Post on 29-Oct-2019
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 20
d f df d g dga 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 0mxdy
y Ae A m b cydx
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 Ame
dx
22
2mxd y Am 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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 6
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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 7
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
22 ( )
d g dga b cg Q x
dx dx
Non-homogeneousSecond-order lineardifferential equations
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 8
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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 9
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 21 21
(3) If 1 = 2 (if b2 - 4ac = 0), then the general solution is:
xx xececy 11 21 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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 10
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
422,1
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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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 12
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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 13
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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 14
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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 15
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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 16
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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 18
(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 =
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 19
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
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 20
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-)(e2
xx
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 29
)(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 DDSince
DD
exPI
x
)22212(
cos21
DDD
xePI
x
2
)sin(x)esin Bx (Acos x
xxePIisCFsolutiongeneralThe
x
2
)sin( xxePI
x
)1(cos21
D
xePI
x
12/23/2014SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR 33
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 2xx21
)()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- xexePICFGS
xx
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 xxeBeAePICFGSx
xx
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
xxePICFGS
x
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(sin1sinsin
11
sin
)1(
sin 2222
Dcexe
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
42sinh
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 log2
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)
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UNIT 2
-P.VEERAIAH
DEPARTMENT OF APPLIED MATHEMATICS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
1
ORDINARY DIFFERENTIAL EQUATIONSMATHEMATICS II(MA6251)
1
UNIT 2 SYLLABUS
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
2
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 constant coefficients
Second-order linear differential equations
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
3
Note that the general solution to such an equation must include two arbitrary constants to be completely general.
Second-order linear differential equations
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4
Adding
Second-order linear differential equations
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
5
The solutions to this quadratic will provide two values of m which will make y = Aemx a solution.
Second-order linear differential equations
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
6
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 and second order equations need two
With a little further searching we find that y = Bxemx is a solution. So a general solution is
Roots are complex conjugates
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
7
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
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
8
Non homogeneous equations take the form
Suppose g(x) is a particular solution to this equation. Then
Now suppose that g(x) + k(x) is another solution. Then
Non-homogeneous Second-order linear differential equations
Non homogeneoussecond order differential equations
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
9
Giving
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
2nd Order DE – Homogeneous LE with Constant Coefficients
(2) If 1 and 2 are distinct real numbers (if b2 - 4ac > 0), then the general solution is:
(3) If 1 = 2 (if b2 - 4ac = 0), then the general solution is:
(4) If 1 and 2 are complex numbers (if b2 - 4ac < 0), then the general solution is:
Where:
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
10
2nd Order DE – Homogeneous LE with Constant Coefficients
Homogeneous Linear Equations with Constant Coefficients
A second order homogeneous equation with constant coefficients is written as:
where a, b and c are constant
The steps to follow in order to find the general solution is as follows:
(1) Write down the characteristic equation
This is a quadratic. Let 1 and 2 be its roots we have
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
11
Type-1 particular integrals f(x) = eax+b
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
12
Solve the equation
The auxiliary equation is
m2-2m + 1=0
i.e. (m-1)2 =0
i.e. m = 1,1 . The roots are real and equal.
The complementary function is
CF =(A+Bx)ex
The given differential equation is (
=Cosh x
Type-1 particular integrals f(x) = eax+b
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
13
Now the general solution of the DE is GS = CF + PI1 + PI2 =
(A+Bx)ex +
Type-1 particular integrals
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
14
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
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
Type-1 particular integrals
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
15
Type-2 particular integralsf(x) = cos(ax+b) or sin (ax+b)
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
16
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
i.e. m = 1 ,2
The roots are real and distinct
The complementary function is
CF = (A ex + B e2x )
Type-2 particular integrals
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
17
Now we have to find the particular integral
PI =
Type-2 particular integrals
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
18
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
(m2+1) (m2+1) = 0
i.e. m2 = -1 = i2 twice
Therefore m = ±i, ±i
The roots are pair of complex conjugates
Type-2 particular integrals
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
19
The complementary function is
CF = (A+Bx)cos x +(C+Dx) sin x
Now we have to find the particular integral
Type-2 particular integrals
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
20
Type-3 particular integrals f(x) = xm ( m a positive integer)
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
21
Solve the DE
Type-3 particular integrals
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
22
2.Solve the DE
Type-4 particular integrals
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
23
1.Solve (D2-2D+2) y = ex x2
Type-4 particular integrals f(x) = eaxV(x)
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
24
2.
Type-5 particular integrals f(x) = xV(x)
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
25
Solve (D3-1)y=x sinx
Solution: The given differential equation is (D3-1)y = x sinx
The auxiliary equation is
m3-1= 0
i.e (m-1)(m2 +m+1)=0
Solving for m , we get m =
i.e. m = 1,
Type-5 particular integrals
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
26
Type-5 particular integrals
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
27
Type-6 particular integrals
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28
Type-6 particular integrals
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
29
MISCELLANEOUS MODEL (using Euler’s formula)
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
30
MISCELLANEOUS MODEL
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
31
Contd.,
EXTRA PROBLEMS
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SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
32
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
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
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
33
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
34
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
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 )
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
35
Now we have to find the particular integral
PI1 =
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
36
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
The complementary function is CF = Ae-x +Be-3x
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
37
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.
The complementary function is
CF = Ae-x+Be-4x
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
38
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
39
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
40
6. Solve the differential equation
(D2 -4D+3)y= ex cos 2x
Solution The given DE is (D2 -4D+3)y= ex cos 2x
The AE is m2 +4m +3 =0 i.e. m=1, 3
The complementary function is CF = Aex +Be3x
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
41
7. Solve the differential equation
(D2 +3D+2)y= sin x + x2
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
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
42
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
43
8. Solve the differential equation
(D2 +16)y = cos 3 x
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
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
44
EXTRA PROBLEMS-PARTA
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
45
1.Solve the DE
(D3+1)y =0
2.Find the particular integral of (D2-4D+4)y =x2e2x
EXTRA PROBLEMS-PARTA
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
46
3.Find the particular integral of
(D2+4)y =sin2x
4.Find the particular integral of
(D-1)2y = exsinx
5.Find the particular integral of
(D-1)2y = coshx
EXTRA PROBLEMS-PARTA
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
47
6.Find the particular integral of
(D2-4)y = cosh2x
7.Find the particular integral of
(D2-4)y =1
8. Find the particular integral of
(D-2)2y = 2x
9.Find the particular integral of
(D+1)2y =e-xcosx
EXTRA PROBLEMS-PARTA
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
48
EXTRA PROBLEMS
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
49
Solve 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
SOLUTIONS(PART-A & B)
12/23/2014
SVCE, DEPARTMENT OF APPLIED MATHEMATICS, SRIPERUMBUDUR
50
2
2
Differential equations of the form ()
are called second order linear different
ial equations.
dydy
abcyQx
dxdx
++=
When ()0 then the equations are referred
to as homogeneous,
Qx
=
When ()0 then the equations are non-homo
geneous.
Qx
¹
Theorem
If () and () are two solutions then so i
s ()()
yfxygxyfxgx
===+
2
2
we have 0
dfdf
abcf
dxdx
++=
2
2
and 0
dgdg
abcg
dxdx
++=
22
22
0
dfdfdgdg
abcfabcg
dxdxdxdx
+++++=
(
)
22
22
0
dfdgdfdg
abcfg
dxdxdxdx
æö
æö
+++++=
ç÷
ç÷
èø
èø
And so ()() is a solution to the differe
ntial equation.
yfxgx
=+
, for and , is a solution to the equati
on0
mx
dy
yAeAmbcy
dx
=+=
2
0
ambmc
++=
It is reasonable to consider it as a pos
sible solution for
2
2
0
dydy
abcy
dxdx
++=
mx
yAe
=
mx
dy
Ame
dx
Þ=
2
2
2
mx
dy
Ame
dx
Þ=
If is a solution it must satisfy
mx
yAe
=
2
0
mxmxmx
aAmebAmecAe
++=
assuming 0, then by division we get
mx
Ae
¹
mxmx
yAeBxe
=+
Where and ()
CABDABi
=+=-
()()
piqxpiqx
yAeBe
+-
=+
pxiqxpxiqx
AeeBee
-
=+
(
)
pxiqxiqx
eAeBe
-
=+
We know that cossin
i
ei
q
=+
(
)
(
)
(
)
cossincos()sin()
px
eAqxiqxBqxiqx
=++-+-
(
)
(
)
(
)
cossincossin
px
eAqxiqxBqxiqx
=++-
(
)
(
)
(
)
cossin
px
eABqxABiqx
=++-
(
)
cossin
px
eCqxDqx
=+
2
2
()
dydy
abcyQx
dxdx
++=
2
2
()
dgdg
abcgQx
dxdx
++=
2
2
()()
()()
dgkdgk
abcgkQx
dxdx
++
+++=
22
22
()
dgdkdgdk
aabbcgckQx
dxdxdxdx
+++++=
22
22
()
dgdgdkdk
abcgabckQx
dxdxdxdx
æöæö
Þ+++++=
ç÷ç÷
èøèø
2
2
()()
dkdk
QxabckQx
dxdx
æö
Þ+++=
ç÷
èø
2
2
0
dkdk
abck
dxdx
Þ++=
x
x
e
c
e
c
y
2
1
2
1
l
l
+
=
x
x
xe
c
e
c
y
1
1
2
1
l
l
+
=
(
)
(
)
x
e
c
x
e
c
y
x
x
b
b
a
a
sin
cos
2
1
+
=
a
b
ac
a
b
2
4
and
2
2
-
=
-
=
b
a
(
)
0
0
¹
=
+
¢
+
¢
¢
a
cy
y
b
y
a
(
)
0
0
2
¹
=
+
+
a
c
b
a
l
l
a
ac
b
b
2
4
2
2
,
1
-
±
-
=
l
coshx
=
1)y
+
2D
-
(D
2
2
1
2
2
1
2
2
1
2
cosh
integral
particular
the
find
to
have
Now we
+ PI
= PI
)
D+
-
(D
+e
e
=
)
D+
-
(D
x
PI =
-x
x
twice)
0
r
denominato
the
makes
1
=
D
Since
(
4
)
e
(x
=
1)
+
2D
-
2(D
e
=
Now PI
x
2
2
x
1
)
-
ting D =
( Substitu
e
=
)
D+
-
(D
e
=
PI
Similarly
-x
-x
1
8
1
2
2
2
2
2
4.2)
-
(4
±
2
)
1
(
1
2)
+
2D
-
(D
e
2
x
1
=
=
=
D
Since
e
PI
x
)
uting D =
(Substit
=
)
D+
-
D
e
=
PI
Similarly
x
0
2
5
2
2
(
5
2
0
2
2)
=
D
ing
(Substitut
/
2
e
=
2)
+
2D
-
(D
e
=
PI
2x
2
2x
3
DE.
given
the
of
solution
the
is
2
e
+
2
5
+
1
e
+
x)e
sin
B
+
x
(Acos
=
PI
+
PI
+
PI
+
CF
=
GS
solution
general
Now the
2x
x
x
3
2
1
2)
+
3D
-
(D
2
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+
(
2
9
4
(
3
2
cos
2
3
2
3
2
3
2
3
2
cos
2
)
3
2
3
2
3
2
cos
2
=
?)/10
]
3)
+
sin??(2x
3
-
3)
+
([-cos?(2x
=
?)/40
3)
+
sin??(2x
12
-
3)
+
cos?(2x
((-4
?)/10
]
3)
+
sin??(2x
3
-
3)
+
([-cos?(2x
=
?)/40
3)
+
sin??(2x
12
-
3)
+
cos?(2x
((-4
=
)
9.(-4)
-
?)/((4
3)
+
cos??(2x
3D)2
+
((-2
==
)
9.(-4)
-
?)/((4
3)
+
cos??(2x
((2
3D)
+
-2
=
)
9.(-4)
-
((4
3)
+
cos(2x
3D)(2
+
(-2
40
3)
+
sin(2x
12
-
3)
+
cos(2x
(-4
10
]
3)
+
sin(2x
3
-
3)
+
[-cos(2x
DE.
given
the
of
solution
the
is
10
]
3)
+
sin(2x
3
-
3)
+
[-cos(2x
+
)
e
B
+
e
(A
=
PI
+
CF
=
GS
2x
x
2
2
1)
+
(D
2
1
2
2
2
2
1
2
sin
4
sin
1
)
3
cos
sin
2
+ PI
PI
=
)
+
(D
x
x-
=
)
+
(D
x
x
(
PI =
)
16
-
=
D
(since
225
sin4x
=
1)
+
(-16
sin4x
=
1)
+
(D
sin4x
=
PI
2
2
2
2
1
)
4
-
=
D
(since
9
sin4x
=
1)
+
(-4
sin4x
=
PI
2
2
2
DE.
given
the
of
solution
the
is
9
sin2x
+
225
sin4x
+
x
sin
Dx)
+
(C
+
x
Bx)cos
+
(A
=
PI
+
PI
+
CF
=
GS
2
1
4
2
)
4
(
x
y
D
=
+
4
2
)
4
(
x
y
D
=
+
x
B
x
A
F
C
2
sin
2
cos
.
.
+
=
2
4
4
D
x
+
4
2
)
4
1
(
1
4
1
x
D
+
4
1
2
)
4
1
(
4
1
x
D
-
+
÷
ø
ö
ç
è
æ
+
-
=
÷
ø
ö
ç
è
æ
+
-
=
÷
ø
ö
ç
è
æ
+
-
=
2
3
3
4
1
16
24
4
12
4
1
16
4
1
4
1
2
4
2
4
4
4
2
x
x
x
x
x
D
D
÷
ø
ö
ç
è
æ
+
-
+
+
=
+
=
2
3
3
4
1
2
sin
2
cos
2
4
x
x
x
B
x
A
PI
CF
GS
x
=
y
2)
+
3D
+
(D
2
2
x
=
2)y
+
3D
+
(D
is
equation
al
differenti
given
The
:
Solution
2
2
distinct.
and
real
are
roots
The
.
,-2
1
-
=
m
i.e.
0
=
2)
+
(m
1)
+
(m
get
we
,
m
for
Solving
0
=
2
+
3m
+
m
is
equation
auxiliary
The
2
)
e
B
+
e
(A
=
CF
is
function
ary
complement
The
.
2x
-
x
-
]
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
1
+
6x)
+
(2
2
1
-
[x
2
1
=
PI
)
....]x
)
2
3D)
+
(D
(
+
2
3D)
+
(D
-
([1
2
1
=
2
2
2
2
2
2
2
+
2
7)
-
6x
-
(x
2
1
+
)
e
B
+
e
(A
=
PI
+
CF
=
GS
2
2x
-
x
-
x
e
=
2)y
+
2D
-
(D
is
equation
al
differenti
given
The
2
x
2
.
conjugates
complex
are
roots
The
.
i
±
1
=
m
i.e.
2
4.2)
-
(4
±
(2
=
m
get
we
,
m
for
Solving
0
=
2
+
2m
-
m
is
equation
auxiliary
The
2
x)e
sin
B
+
x
(Acos
=
CF
is
function
ary
complement
The
x
1)
+
D
=
D
(Since
1)
+
2
-
2D
-
1
+
2D
+
(D
e
=
2)
+
2D
-
(D
)
x
(e
=
PI
integral
particular
the
find
to
have
Now we
2
2
x
2
2
x
x
12
x
e
=
dx
3
x
e
=
3
x
D
1
e
=
dx
x
D
1
e
=
D
x
e
4
x
3
x
3
x
2
x
2
2
x
ò
ò
equation.
al
differenti
the
of
solution
required
the
is
12
x
e
+
x)e
sin
B
+
x
(Acos
=
PI
+
CF
=
GS
4
x
x
sinx
e
=
y
3)
+
4D
+
(D
-x
2
0
=
3
+
4m
+
m
is
equation
auxiliary
The
sinx
e
=
3)y
+
4D
+
(D
is
equation
al
differenti
given
The
:
Solution
2
x
-
2
)
e
B
+
e
(A
=
CF
is
function
ary
complement
The
distinct.
and
real
are
roots
The
.
,-3
1
-
=
m
i.e.
0
=
3)
+
(m
1)
+
(m
get
we
,
m
for
Solving
3x
-
x
-
3)
+
4
-
4D
+
1
+
2D
-
(D
sinx)
(e
=
1)
-
D
=
D
(since
3)
+
1)
-
4(D
+
1)
-
((D
sinx)
e
=
3)
+
4D
+
(D
sinx)
(e
=
PI
integral
particular
the
find
to
have
Now we
2
(-x)
2
(-x)
2
(-x)
-1)
=
D
Since
(
5
2cosx))
-
(-sinx
e
=
2D)sinx
-
(-1
)
4D
-
(1
e
=
2D))
-
2D)(-1
+
((-1
2D)sinx
-
)(-1
(e
=
-1)
=
D
Since
(
2D)
+
(-1
sinx)
(e
=
2D)
+
(D
sinx)
(e
=
2
(-x)
2
(-x)
(-x)
2
(-x)
2
(-x)
equation.
al
differenti
the
of
solution
required
the
is
5
2cosx))
-
(-sinx
(e
+
)
e
B
+
e
(A
=
PI
+
CF
=
GS
(-x)
3x
-
x
-
2
)
4.1)
-
(1
±
(-1
2
)
4.1)
-
(1
±
(-1
x/2
-
)e
)x
(3/2
sin(
B
+
)x
(3/2
(Acos(
=
CF
is
function
ary
complement
The
root
real
a
and
conjugates
complex
of
pair
a
are
roots
The
4
cos
2
3
2
sin
cos
4
cos
2
sin
sin
3
2
sin
cos
sin
4
1
3
2
sin
1
sin
1
1
3
1
sin
1
sin
1
1
1
3
1
1
sin
1
1
3
1
sin
2
2
2
2
2
2
2
2
2
2
2
2
2
x)))
(
(
(
-
x))
x-
(x(
=
x))
(-
x-
x+
(-
(
-
x))
x-
(x(
=
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-
x
x
-
-
equation.
al
differenti
the
of
solution
required
the
is
2
3cosx)
-
sinx)
-
(x(cosx
+
)e
x
3/2
sin
B
+
x
3/2
(Acos(
=
PI
+
CF
=
GS
x/2
-
)e
CF =(A+Bx
nction is
mentary fu
The comple
l.
l and equa
ts are rea
. The roo
,
i.e. m =
=
)
i.e. (m-
=
m +
-
m
on is
ary equati
The auxili
x
)y = x e
D+
-
on is (D
ial equati
different
The given
Solution:
x
)y = x e
D+
Solve (D
x
x
x
-
1
1
0
1
0
1
2
sin
1
2
sin
1
2
2
2
2
2
)
xdx
x
(
D
e
=
x])
([x
D
e
=
)
-
D+
x]
[x
e
=
)
D-
x
xe
PI =
egral
ular
the partic
e to find
Now we hav
x
x
x
x
ò
sin
sin
1
1
(
sin
1
(
sin
int
2
2
2
uation
rential eq
the diffe
olution of
required s
x) is the
x+
(-x
+e
(A+Bx)e
F +PI =
GS = C
x)
x+
(-x
e
=
dx
x)
x+
(
x
[-x
e
=
)dx
x
x+
(-x
e
=
x))
x+
(-x
(e
D
=
x
x
x
x
x
x
cos
2
sin
cos
2
sin
]
sin
sin
sin
sin
cos
sin
cos
1
ò
ò
-
Bx)e
+
(A
=
CF
is
function
ary
complement
The
equal.
and
real
are
roots
The
.
1,1
=
m
i.e.
0
=
1)
-
(m
i.e.
0
=
1
+
2m
-
m
is
equation
auxiliary
The
x
log
e
=
1)y
+
2D
-
(D
is
equation
al
differenti
given
The
:
Solution
x
log
e
=
1)y
+
2D
-
(D
1.
x
2
2
x
2
x
2
dx
x)]
-
[(xlogx
e
=
dx
]
e
x)
-
(xlogx
[e
e
=
1)
-
(D
x))
-
(xlogx
(e
=
1)
-
(D
dx)
e
logx
e
e
=
1
-
D
1
-
D
logx
e
=
1)
-
(D
logx)
(e
=
PI
x
(-x)
x
x
x
(-x)
x
x
x
2
x
ò
ò
ò
3]
-
[2logx
4
e
x
=
]
3x
-
logx
[2x
4
e
=
]
4
3x
-
logx
2
x
[
e
=
]
2
x
-
4
x
-
logx
2
x
[
e
=
]
2
x
-
dx
x
2
1
logx
2
x
[
e
=
x
2
2
2
x
2
2
x
2
2
2
x
2
2
2
x
ò
-
x
ò
dx
)
Xe(
e
=
X
a
-
D
1
ax
-
ax
)
e
B
+
e
(A
=
CF
is
function
ary
complement
The
distinct.
and
real
are
roots
The
2,-1
-
=
m
i.e.
0
=
1)
+
2)(m
+
(m
get
we
,
m
for
Solving
0
=
2
+
3m
+
m
is
equation
auxiliary
The
e
=
2)y
+
3D
+
(D
is
equation
al
differenti
given
The
:
Solution
e
=
2)y
+
3D
+
2.(D
x
-
2x
-
2
)
(e
2
)
(e
2
x
x
2
1
2
2
1
1
1
2
3
int
-PI
=PI
e
D+
e
D+
D+
+
D
e
PI =
egral
ular
the partic
e to find
Now we hav
x
x
x
e
e
e
-
)
(e
(-x)
t
(-x)
x
t
(-x)
x
e
x
-
e
1
x
x
x
e
e
=
e
e
=
e
=
t
where
dt
e
e
=
dx
e
e
e
=
e
1
D
1
=
PI
ò
ò
+
1)
-
.(e
.e
e
=
))
e
-
.(e
(e
e
=
)
e
-
(te
e
=
e
=
t
where
dt
te
e
=
dx
.e
.e
e
e
=
e
2
+
D
1
=
PI
x
)
(e
2x
-
)
(e
x
)
(e
(-2x)
t
t
(-2x)
x
t
(-2x)
x
x
e
2x
-
)
(e
2
x
x
x
x
x
ò
ò
equation.
al
differenti
the
of
solution
required
the
is
tion)
simplifica
(after
.e
e
+
)
e
B
+
e
(A
=
1)
-
.(e
.e
e
-
e
e
+
)
e
B
+
e
(A
=
PI
+
CF
=
GS
)
(e
(-2x)
x
-
2x
-
x
)
(e
(-2x)
)
(e
(-x)
x
-
2x
-
x
x
x
Bsin2x)
+
cos2x
(A
=
CF
is
function
ary
complement
The
conjugates
complex
are
roots
The
2i
±
=
m
i.e.
0
=
2i)
-
2i)(m
+
(m
get
we
,
m
for
Solving
4i
=
4
-
=
m
0,
=
4
+
m
is
equation
auxiliary
The
cos2x
x
=
4)y
+
(D
is
equation
al
differenti
given
The
:
Solution
cos2x
x
=
4)y
+
(D
equation
the
Solve
2
2
2
2
2
2
2
2
(-1)
2ix
2
2
2ix
2
2
2
2ix
2
2
2ix
2
i2x
2
2
2
x
)
4i
D
+
1
1
(
4iD
e
of
part
Real
=
4iD)
+
(D
x
e
of
part
Real
=
4)
+
4i
+
4iD
+
(D
x
e
of
part
Real
=
4)
+
2i)
+
((D
x
e
of
part
Real
=
4)
+
(D
e
x
of
part
Real
=
4)
+
(D
cos2x
x
=
PI
integral
particular
the
find
to
have
Now we
equation.
al
differenti
the
of
solution
required
the
is
cos2x
4
x
-
)
8
x
-
3
x
(
4
sin2x)
(
+
Bsin2x)
+
cos2x
(A
=
PI
+
CF
=
GS
cos2x
4
x
-
)
8
x
-
3
x
(
4
sin2x)
(
=
x/8)
-
4i
x
-
3
x
(
4i
e
of
part
Real
=
)dx
8
1
-
2i
x
-
(x
(
4i
e
of
part
Real
=
)
8
1
-
2i
x
-
(x
4iD
e
of
part
Real
=
)
16i
2
+
4i
2x
-
(x
4iD
e
of
part
Real
=
x
)
16
D
+
4i
D
-
(1
4iD
e
of
part
Real
=
2
3
2
3
2
3
2ix
2
2ix
2
2ix
2
2
2ix
2
(-1)
2
2
2ix
ò
i
)
1
(
2
)
1
(
2
)
1
(
2)
+
2D
-
(D
cos
e
2
2
x
1
+
=
+
+
-
+
=
=
D
D
Since
D
D
e
x
PI
x
)
2
2
2
1
2
(
cos
2
1
+
-
-
+
+
=
D
D
D
x
e
PI
x
2
)
sin
(
x)e
sin
B
x
(Acos
x
x
x
e
PI
isCF
solution
general
The
x
+
+
=
+
2
)
sin
(
x
x
e
PI
x
=
)
1
(
cos
2
1
+
=
D
x
e
PI
x
4)
-
=
D
(since
=
2)
+
3D
-
(-4
3)
+
cos(2x
(2
=
2)
+
3D
-
(D
3)
+
cos(2x
(2
=
2
2
DE.
given
the
of
solution
the
is
10
]
3)
+
sin(2x
3
-
3)
+
[-cos(2x
+
)
e
B
+
e
(A
=
PI
PI
+
CF
=
GS
2x
x
2
1
+
)
(
)
2
1
(
2
)
2
)(
1
(
2
2
case
exception
xe
D
D
e
PI
x
x
-
=
-
-
=
1
)
cos
3
(cos
3
3
8
4
4
4
)
2
sin
sin
2
(
3
3
)
2
(
4
)
2
(
)
2
sin
sin
2
(
3
)
3
4
(
2
sin
sin
6
2
2
2
2
2
2
2
2
-
-
-
=
+
-
+
+
-
-
-
=
+
-
+
-
-
-
=
+
+
=
-
-
-
-
D
x
x
e
D
D
D
x
x
e
D
D
x
x
e
D
D
x
x
e
PI
x
x
x
x
1
1
cos
3
1
9
3
cos
3
1
)
cos
3
(cos
3
2
2
2
2
-
-
+
-
-
-
=
-
-
-
=
-
-
-
x
e
x
e
D
x
x
e
PI
x
x
x
2
cos
3
10
3
cos
3
Be
Ae
2
2
3x
-
x
-
x
e
x
e
PI
CF
GS
x
x
-
-
-
+
+
=
+
=
D
x
e
D
D
x
e
D
D
D
x
e
D
D
x
e
D
D
x
e
PI
x
x
x
x
x
5
4
2
sin
5
2
sin
1
4
5
5
2
2
sin
4
)
1
(
5
)
1
(
2
sin
4
5
2
sin
2
2
2
2
+
-
=
+
=
+
+
-
+
-
=
+
-
+
-
=
+
+
=
-
-
-
-
-
116
)
2
cos
10
2
sin
4
(
)
4
(
25
16
)
2
cos
10
2
sin
4
(
25
16
2
sin
)
5
4
(
)
5
4
)(
5
4
(
2
sin
)
5
4
(
2
x
x
e
x
x
e
D
x
D
e
D
D
x
D
e
PI
x
x
x
x
-
-
=
-
-
-
-
=
-
-
-
=
-
-
+
-
-
-
=
-
-
-
-
116
)
2
cos
10
2
sin
4
(
4
x
x
e
Be
Ae
PI
CF
GS
x
x
x
-
-
+
+
=
+
=
-
-
-
)e
CF =(A+Bx
nction is
mentary fu
The comple
l.
l and equa
ts are rea
. The roo
,
i.e. m =
=
)
i.e. (m-
=
m +
-
m
on is
ary equati
The auxili
x
x e
)y =
D+
-
on is (D
ial equati
different
The given
Solution:
x
x e
)y =
D+
Solve (D
x
x
x
-
1
1
0
1
0
1
2
sin
8
1
2
sin
8
1
2
.
5
2
2
2
2
)
xdx
x
(
D
e
=
x])
([x
D
e
=
)
-
D+
x]
[x
e
=
)
D-
x
xe
PI =
egral
ular
the partic
e to find
Now we hav
x
x
x
x
ò
sin
8
sin
8
1
1
(
sin
8
1
(
sin
8
int
2
2
2
uation
rential eq
the diffe
olution of
required s
x) is the
x+
(-x
e
+
(A+Bx)e
F +PI =
GS = C
x)
x+
(-x
e
=
dx
x)
x+
(
x
[-x
e
=
)dx
x
x+
(-x
e
=
x))
x+
(-x
e
(
D
=
x
x
x
x
x
x
cos
2
sin
8
cos
2
sin
8
]
sin
sin
sin
8
sin
cos
8
sin
cos
8
1
ò
ò
-
8
)
2
sin
2
(cos
32
)
2
sin
4
2
cos
4
(
)
4
(
4
16
)
2
)(cos
2
4
(
)
4
16
(
)
2
)(cos
2
4
(
2
4
)
2
(cos
2
)
2
(cos
2
2
x
x
e
x
x
e
x
D
e
D
x
D
e
D
x
e
D
D
x
e
PI
x
x
x
x
x
x
+
-
=
-
-
=
-
-
+
-
=
-
+
-
=
-
-
=
-
=
D
D
x
e
D
D
D
x
e
D
D
x
e
D
D
x
e
PI
x
x
x
x
2
2
cos
3
4
4
1
2
2
cos
3
)
1
(
4
)
1
(
2
cos
3
4
)
2
(cos
2
2
2
2
-
=
+
-
-
+
+
=
+
+
-
+
=
+
-
=
8
)
2
sin
2
(cos
Be
Ae
3x
x
x
x
e
PI
CF
GS
x
+
-
+
=
+
=
1
9
sin
)
1
3
(
)
1
3
)(
1
3
(
sin
)
1
3
(
1
3
sin
2
3
1
sin
2
3
sin
2
2
1
-
-
=
+
-
-
=
+
=
+
+
-
=
+
+
=
D
x
D
D
D
x
D
D
x
D
x
D
D
x
PI
10
sin
cos
3
1
)
1
(
9
sin
)
1
3
(
1
-
-
=
-
-
-
=
x
x
x
D
PI
]
2
3D)
+
(D
+
[1
2
1
=
)
2
3D)
+
(D
+
(1
2
1
=
3D)
+
D
+
(2
x
=
PI
(-1)
2
1
-
2
2
2
2
]
2
7)
-
6x
-
(x
[
2
1
=
PI
(9)]
2
1
+
3x)
+
(1
-
[x
2
1
=
PI
(9.2)]
4
1
+
6x)
+
(2
2
1
-
[x
2
1
=
PI
)
....]x
)
2
3D)
+
(D
(
+
2
3D)
+
(D
-
([1
2
1
=
2
2
2
2
2
2
2
2
2
2
+
2
7)
-
6x
-
(x
2
1
10
sin
cos
3
+
)
e
B
+
e
(A
=
PI
PI
+
CF
=
GS
2
2x
-
x
-
2
1
+
-
-
+
x
x
7
3
cos
16
9
3
cos
16
3
cos
20
cos
)
15
(
4
cos
3
)
16
1
(
4
cos
3
)
16
(
4
cos
3
)
16
(
4
3
cos
cos
3
)
16
(
cos
2
2
2
1
2
2
3
x
x
D
x
PI
x
x
x
D
x
PI
D
x
x
D
x
PI
=
+
-
=
+
=
=
=
+
-
=
+
=
+
+
=
+
=
7
3
cos
20
cos
4
sin
4
cos
2
1
x
x
x
B
x
A
PI
PI
CF
GS
+
+
+
=
+
+
=
2
/
x
-
3
)
2
3
sin
2
3
(Bcos
Ae
CF
root.
real
a
and
conjugates
complex
are
roots
The
.
2
3
i
±
1
1,
-
=
m
i.e.
-1
m
,
2
4)
-
(1
±
1
=
m
get
we
,
m
for
Solving
0
=
1
m
is
equation
auxiliary
The
x
e
x
C
x
+
+
=
=
+
(
)
12
4
3
1
3
3
2
2
)
2
(
4
2
4
2
3
2
3
2
2
2
2
2
2
2
2
2
2
2
2
x
e
x
e
dx
x
e
x
D
e
dx
x
D
e
D
x
e
D
x
e
D
e
x
PI
x
x
x
x
x
x
x
x
=
=
=
=
=
=
-
+
=
-
=
ò
ò
)
(
4
2
cos
)
4
(
2
sin
2
case
Exception
x
x
D
x
PI
-
=
+
=
(
)
)
1
(sin
1
sin
sin
1
1
sin
)
1
(
sin
2
2
2
2
-
=
-
=
=
-
+
=
-
=
D
ce
x
e
D
x
e
D
x
e
D
e
x
PI
x
x
x
x
(
)
8
4
)
1
(
2
)
1
(
2
1
2
)
1
(
cosh
2
2
2
2
2
x
x
x
x
x
x
e
e
x
D
e
D
e
D
e
e
D
x
PI
-
-
-
+
=
-
+
-
=
-
+
=
-
=
(
)
4
2
sinh
8
8
)
2
)(
2
(
2
)
2
)(
2
(
2
4
2
)
4
(
2
cosh
.
6
2
2
2
2
2
2
2
2
x
x
xe
xe
D
D
e
D
D
e
D
e
e
D
x
PI
x
x
x
x
x
x
=
-
=
+
-
+
+
-
=
-
+
=
-
=
-
-
-
4
1
2
,
2
,
,
0
4
4
1
4
4
)
4
(
1
.
7
2
2
2
2
2
0
2
0
2
-
+
=
+
=
+
=
-
=
=
-
-
=
-
=
-
=
-
=
-
-
x
x
x
x
x
x
Be
Ae
PI
CF
GS
Be
Ae
CF
m
get
we
m
for
solving
m
is
AE
The
e
D
e
D
PI
(
)
)
(
)
2
2
(log
2
)
2
(
2
.
8
log
2
2
log
2
2
log
2
a
x
x
x
x
x
e
a
Since
e
D
e
D
PI
=
-
=
-
=
-
=
)
1
(sin
1
cos
cos
)
1
1
(
cos
)
1
(
cos
.
9
2
2
2
2
-
=
-
=
=
+
-
=
+
=
-
-
-
-
ceD
x
e
D
x
e
D
x
e
D
x
e
PI
x
x
x
x