
coshx=1)y+2D-(D2
The given differential equation is ( =Coshx
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The auxiliary equation ism2-2m + 1=0i.e. (m-1)2 =0 i.e. m = 1,1 . The roots are real and equal.The complementary function isCF =(A+Bx)ex
TYPE-1 PARTICULAR INTEGRALS f(x) = eax+b
2122 12212
cosh
integralparticular thefind tohaveNow we
+ PI = PI)D+-(D
+ee =
)D+-(D
xPI =
-xx
twice)0r denominato themakes1= DSince( )e(x
= e
=Now PIx2x
1
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twice)0r denominato themakes1= DSince( 4
= 1)+2D-2(D
=Now PI21
)-ting D = ( Substitu
e =
)D+-(D
e = PISimilarly
-x-x
1
8122 22
Now the general solution of the DE is GS = CF + PI1 + PI2 =
(A+Bx)ex +
is the solution of the given DE. 2. Solve (D2-2D+2) y = ex + 5 + e-2x
Solution: The given differential equation is(D2-2D+2) y = ex + 5 + e-2x
The auxiliary equation is
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The auxiliary equation is m2-2m + 2= 0 Solving for m , we get m = i.e. m = 1 ± i . The roots are complex conjugates. The complementary function is CF = (Acos x + B sin x)ex
2
4.2)-(4±2
)1(12)+2D-(D
e2
x
1 DSincee
PIx
)uting D = (Substit = )D+-D
e = PISimilarly
x
02
5
22(
52
0
2
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2) = Ding(Substitut/ 2
e=
2)+2D-(D
e=PI
2x
2
2x
3
DE.given theofsolution theis 2
e+
2
5+
1
e+x)esin B+x (Acos
= PI+ PI+ PI+CF=GS
solutiongeneralNow the
2xxx
321
Solve (D2-3D+2) y = 2cos(2x+3) Solution : The given differential equation is y = 2cos(2x+3) The auxiliary equation is m2- 3m + 2= 0 Solving for m , we get (m -1) (m-2) = 0
2)+3D-(D2
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Solving for m , we get (m -1) (m-2) = 0 i.e. m = 1 ,2 The roots are real and distinct The complementary function is CF = (A ex + B e2x )
Now we have to find the particular integral
PI =
4)- = D(since
= 2)+3D-(-4
3)+cos(2x(2 =
2)+3D-(D
3)+cos(2x(2 =
2
2
)x+(D)+(-)x+(D+(-)x+( 32cos23232cos2)3232cos2
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 17
) ) D-
)x+(D)+(-
D) +D)(--(-
)x+(D+(- =
D)-(-
)x+(294(
32cos232
3232
32cos2)32
32
32cos2
= )9.(-4)-((4
3)+cos(2x3D)(2+(-2
40
3)+sin(2x12-3)+cos(2x(-4
10
] 3)+sin(2x3-3)+[-cos(2x
DE.given theofsolution theis 10
] 3)+sin(2x3-3)+[-cos(2x+ )e B+e(A= PI+CF=GS 2xx
Solve (D2+1)2y = 2sinx cos3x
Solution: The given differential equation is
y = 2sinx cos3x
The auxiliary equation is
(m2+1)2 =0 Solving for m , we get
22 1)+(D
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(m2+1)2 =0 Solving for m , we get
(m2+1) (m2+1) = 0
i.e. m2 = -1 = i2 twice
Therefore m = ±i, ±i
The roots are pair of complex conjugates
The complementary function is
CF = (A+Bx)cos x +(C+Dx) sin x
Now we have to find the particular integral
2222 1
2sin4sin
1
)3cossin2 =
)+(D
xx- =
)+(D
xx(PI =
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21
2222 11
+ PIPI
)+(D)+(D
)16-= D(since
225
sin4x =
1)+(-16
sin4x=
1)+(D
sin4x = PI
2
2221
)4-= D(since 9
sin4x=
1)+(-4
sin4x =
PI
22
2
DE.given theof solution theis
9
sin2x+
225
sin4x+x sinDx)+(C+x Bx)cos+(A=PI+ PI+CF=GS 21
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42 )4( xyD
42 )4( xyD
xBxAFC 2sin2cos.. 11
24121
1641
4
1
2
4
42
x
xDD
Solve the DE
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 21
2
4
4 Dx
4
2
)4
1(
141
xD
41
2
)4
1(41
xD
23
341
1624
412
41
24
2
4
xx
xx
23
341
2sin2cos 24 xxxBxAPICFGS
2.Solve the DE x = y 2)+3D+(D 22
x =2)y +3D+(D
isequationaldifferentigivenThe:Solution22
distinct.andrealarerootsThe
. ,-21-=m i.e.
0= 2)+(m1)+(mget we,mfor Solving
0=2+3m+m
isequationauxiliary The2
)e B+e(A=CF
isfunctionary complementThe
.
2x-x-
x2 (9.2)]1
+6x)+(21
-[x1
=PI
)....]x)2
3D)+(D(+
2
3D)+(D-([1
2
1 =
2
2222
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 22
]2
3D)+(D+[1
2
1=
)2
3D)+(D+(1
2
1 =
3D)+D+(2
x=PI
(-1)2
1-2
2
2
]2
7)-6x-(x[
2
1=PI
(9)]2
1+3x)+(1-[x
2
1=PI
(9.2)]4
+6x)+(22
-[x2
=PI
2
2
2
7)-6x-(x
2
1+ )e B+e(A=
PI+CF=GS2
2x-x-
TYPE-4 PARTICULAR INTEGRALS
1.Solve (D2-2D+2) y = ex x2
xe=2)y +2D-(DisequationaldifferentigivenThe 2x2
2
4.2)-(4±(2=mget we,mfor Solving
0=2+2m-m
isequationauxiliary The2
x)esin B+x (Acos=CF
isfunctionary complementThex
1)+ D= D(Since
1)+2-2D-1+2D+(D
e =
2)+2D-(D
)x(e=PI
integralparticular thefind tohaveNow we
2
2x
2
2x x
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 23
.conjugatescomplex arerootsThe
. i ±1=m i.e.
2
=mget we,mfor Solving1)+ D= D(Since
12
xe=dx
3
xe=
3
x
D
1 e= dxx
D
1 e=
D
x e
4x
3x
3x2x
2
2x
equation.aldifferenti theofsolutionrequired theis 12
xe+x)esin B+x (Acos =PI+CF=GS
4xx
2. sinx e = y 3)+4D+(D -x2
0=3+4m+m
isequationauxiliary The
sinx e =3)y +4D+(D
isequationaldifferentigivenThe:Solution
2
x-2
)e B+e(A=CF
isfunctionary complementThe
distinct.andrealarerootsThe
. ,-31-=m i.e.
0= 3)+(m1)+(mget we,mfor Solving
3x-x-
1)- D= D(since 3)+1)-4(D+1)-((D
sinx)e =
3)+4D+(D
sinx)(e= PI
integralparticular thefind tohaveNow we
2
(-x)
2
(-x)
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 24
3)+4-4D+1+2D-(D
sinx)(e =
1)- D= D(since 3)+1)-4(D+1)-((D
=
2
(-x)
2
-1)= DSince( 5
2cosx))-(-sinxe=
2D)sinx -(-1)4D-(1
e=
2D))-2D)(-1+((-1
2D)sinx-)(-1(e =
-1)= DSince( 2D)+(-1
sinx)(e=
2D)+(D
sinx)(e=
2(-x)
2
(-x)(-x)
2(-x)
2
(-x)
equation.aldifferenti theofsolutionrequired theis 5
2cosx))-(-sinx(e+)e B+e(A
= PI+CF=GS(-x)
3x-x-
Solve (D3-1)y=x sinx
Solution: The given differential equation is (D3-1)y = x sinx
The auxiliary equation is
m3-1= 0
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 25
m3-1= 0
i.e (m-1)(m2 +m+1)=0
Solving for m , we get m =
i.e. m = 1,
2
)4.1)-(1±(-1
2
)4.1)-(1±(-1
x/2-)e)x (3/2sin( B+)x (3/2(Acos(=CF
isfunctionary complementThe
rootreala andconjugatescomplex ofpair a arerootsThe
sin4
13
2
sin1
sin1
13
1
sin1
sin11
13
11
sin1
13
1
sin
22
22
22
2
22
22
2
2
x)-D) (D(
)(-
x))-x(D-=
x )-D(
)-D) (D(
))-((D
x))(-x(D-=
x) (D+-D)(
)-D) (D(-
)))(D-(D+
x))(-x(D-=
)(-D-)D(
)-(-D-
xx
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 26
4
cos23
2
sincos4
cos2sinsin3
2
sincos
42
x)))(((-
x))x-(x(=
x))(-x-x+(-(-
x))x-(x( =
)(-
equation.aldifferenti theofsolutionrequired theis2
3cosx)-sinx)-(x(cosx+)ex 3/2sin B+x 3/2(Acos(
= PI+CF=GS
x/2-
)e CF =(A+Bx
nction ismentary fuThe comple
l.l and equats are rea . The roo,i.e. m =
=)i.e. (m-
=m + - m
on isary equatiThe auxili
x )y = x e D+-on is (Dial equati different The givenSolution:
x )y = x e D+Solve (D
x
x
x-
11
01
012
sin12
sin12
2
2
2
2
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 27
) xdxx(D
e =
x]) ([xD
e =
)-D+
x] [xe =
)D-
xxePI =
egralular the partice to find Now we hav
x
x
x
x
sin
sin
11(
sin
1(
sin
int
2
2
2
uationrential eq the diffeolution ofrequired sx) is the x+ (-x +e(A+Bx)e
F +PI = GS = C
x) x+ (-x e=
dx x)x+(x [-x e=
)dx xx+(-xe =
x)) x+ (-x(eD
=
xx
x
x
x
x
cos2sin
cos2sin
]sinsinsin
sincos
sincos1
equal.andrealarerootsThe.1,1=m i.e.
0=1)-(m i.e.
0=1+2m-m
isequationauxiliary The
x loge=1)y +2D-(DisequationaldifferentigivenThe:Solution
x loge=1)y +2D-(D1.
2
2
x2
x2
1)-(D
dx)elogx ee=
1-D1-D
logxe
=
1)-(D
logx)(e=PI
(-x)xx
x
2
x
dx)Xe(e=Xa-D
1 ax-ax
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 28
Bx)e+(A=CF
isfunctionary complementThe
equal.andrealarerootsThe.1,1=m i.e.
x
dxx)]-[(xlogxe=
dx]ex)-(xlogx[ee=
1)-(D
x))-(xlogx(e=
x
(-x)xx
x
3]-[2logx4
ex=
]3x-logx[2x4
e=
]4
3x-logx
2
x[e=
]2
x-
4
x-logx
2
x[e=
]2
x-dx
x
2
1logx
2
x[e=
x2
22x
22x
222x
222x
x
)e B+e(A=CF
isfunctionary complementThe
distinct.andrealarerootsThe
2,-1-=m i.e.
0=1)+2)(m+(mget we,mfor Solving
0=2+3m+ m
isequationauxiliary The
e=2)y +3D+(D
isequationaldifferentigivenThe :Solution
e=2)y +3D+2.(D
x-2x-
2
)(e2
)(e2
x
x
21
2
2
1
1
123
int
-PI=PI
eD+
eD+
D+ +D
ePI =
egralular the partice to find Now we hav
xx
x
ee
e
xt(-x)
xex-e1
xx
e= t wheredtee =
dxeee= e1D
1=PI
dx.e.eee=e2+D
1 =PI xxe2x-)(e
2
xx
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)(e(-x)t(-x)
xt(-x)
x
ee=ee=
e= t wheredtee =
1)-.(e.ee=
))e-.(e(ee=)e-(tee=
e= t wheredttee=2+D
x)(e2x-
)(ex)(e(-2x)tt(-2x)
xt(-2x)
x
xx
equation.aldifferenti theofsolutionrequired theistion)simplifica(after .ee+)e B+e(A=
1)-.(e.ee-ee+)e B+e(A
= PI+CF=GS
)(e(-2x)x-2x-
x)(e(-2x))(e(-x)x-2x -
x
xx
)
isfunctionary complementThe
conjugatescomplex arerootsThe
2i ±=m i.e.
0=2i)-2i)(m+(mget we,mfor Solving
4i=4- =m 0,=4+m
isequationauxiliary The
cos2xx =4)y +(D
isequationaldifferentigivenThe:Solution
cos2xx =4)y + (Dequation theSolve
222
22
22
2ix
2
22ix
22
22ix
2
22ix
2
i2x2
2
2
1e
4iD)+(D
xe ofpart Real=
4)+4i+4iD+(D
xe ofpart Real=
4)+2i)+((D
xeofpart Real=
4)+(D
exofpart Real=
4)+(D
cos2xx=PI
integralparticular thefind tohaveNow we
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 30
Bsin2x)+cos2x (A=CF 2(-1)2ix
x)
4iD
+1
1(
4iD
e ofpart Real=
Contd.,
)dx8
1 -
2i
x-(x(
4i
e ofpart Real=
)8
1 -
2i
x-(x
4iD
eofpart Real=
)16i
2+
4i
2x-(x
4iD
eofpart Real=
x)16
D+
4i
D-(1
4iD
e ofpart Real=
22ix
22ix
2
22ix
2(-1)
2
22ix
i
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 31
equation.aldifferenti theofsolutionrequired theiscos2x 4
x - )
8
x -
3
x(
4
sin2x)( +
Bsin2x)+cos2x (A= PI+CF=GS
cos2x4
x - )
8
x -
3
x(
4
sin2x)( =
x/8) - 4i
x-
3
x(
4i
e ofpart Real=
)dx8
- 2i
-(x(4i
ofpart Real=
23
23
232ix
1. Solve (D2-2D+2) y = ex cosx
Solution: The given differential equation is
(D2-2D+2) y = ex cos x
The auxiliary equation is
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 32
m2-2m + 2= 0
Solving for m , i.e. m = 1 ± i .
The roots are complex conjugates.
The complementary function is
CF = (Acos x + B sin x)ex
EXTRA PROBLEMS)1(
2)1(2)1(2)+2D-(D
cose22
x
1
DDSinceDD
exPI
x
)22212(
cos21
DDD
xePI
x
2
)sin(x)esin Bx (Acos x xxe
PIisCFsolutiongeneralThex
2
)sin( xxePI
x
)1(
cos21
D
xePI
x
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2. Solve (D2-3D+2) y = 2cos(2x+3) + 2 ex
Solution : The given differential equation is (D2-3D+2 ) y = 2cos(2x+3) + 2ex
The auxiliary equation is
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 34
The auxiliary equation is m2- 3m + 2= 0 Solving for m , we get (m -1) (m-2) = 0 i.e. m = 1 ,2 The roots are real and distinct The complementary function is CF = (A ex + B e2x )
Now we have to find the particular integralPI1 =
4)- = D(since
= 2)+3D-(-4
3)+cos(2x(2 =
2)+3D-(D
3)+cos(2x(2 =
2
2
) ) D-
)x+(D)+(-
D) +D)(--(-
)x+(D+(- =
D)-(-
)x+(294(
32cos232
3232
32cos2)32
32
32cos2
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 35
)9.(-4)-((4
3)+cos(2x3D)(2+(-2
40
3)+sin(2x12-3)+cos(2x(-4
10
] 3)+sin(2x3-3)+[-cos(2x
DE.given theofsolution theis 10
] 3)+sin(2x3-3)+[-cos(2x+ )e B+e(A=PI PI+CF=GS 2xx
21
)()21(
2
)2)(1(
22 caseexception
xe
DD
ePI
xx
3. Solve the differential equation
(D2 +4D+3)y= 6e-2x sinx sin 2x
Solution The given DE is (D2 +4D+3)y= 6e-2x
sinx sin 2x
The AE is m2 +4m +3 =0 i.e. m=-1, -3
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 36
The AE is m2 +4m +3 =0 i.e. m=-1, -3
The complementary function is CF = Ae-x +Be-3x
1
)cos3(cos3
38444
)2sinsin2(3
3)2(4)2(
)2sinsin2(3
)34(
2sinsin62
2
2
2
2
2
2
2
D
xxe
DDD
xxe
DD
xxe
DD
xxePI
xxxx
11
cos3
19
3cos3
1
)cos3(cos3 22
2
2
xexe
D
xxePI
xxx
2
cos3
10
3cos3 BeAe
223x-x- xexe
PICFGSxx
4.Solve the equation (D2 +5D+4)y = e-x sin2x
Solution: The differential equation is
(D2 +5D+4)y = e-x sin2x
The auxiliary equation is m2 + 5m +4 =0
Solving for m, we get m = -1, -4.
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 37
Solving for m, we get m = -1, -4.
The complementary function is
CF = Ae-x+Be-4x
D
xe
DD
xe
DDD
xe
DD
xe
DD
xePI
xxxxx
54
2sin
5
2sin
14552
2sin
4)1(5)1(
2sin
45
2sin2222
116
)2cos102sin4(
)4(2516
)2cos102sin4(
2516
2sin)54(
)54)(54(
2sin)54(2
xxexxe
D
xDe
DD
xDePI
xx
xx
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 38
116)4(2516
116
)2cos102sin4(4 xxeBeAePICFGS
xxx
EXTRA PROBLEMS
)e CF =(A+Bx
nction ismentary fuThe comple
l.l and equats are rea . The roo,i.e. m =
=)i.e. (m-
=m + - m
on isary equatiThe auxili
x x e )y = D+-on is (Dial equati different The givenSolution:
x x e )y = D+Solve (D
x
x
x-
11
01
012
sin812
sin812.5
2
2
2
2
) xdxx(D
e =
x]) ([xD
e =
)-D+
x] [xe =
)D-
xxePI =
egralular the partice to find Now we hav
x
x
x
x
sin8
sin8
11(
sin8
1(
sin8
int
2
2
2
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 39
uationrential eq the diffeolution ofrequired sx) is the x+ (-xe +(A+Bx)e
F +PI = GS = C
x) x+ (-xe =
dx x)x+(x [-xe =
)dx xx+(-xe =
x)) x+ (-xe(D
=
xx
x
x
x
x
cos2sin8
cos2sin8
]sinsinsin8
sincos8
sincos81
6. Solve the differential equation (D2 -4D+3)y= ex cos 2xSolution The given DE is (D2 -4D+3)y= ex cos 2xThe AE is m2 +4m +3 =0 i.e. m=1, 3The complementary function is CF = Aex +Be3x
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 40
8
)2sin2(cos
32
)2sin42cos4(
)4(416
)2)(cos24(
)416(
)2)(cos24(
24
)2(cos
2
)2(cos22
xxe
xxexDe
D
xDe
D
xe
DD
xePI
x
xx
xxx
DD
xe
DDD
xe
DD
xe
DD
xePI
xxxx
2
2cos
34412
2cos
3)1(4)1(
2cos
34
)2(cos2222
8
)2sin2(cos BeAe 3xx xxe
PICFGSx
The complementary function is CF = Ae +Be
7. Solve the differential equation
(D2 +3D+2)y= sin x + x2
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 41
(D +3D+2)y= sin x + x
Solution The given DE is (D2 +3D+2)y= sin x + x2
The AE is m2 +3m +2=0 i.e. m=-1, -2
The complementary function is CF = Ae-x +Be-2x
EXTRA PROBLEMS
19
sin)13(
)13)(13(
sin)13(
13
sin
231
sin
23
sin221
D
xD
DD
xD
D
x
D
x
DD
xPI
10
sincos3
1)1(9
sin)13(1
xxxD
PI
)2
3D)+(D+(1
2
1 =
3D)+D+(2
x=PI
1-2
2
2
2
11
(9.2)]4
1+6x)+(2
2
1-[x
2
1=PI
)....]x)2
3D)+(D(+
2
3D)+(D-([1
2
1 =
2
2
2222
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 42
]2
3D)+(D+[1
2
1=
)2
+(12
=
(-1)2
]2
7)-6x-(x[
2
1=PI
(9)]2
1+3x)+(1-[x
2
1=PI
2
2
2
2
2
7)-6x-(x
2
1
10
sincos3+ )e B+e(A=
PI PI+CF=GS2
2x-x-
21
xx
8. Solve the differential equation
(D2 +16)y = cos 3 x
Solution The given DE is (D2 +16)y= cos 3 x
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 43
Solution The given DE is (D2 +16)y= cos 3 x
The AE is m2 +16=0 i.e. m = ±4i
The complementary function is CF = Acos4x +Bsin4x
7
3cos
169
3cos
16
3cos
20
cos
)15(4
cos3
)161(4
cos3
)16(4
cos3
)16(4
3coscos3
)16(
cos
22
21
22
3
xx
D
xPI
xxx
D
xPI
D
xx
D
xPI
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 44
7
3cos
20
cos4sin4cos21
xxxBxAPIPICFGS
2/x-
3
)2
3sin
2
3(BcosAe CF
root.reala andconjugatescomplex arerootsThe
. 2
3i±11,-=m i.e.
-1m,2
4)-(1±1=mget we,mfor Solving
0=1m
isequationauxiliary The
xexCx
1.Solve the DE(D3+1)y =0
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 45
1243133
22)2(42423232
22
2
22
2
22
2
22
xexedxxex
D
e
dxxD
e
D
xe
D
xe
D
exPI
xxxx
xxxx
2.Find the particular integral of (D2-4D+4)y
=x2e2x
EXTRA PROBLEMS-PARTA
3.Find the particular integral of (D2+4)y =sin2x
)(4
2cos
)4(
2sin2
caseExceptionxx
D
xPI
4.Find the particular integral of )1(sin1
sinsin
11
sin
)1(
sin 2
222
Dce
xe
D
xe
D
xe
D
exPI
xxxx
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 46
4.Find the particular integral of (D-1)2y = exsinx
111)1( 222 DDD
5.Find the particular integral of (D-1)2y = coshx
84)1(2)1(212
)1(
cosh 2
2222
xxxx
xx
eex
D
e
D
e
D
ee
D
xPI
EXTRA PROBLEMS-PARTA
6.Find the particular integral of (D2-4)y = cosh2x
7.Find the particular integral of
8. Find the particular integral of
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 47
integral of (D2-4)y =1
integral of (D-2)2y = 2x
9.Find the particular integral of
(D+1)2y =e-xcosx
EXTRA PROBLEMS-PARTA
4
2sinh
88)2)(2(2)2)(2(242
)4(
2cosh.6
2222
2
22
2
xxxexe
DD
e
DD
e
D
ee
D
xPI
xxxx
xx
4
1
2,2,,04
4
1
44)4(
1.7
22
22
2
0
2
0
2
xx
xx
xx
BeAePICFGS
BeAeCF
mgetwemforsolvingmisAEThe
e
D
e
DPI
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 48
4 BeAePICFGS
)()22(log2)2(
2.8 log
2
2log
2
2log
2
axxxxx
eaSincee
D
e
DPI
)1(sin1
coscos
)11(
cos
)1(
cos.9 2
222
ceDxe
D
xe
D
xe
D
xePI
xxxx
EXTRA PROBLEMSSolve the equation (D2+2D+5)y = ex cos3 x
Solve the equation (D2+2D-1)y = (x+ex )2
Find the particular integral of (D2+a2 )y = b cos ax + c sin ax
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 49
SOLUTIONS(PART-A & B)
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 50