Download UNIT 2(PART I)

ORDINARY DIFFERENTIAL
EQUATIONS
MATHEMATICS II(MA6251)
UNIT 2
-P.VEERAIAH
DEPARTMENT OF APPLIED
MATHEMATICS
1
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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
constant coefficients
2
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Second-order linear differential
equations
d2y
dy
Differential equations of the form a 2  b  cy  Q( x)
dx
dx
are called second order linear differential equations.
When Q( x)  0 then the equations are referred to as homogeneous,
When Q( x)  0 then the equations are non-homogeneous.
Note that the general solution to such an equation
must include two arbitrary constants to be
completely general.
3
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Second-order linear differential
equations
Theorem
If y  f ( x) and y  g ( x) are two solutions then so is y  f ( x)  g ( x)
2
d2 f
df
d
we have a 2  b  cf  0 and a g  b dg  cg  0
dx
dx
dx 2
dx
Adding
d2 f
df
d 2g
dg
a 2  b  cf  a 2  b  cg  0
dx
dx
dx
dx
 d 2 f d 2 g   df dg 
a 2  2   b    c f  g   0
dx   dx dx 
 dx
And so y  f ( x)  g ( x) is a solution to the differential equation.
4
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Second-order linear differential
equations
dy
y  Ae , for A and m, is a solution to the equation b  cy  0
dx
It is reasonable to consider it as a possible solution for
mx
d2y
dy
a 2  b  cy  0
dx
dx
2
d
y
mx
dy
2 mx
mx
y  Ae


Am
e

 Ame
2
dx
dx
If y  Ae mx is a solution it must satisfy aAm 2 e mx  bAme mx  cAemx  0
assuming Aemx  0, then by division we get am 2  bm  c  0
The solutions to this quadratic will provide two values of
m which will make y = Aemx a solution.
5
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Second-order linear differential
equations
 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.mx So a mxgeneral solution is
y  Ae  Bxe
6
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Roots are complex conjugates
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
 Ae px eiqx  Be px e  iqx
 e px  Aeiqx  Be  iqx 
y  Ae( piq ) x  Be( piq ) x
We know that ei  cos  i sin 
 e px  A  cos qx  i sin qx   B  cos( qx)  i sin(  qx)  
 e px  A  cos qx  i sin qx   B  cos qx  i sin qx  
 e px   A  B  cos qx   A  B  i sin qx 
 e px  C cos qx  D sin qx 
Where C  A  B and D  ( A  B )i
7
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Non-homogeneous
Second-order linear
differential equations
d2y
dy
a 2  b  cy  Q( x)
dx
dx
d 2g
dg
a 2  b  cg  Q( x)
dx
dx
 Non homogeneous equations take the form
8
Suppose g(x) is a particular solution to this
equation. Then
Now suppose that g(x) + k(x) is another
solution. Then
d 2 (g  k)
d (g  k)
a
b
 c( g  k )  Q( x)
2
dxDEPARTMENT OF
dxAPPLIED
SVCE,
MATHEMATICS, SRIPERUMBUDUR
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Non homogeneous
second order differential equations
Giving
d 2g
d 2k
dg
dk
a 2  a 2  b  b  cg  ck  Q ( x)
dx
dx
dx
dx
 d 2k

dk
 Q ( x)   a 2 2  b  ck   Q( x2)
dx
 ddxg
  d k

dg
dk
  a 2  b  cg    a 2  b  ck   Q ( x)
dx
dx
 dx
  dx

d 2k
dk
 a 2  b  ck  0
dx
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)
9
General
Solution
= CF +
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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:
y  c1e
1x
 c 2e
2 x
(3) If 1 = 2 (if b2 - 4ac = 0), then the general solution is:
y  c1e 1x  c 2 xe 1x
(4) If 1 and 2 are complex numbers (if b2 - 4ac < 0), then the general solution is:
y  c1ex cos x   c 2ex sin x 
Where:
10
b
and  

2a
SVCE, DEPARTMENT OF APPLIED
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4ac  b
2a
2
12/23/2014
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:
ay   by   cy  0
a  0
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
a2  b  c  0
a  0
This is a quadratic. Let 1 and 2 be its roots we have
1,2
11
 b  b 2  4ac

2a
SVCE, DEPARTMENT OF APPLIED
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TYPE-1 PARTICULAR INTEGRALS f(x) = eax+b
 Solve the equation
2
(D - 2D+1)y = coshx
The given differential equation is (
=Cosh
x
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
12
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12/23/2014
TYPE-1 PARTICULAR INTEGRALS f(x) = eax+b
Now we
have to find the particular
PI =
cosh x
=
2
(D - 2 D+ 1 )
ex
Now PI1 =
2(D 2 - 2D + 1)
Similarly
( Substitu
(x 2 e x )
=
4
PI
2
integral
e x +e -x
= PI
2
2 (D - 2 D+ 1 )
1
+ PI
2
( Since D = 1 makes the denominator 0 twice)
=
e
2
-x
2 (D - 2 D+ 1 )
ting D =
-1 )
=
e -x
8
Now the general solution of the DE is
GS = CF + PI1 + PI2 =
(A+Bx)ex +
13
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 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 =
2 ±
(4
 i.e.
m=1±i .
2
 The roots are complex conjugates.
 The complementary function is
 CF = (Acos x + B sin x)ex
14
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- 4.2)
12/23/2014
ex
ex
PI1  2

( Since D  1)
(D - 2D + 2) 1
5e0x
5
SimilarlyPI2 = 2
=
(Substituting D =0 )
2
( D -2D+2 )
PI 3
e 2x
e 2x
=
=
/ (Substitut ing D = 2)
2
2
(D - 2D + 2)
Now the general solution
GS = CF + PI 1 + PI 2 + PI
15
3
=
x
2x
e
5
e
(Acos x + B sin x)e x +
+
+
1
2
2
SVCE,isDEPARTMENT
OF APPLIED
the solution
of the given DE.
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
 Solve (D2-3D+2) y = 2cos(2x+3)
 Solution :
 The given differential equation is
2
(D - 3D + 2)
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 )

16
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 Now we have to find the particular integral
 PI = = (2 cos(2x + 3) = (2 cos(2x + 3) =
(D 2 - 3D + 2)
(-4 - 3D + 2)
(since D 2 = - 4)
2 cos( 2x+3)
(-2-3D)
=
(-2+3D)2 cos( 2x+3)
(-2+3D)2 cos( 2x+3)
=

(-2-3D)(-2+3D)
(4-9D2 ) )
(-2 + 3D)(2 cos(2x + 3)
((4 - 9.(-4) )
[-cos(2x+ 3) - 3 sin(2x+ 3) ]
10
(-4 cos(2x + 3) - 12 sin(2x + 3)
40
[-cos(2x + 3) - 3 sin(2x + 3) ]
10
is the solution of the given DE.
GS = CF + PI = (A e x + B e 2x ) +
17
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 Solve (D2+1)2y = 2sinx cos3x
 Solution: The given differential equation is







18
2
2
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
(D +1)
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 The complementary function is
 CF = (A+Bx)cos x +(C+Dx) sin x
 Now we have to find the particular integral
PI =
( 2 sin x cos 3 x )
(D 2+1 ) 2
PI 1 + PI
2
PI1 =
=
sin4x
(D2 + 1)2
sin 4 x- sin 2 x
=
2
2
(D +1 )
=
sin4x
(-16 + 1)2
=
sin4x
225
(since D2 = - 16 )
PI
=
19
2
sin4x
(-4 + 1)
2
=
sin4x
9
SVCE, DEPARTMENT OF APPLIED
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(since
D
2
= -4 )
12/23/2014
sin4x
sin2x
GS = CF + PI1 + PI 2 = (A + Bx)cos x + (C + Dx) sin x +
+
225
9
is the solution of
20
the given DE.
SVCE, DEPARTMENT OF APPLIED
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Solve
the DE
(D
2
 4) y  x4
( D2  4) y  x4
D D 
1
 1 
 x
4
4 16 
1
12 x 24 
  x 
 
16 
4
4
1
3
  x  3x  
4
2
2
4
4
C . F .  A cos 2 x  B sin 2 x
1
1
x
x
D
4 (1  )
4D
4
1
D
(1 
) x
4
4
4
2
4
2
2
4
2
1
4
4
2
1
3
GSCF PI  Acos2x Bsin2x   x 3x  
4
2
4
21
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2
12/23/2014
 2.Solve the DE
(D 2 + 3D + 2) y = x 2
Solution : The given differential equation is
(D 2 + 3D + 2)y = x 2
.
The auxiliary equation is
m 2 + 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 e-x + B e-2x )
x2
PI =
(2 + D 2 + 3D)
=
1
(D 2 + 3D) -1
(1 +
)
2
2
1
(D 2 + 3D) (-1)
= [1 +
]
2
2
1
(D 2 + 3D) (D 2 + 3D) 2
([1 +(
)  ....]x 2 )
2
2
2
1
1
1
PI = [x 2 - (2 + 6x) + (9.2)]
2
2
4
1
1
PI = [x 2 - (1 + 3x) + (9)]
2
2
2
1 (x - 6x - 7)
PI = [
]
2
2
=
GS = CF + PI
-x
= (A e + B e
22
SVCE, DEPARTMENT OF APPLIED
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- 2x
1 (x 2 - 6x - 7)
) +
2
2
12/23/2014
TYPE-4 PARTICULAR INTEGRALS
 1.Solve (D2-2D+2) y = ex x2
The complementary function is
The given differential equation is (D 2 - 2D + 2)y = e x x 2
The auxiliary equation is
Now we have to find the particular integral
m - 2m + 2 = 0
2
Solving for m , we get m =
CF = (Acos x + B sin x)e x
(e x x 2 )
ex
x2
=
(D 2 - 2D + 2)
(D 2 + 2D + 1 - 2D - 2 + 1)
(Since D = D + 1)
PI =
(2 ± (4 - 4.2)
2
i.e. m = 1 ± i .
The roots are complex conjugates.
4
x2
1 2
1 x3
x3
x
x
x
x x
e
=e
= e  dx = e
 x dx = e
D2
D
D 3
3
12
x
x4
is the required solution of the differential equation.
GS = CF + PI = (Acos x + B sin x)e + e
12
x
23
x
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 2. (D2 + 4D+ 3) y = e-x sinx
Solving for m , we get (m + 1) (m + 3) = 0
i.e.
m = - 1 ,-3 .
Solution : The given differential equation is
(D 2 + 4D + 3)y = e-x sinx
The auxiliary equation is
m 2 + 4m + 3 = 0
Now we have to find the particular integral
The roots are real and distinct.
PI =
The complementary function is
CF = (A e-x + B e-3x )
=
(e (-x) sinx) (e (-x) sinx)
=
(D 2 + 2D) (-1 + 2D)
(-x)
( Since D 2 = -1)
(-x)
=
(e )(-1 - 2D)sinx
e
=
(-1 - 2D)sinx
((-1 + 2D)(-1 - 2D))
(1 - 4D 2 )
=
e(-x) (-sinx - 2cosx))
5
=
e(-x) sinx)
((D - 1) 2 + 4(D - 1) + 3)
=
(e (-x) sinx)
(D 2 - 2D + 1 + 4D - 4 + 3)
(since D = D - 1)
( Since D2 = -1)
GS = CF + PI =
-x
(A e + B e
24
(e (-x) sinx)
(D 2 + 4D + 3)
-3x
(e (-x) (-sinx - 2cosx))
)+
is the required solution of the differential equation.
5
SVCE, DEPARTMENT OF APPLIED
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 Solve (D3-1)y=x sinx
 Solution:





The given differential equation is (D31)y = x sinx
The auxiliary equation is
m3-1= 0
i.e (m-1)(m2 +m+1)=0 (-1 ± (1 - 4.1) )
2
± (1 - 4.1)
)
Solving for m ,(-1we
get
m=
2
The roots are
i.e.
ma pair
= of1,complex conjugates and a real root
The complementary function is

25
CF = (Acos( (3/2 )x + B sin( (3/2 )x )e -x/2
SVCE, DEPARTMENT OF APPLIED
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x sin
(-D(-x(D=
(D+
1 ) sin
x)
1 )(D1 ))
(-x(D((D
=
( 3 D
( 1 -D)
1 ) sin
2
- 1 ))
=
(x(
x)
2
( 1 -D) 2 )
2
(D+
1 )2
( 3 D 2 ( 1 -D)

( 1 -D 2 ) 2
1 ) sin
x)
( 3 D 2 ( 1 -D) 2 )

(- 2 )
4
(x(
cos
x- sin
x))
( 3 (- sin
x+
2
cos
x- sin
x))
( 3 ( 2 ( cos
x)))
2
4
-x(D-
=
=
x
( 3 D 2 )
1 )(-D1 )2
sin
2
x
)
sin
x
sin
x
sin
x- 2 (- cos
x))
4
GS = CF + PI =
(Acos( 3/2 x + B sin 3/2 x )e -x/2 +
26
(x(cosx - sinx) - 3cosx)
is the required solution of the differential equation.
2
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Solve (D 2 - 2 D+1 )y = x e
x
sin x
Solution: The given differential equation is (D 2 -2 D+1 )y = x e
x
sin x
The auxiliary equation is
m 2 -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)e x
Now we have to find the particular int egral
PI =
=
xe x sin x
( D-1 )2
e x [x sin x]
( D+1-1 )2
ex
=
([x sin x])
D2
ex
=
(  x sin xdx )
D
27
=
=
1 x
(e (-x cos x+ sin x))
D
x
 e (-x cos x+ sin x )dx
=
e x [-x sin x   ( sin x+ sin x)]dx
=
e x (-x sin x+2 cos x)
GS = CF +PI =
(A+Bx)e x +e x (-x sin x+2 cos x) is the required solution of the differential equation
SVCE, DEPARTMENT OF APPLIED
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12/23/2014
1
X = eax  Xe( -ax ) dx
D-a
1. (D 2 - 2D + 1)y = e x log x
Solution : The given differential equation is (D 2 - 2D + 1)y = e x log x
The auxiliary equation is
PI =
m 2 - 2m + 1 = 0
e x logx
x
(-x)
dx)
x  e logx e
D
1
=e
=
D -1
(D - 1)
i.e. (m - 1) 2 = 0
i.e. m = 1,1 . The roots are real and equal.
=
The complementary function is
CF = (A + Bx)e
= e
x
=
e
x
= e
x
= e x  [(xlogx - x)]dx
1
x 2
dx

x
2
x 2
x 2
]
4
2
3x 2
]
4

e x
2
[2x
logx
- 3x
4
x 2 e x
=
[2logx
- 3]
SVCE, DEPARTMENT
OF APPLIED
4
=
28
(e x (xlogx - x))
(D - 1)
= e x  [e x (xlogx - x) e(-x) ]dx
x
x 2
logx
2
x 2
[
logx
2
x 2
[
logx
2
[
(e x logx)
(D - 1) 2
MATHEMATICS, SRIPERUMBUDUR
2
-
x 2
]
2
]
12/23/2014
2.(D
2
+ 3D
+ 2)y
:
The
Solution
(D
The
m
2
= e
+ 3m
Solving
i.e.
The
m
, we
complement
CF
= (A
)
x
differenti
al
e
get
x
= e(-x) e t = e(-x) e(e
Now we have to find the particular int egral
(m
+ 2)(m
+ 1)
=
0
ee
PI =
D 2 +3D+2
1 e
1 e
e 
e
D+1
D+2
=PI1-PI 2
x
and
ary
+ B
- 2x
is
x
is
e
- x
distinct.
function
is
)
1 e
e = e-x  ee e x dx
D 1
(-x)
= e  e t dt where t = e x
PI1 =
equation
)
+ 2 = 0
for
m = - 2,-1
roots
are
real
The
x
given
+ 3D
+ 2)y
= e (e
auxiliary
equation
2
(e
x
x
x
1
e(e ) = e-2x  ee .e x .e x dx
D+2
(-2x)
= e  te t dt where t = e x
x
PI 2 =
)
x
x
x
= e(-2x) (te t - e t ) = e(-2x) (e (e ) .(e x - e(e ) ))
x
= e-2x .e(e ) .(e x - 1)
GS = CF + PI =
x
x
(A e -2x + B e - x ) + e (-x) e (e ) - e (-2x) .e (e ) .(e x - 1)
x
= (A e -2x + B e - x ) + e (-2x) .e (e ) (after simplifica tion) is the required solution of the differenti al equation.
29
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
)
Solve the equation (D 2 + 4)y = x 2cos2x
Now we have to find the particular integral
Solution : The given differential equation is
(D 2 + 4)y = x 2 cos2x
x 2 cos2x
x 2ei2x
PI = 2
= Real part of 2
(D + 4)
(D + 4)
The auxiliary equation is
= Real part of
m 2 + 4 = 0, m 2 = - 4 = 4i 2
Solving for m , we get (m + 2i)(m - 2i) = 0
i.e.
m = ± 2i
The roots are complex conjugates
The complementary function is
CF = (A cos2x + Bsin2x)
30
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
e2ix x 2
((D + 2i)2 + 4)
e 2ix x 2
= Real part of
(D 2 + 4iD + 4i 2 + 4)
e 2ix x 2
= Real part of
(D 2 + 4iD)
= Real part of
e 2ix
1 (-1) 2
(
) x
4iD 1 + D
4i
12/23/2014
 Contd.,
e2ix
D D2 (-1) 2
= Real part of
(1 - + 2 ) x
4iD
4i 16i
2ix
e
2x
2
= Real part of
(x 2 + 2)
4iD
4i 16i
2ix
e
x 1
= Real part of
(x 2 - )
4iD
2i 8
e2ix
x 1
= Real part of
(  (x 2 - )dx
4i
2i 8
e2ix x 3 x 2
= Real part of
( - x/8)
4i 3 4i
3
sin2x) x
x
x2
= (
( - ) cos2x
4
3
8
4
GS = CF + PI = (A cos2x + Bsin2x)
3
+ (
31
sin2x) x
x
x2
( - ) cos2x is the required solution of the differential equation.
4
4
3
8
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
 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
32
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
EXTRA PROBLEMS
PI1 
e x cos x
ex

( Since D  D  1)
(D 2 - 2D + 2) ( D  1) 2  2( D  1)  2
e x cos x
PI1  2
( D  2 D  1  2 D  2  2)
The general solution isCF  PI  (Acos x  B sin x)e x 
33
e x cos x
PI1  2
( D  1)
PI 
e x ( x sin x )
2
e x ( x sin x )
2
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
 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 )
34
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014

Now we have to find the particular integral
PI1 = = (2 cos(2x + 3) = (2 cos(2x + 3) =
(D 2 - 3D + 2)

(-4 - 3D + 2)
(since D2 = - 4)
 2 cos( 2x+3)
(-2-3D)
(-2+3D)2 cos( 2x+3)
(-2+3D)2 cos( 2x+3)
=

(-2-3D)(-2+3D)
(4-9D2 ) )
(-2 + 3D)(2 cos(2x + 3)
((4 - 9.(-4) )
[-cos(2x+ 3) - 3 sin(2x+ 3) ]
10
(-4 cos(2x + 3) - 12 sin(2x + 3)
40
2e x
2 xe x

PI 2 
( exception case)
( D  1)( D  2) (1  2)
[-cos(2x + 3) - 3 sin(2x + 3) ]
GS = CF + PI1  PI 2 = (A e + B e ) +
10
is the solution of the given DE.
x
35
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
2x
12/23/2014
 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
-x
-3x
 The
6e complementary
sin x sin 2 x  3e ( 2 sin x sin 2 x ) function
2 x ) CF
 3e =
(cos 3xAe
 cos x ) +Be
 3e ( 2 sin x sinis
PI 



2 x
2 x
( D 2  4 D  3)
2 x
( D  2)2  4( D  2)  3
D2  4D  4  4D  8  3
2 x
D2  1
 3e 2 x (cos 3x  cos x )  3e 2 x cos 3x 3e 2 x cos x

PI 

D2  1
 9 1
11
GS  CF  PI  Ae  Be
-x
36
-3x
3e 2 x cos 3x 3e 2 x cos x


10
2
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
 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
e  x sin 2 x
e  x sin 2 x
e  x sin 2 x
e  x sin 2 x e  x sin 2 x




PI  2
D  5D  4 ( D  1) 2  5( D  1)  4 D 2  2 D  5D  5  4  1 D 2  5D  4  5D
37
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
e x (4  5D) sin 2 x e x (4  5D) sin 2 x
PI 

(4  5D)(4  5D)
16  25D2
e x (4 sin 2 x  10cos2 x) e x (4 sin 2 x  10cos2 x)


16  25(4)
116
x
e
(4 sin 2 x  10cos2 x)
x
4 x
GS  CF  PI  Ae  Be 
116
38
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
EXTRA PROBLEMS
5.Solve (D 2 - 2 D+1 )y = 8 x e
x
Solution: The given differential equation is (D 2 -2 D+1 )y = 8 x e
The auxiliary equation is
i.e. (m-1 )2 =0
i.e. m = 1,1 . The roots are real and equal.
The complementary function is
CF =(A+Bx)e x
=
=
=
x
sin x
PI =
8 xe x sin x
( D-1 )2
8e x [x sin x]
=
( D+1-1 )2
m 2 -2m + 1=0
=
Now we have to find the particular int egral
sin x
8e x
=
([x sin x])
D2
8e x
=
(  x sin xdx )
D
1
( 8e x (-x cos x+ sin x))
D
x
 8e (-x cos x+ sin x )dx
8e x [-x sin x   ( sin x+ sin x)]dx
8e x (-x sin x+2 cos x)
GS = CF +PI =
(A+Bx)e x +8e x (-x sin x+2 cos x) is the required solution of the differential equation
39
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
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
e x (cos 2 x )
e x cos 2 x
e x cos 2 x
e x cos 2 x



PI  2
D  4 D  3 ( D  1) 2  4( D  1)  3 D 2  2 D  1  4 D  4  3 D 2  2 D
PI 

e x (cos 2 x ) e x (cos 2 x ) e x (  4  2 D )(cos 2 x )


D2  2D
 4  2D
(16  4 D 2 )
e x (  4  2 D )(cos 2 x ) e x (  4 cos 2 x  4 sin 2 x )

16  4 (  4 )
32
 e x (cos 2 x  sin 2 x )

8
40
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
GS  CF  PI  Ae  Be
x
3x
 e x (cos 2 x  sin 2 x )
8
12/23/2014
 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
41
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
EXTRA PROBLEMS
PI1 
sin x
sin x
sin x
(3D  1) sin x
(3D  1) sin x




D 2  3D  2  1  3D  2 3D  1 (3D  1)(3D  1)
9D2  1
PI1 
(3D  1) sin x 3 cos x  sin x

9( 1)  1
 10
x
(2 + D 2 + 3D)
2
PI 2 =
=
1
(D 2 + 3D) -1
(1 +
)
2
2
1
(D 2 + 3D) (-1)
= [1 +
]
2
2
1
(D 2 + 3D) (D 2 + 3D) 2
([1 +(
)  ....]x 2 )
2
2
2
1
1
1
PI 2 = [x 2 - (2 + 6x) + (9.2)]
2
2
4
1
1
PI 2 = [x 2 - (1 + 3x) + (9)]
2
2
2
1 (x - 6x - 7)
PI 2 = [
]
2
2
=
GS = CF + PI1  PI 2
-x
= (A e + B e
42
- 2x
3 cos x  sin x 1 (x 2 - 6x - 7)

) +
 10
2
2
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
 8. Solve the differential equation
 (D2 +16)y = cos
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
43
3
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
cos3 x
3 cos x  cos 3x
PI  2

( D  16)
4( D 2  16)
3 cos x
3 cos x
3 cos x cos x



PI1 
4( D 2  16) 4( 1  16) 4(15)
20
cos 3x
cos 3x
cos 3x


PI 2  2
D  16  9  16
7
GS  CF  PI1  PI 2  A cos 4 x  B sin 4 x 
44
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
cos x cos 3x

20
7
12/23/2014
The auxiliary equation is
m3  1 = 0
Solving for m , we get m =
1±
(1 - 4)
, m  -1
2
1±i 3
.
2
The roots are complex conjugates and a real root.
i.e.
m = - 1,
3
3
x  C sin
x )e
2
2
CF  Ae - x  (Bcos
x /2
x 2e2 x
e2 x x 2
e2 x x 2 e 2 x 2


PI 
 x dx
2 
D2
D
( D  2 ) 2 D  2  2 

45
2x
3
2x
e x
e

D 3
3

3
2x
4
2x
1.Solve the DE
(D3+1)y =0
x dx e x
e x


1
3 4
12
4
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
2.Find the particular
integral of (D2-4D+4)y
=x2e2x
12/23/2014
EXTRA PROBLEMS-PARTA
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
46
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
PI 
sin 2 x
 x cos 2 x

( Exception case)
2
( D  4)
4
sin x e x
e x sin x
e x sin x e x sin x


(sin ce D 2  1)
PI 
2 
2
2
D  1  1
( D  1)
D
1
e x  e x
cosh x
ex
e x
x 2e x e x
2
PI 





2
( D  1) 2 D  1
2( D  1) 2 2( D  1) 2
4
8
12/23/2014
EXTRA PROBLEMS-PARTA
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
47
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
EXTRA PROBLEMS-PARTA
cosh 2 x

6 . PI 
(D 2  4)
7. PI 
e 2 x  e 2 x
e2x
e 2 x
xe 2 x
xe  2 x
x sinh 2 x
2





2
D  4  2 ( D  2 )( D  2 ) 2 ( D  2 )( D  2 ) 8
8
4
e0 x
e0 x  1
1



( D 2  4) D 2  4  4 4
The AE is m 2  4  0, solving for m, we get m  2,2
CF  Ae 2 x  Be 2 x
GS  CF  PI  Ae 2 x  Be 2 x 
1
4
2x
e x log 2
e x log 2
8. PI 

( Since a x  e x log a )
2 
2
2
D  2  (log 2  2)
( D  2)
e  x cos x
e  x cos x
e  x cos x e  x cos x
9. PI 



(sin ceD 2  1)
2
2
2
( D  1)
( D  1  1)
D
1
48
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
EXTRA PROBLEMS
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
49
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014
SOLUTIONS(PART-A & B)
50
SVCE, DEPARTMENT OF APPLIED
MATHEMATICS, SRIPERUMBUDUR
12/23/2014