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Part II:

65

Equations with Constant Coefficients

Linear 2-nd order homogenous equations

(basic case)

Methods We Need to Learn:

2

1 22

d y dya a y 0

dxdx+ + =

66

Linear 2-nd order non-homogenous equations:Variation of parameters

Undetermined coefficients (special cases)

D-operator method

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2

1 22

d y dya a y F(x)

dxdx+ + =

F(x) 0 : Non-homogeneous

F(x) = 0 : Homogeneous

The ODE is 'linear' even if a1, a2, ..., an are functions of x.

Linear 2nd-order Homogeneous Eqns with Constant Coefficients :

67

This is because :

degree of

2

2

d y1

dx=

dy1

dx=degree of

degree of y = 1

and a1, a2, ..., an are at most functions of x only.

Linear Differential Operator (D-Operator)

D f(x) : meansd

f(x)dx

D : Operation of differentiation w.r.t. x

= =2

2 d f(x)D f(x) D (Df(x))3

3 )x(fd)x(fD =

68

)x(f2dx

)x(df

dx

)x(fd)x(f)2DD(

2

22 +=+

dxdx

Linear D operator satisfies basic algebraic laws.

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2 D2 + bD + c f = D r D r f

(1) (D2 + D 2) f = (D + 2)(D 1) f

Examples :

69

2

c4bbr,

2

c4bbr

2

2

2

1

=+

=

Linear 2nd-order Homogeneous Equation:

(D2 + 2aD + b) y = 0 (1)

Characteristic Equation : r2 + 2ar + b = 0

(D2 + 2aD + b) = (D r1) (D r2)

70

1du

r u 0dx

=

(D r1) u = 0

1 2

Let (D r2) y = u

dur dx1u

=

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= 1x12rD c( er )y

1r x12dy

ri.e., = c ey

= =1 11r x r xc u e e c e1 ln(u) r x c= +

71

Recall Equation (9):1

y Qdx=

2r x e=Integrating factor:

=

=

=

2 2 1

2 1 2

r x -r x r x1

r x (r r )x1

y ( e e c e dx

e

1 Qd

c e

x )

dx

Case 1 : r1 r2

c

72

21 2

y e e c

r r

= +

1 2r x r x12

1 2

ce c e

r r= +

1 2r x r x3 2c e c e= +

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1 2(r r )xe 1 =

Case 2 : r1 = r2

2 1 2r x (r r )x1y e c e dx

=

73

2r x1 2e (c x c )= +

= + 2r x1 2(c x c ) e

2r x1e c dx=

(1) Find the general solution of

0y6dx

dy5

dx

yd2

2=++

Example:

74

+ + = = 2(r 5r 6) 0 r 2 , 3

2x 3x1 2y c e c e

= +

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4x x=

given that y(0) = 2 and y'(0) = 3.

0y4dx

dy3

dx

yd2

2=

(2) Find the solution of

1 2( 4, 1 )r r= =

75

3 = y'(0) = 4 c1e40 c2e

-0

2 = y(0) = c1 + c2

3 = 4 c1 c2

2 = c1 + c2 ; 3 = 4 c1 c2

1 21 11

c c5 5

= =

4x x1 11y e e5 5

= +

76


+ =23

3 2

y d y dy6 11 6y 0

dxdx dx

d

3 2(r 6 r 11 r 6) 0 + =

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Case 3: For Imaginary Roots

( i )x ( i )x1 2y c e c e

+ = +

i x i2

x)x e c( ce e = +

1 2 i ir r= + =

79

x1 2( ( )cos x + i sin x cos x - i sin x( ) )e c c

= +

x1 2 1 2( ( + ) cos x + i ( ) sin x) )e c c c c

=

x 3 4( cos x + sin x )e c c=

ix ix

2

2

d yy 0

d x+ =

1 2r i ; r - i= + =

r i ( =0, =1)=

2 1 0r + =(5) Find the solution of:

80

1 2c (co s x i s in x ) c (co s x i s in x )= + +

( )x 3 4e c cos x c s in x= +

3 4c cos x c s in x= +

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1x0

1x0xy2

dx

dy3

dx

yd2

2

>

=+

1 y(0) 2, y '(0)= =

(6) Solve the following initial value problem:

81

(a) Find y(x) in the interval 0 x 1.

dy

dx(b) Find y(x) for x > 1 such that y(x) and

are continuous at x = 1.

Solution:

(a) 1x0

1x0xy2'y3"y

>

=+

and2

1)0('y,2)0(y ==

2

82

1 2 , = = =

x 2xh 1 2y c e c e= +

' "p p pAssum e : y Ax B; y A; y 0= + = =

Find particular solution py :

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x)BAx(2A30 =++

4

3B,

2

1A ==

2x2x1ph ecec43x

21yyy +++=+=

83

x 2x1 21y ' c e 2c e2

= + +

Two arbitrary constants : &1c 2c

21 cc4

3)0(

2

12)0(y +++==

= = + +1 21 1y '(0) c 2c2 2

= = 1 27 9

c , c2 4

84

= + +x 2x1 7 9 3

y(x) x e e2 2 4 4

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x 2xh 1 2y(x) y c e c e= = +

(b) For x > 1, the given D.E. is homogeneous

i.e. y" 3y' + 2y = 0

and the complete solution is given by

It is re uired that x and ' x are continuous at x = 1.

85

21 2y(1) c e c e for x 1

= + >

2

for 0 x 11 7 9 3

y(1) (1) e e2 2 4 4

= + +

Continuity of y(x) at 1 gives:

)1(e4

9e

2

7

4

5ecec 2221 +=+

21 2y '(1) c e 2c e for x 1= + >

x 2xh 1 2y(x) y c e c e for x 1= = + >

86

2 1 7

9

y'(1) e e fo r 0 x 12 2 2

= +

x 2x1 7 9y'(x) e e for 0 x 12 2 2

= +

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(basic case)

Linear 2-nd order non-homogenous equations:

( )+ + = 2

1 22d y dy

a a y F x 0dxdx

89

Variation of parameters


D-operator method

= +

= +

1 2r x r x1 2

1 1 2 2

(i) y c e c e

c u (x) c u (x)

Since 1 2u (x) and u (x) are linearly independent of

another, we have

90

+ + = + + =

+ + = + + =

2 " '1 11 1 1 12

2" '2 2

2 2 2 22

d u dub cu 0 u bu cu 0

dxdx

d u dub cu 0 u bu cu 0

dxdx

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= +

= =

1 1

1 1

r x r x1 2

r x r x1 2

(ii) y c xe c e

u xe ; u e

+ + = + + =" ' " '1 1 1 2 2 2u bu cu 0 ; u bu cu 0

91

x1 2

x x1 2

(iii) y e (c cos x c sin x)

u e cos x ; u e sin x

= +

= =

" ' " '1 1 1 2 2 2u bu cu 0 ; u bu cu 0+ + = + + =

Linear 2nd Order Non-homogeneousEquations with Constant Coefficients :

)x(Fbydx

dya2

dx

yd2

2=++ (2)

The general (complete) solution of (2) is of the form:

92

0bydx

dya2

dx

yd2

2=++

y = yh x + yp x

where yh(x) is the general solution of

yp(x) is a particular solution of (2).

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Also assume:

+ =1 1 2 2v '(x) u (x) v '(x) u (x) 0 (6.1)

= +p 1 1 2 2y v u v u (6)

95

= + ++

p 1 1 2 2 1 1 2 2

0

(v 'y ' v u ' v u ' u v ' u )

= +1 1 2 2v u ' v u ' (7)

= + + +p 1 1 2 2 1 1 2 2y " v u " v u " v 'u ' v 'u ' (8)

Substituting (6), (7), (8) into (5):

0

1 1 1 1v u " 2au ' bu

+ + +

0

2 2 2 2v u " 2au ' bu +

+ +

' ' ' '

96

(6.1)

(9)v1' u1' + v2' u2' = F(x)

v1' u1 + v2' u2 = 0

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121 2 1 2

u F(x)v ' u u ' u ' u=

21

1 2 1 2

u F(x)v '

u u ' u ' u

=

Solve eqns

(6.1)&(9) to obtain:

97

Since u1, u2 and F(x) are known, v1 and v2 can beobtained.

yp = v1u1 + v2u2

The particular solution of (5) is then given by :

Exam le:2d y dy

2 3 6+ =

= (c1 + v1)u1 + (c2 + v2)u2

y = c1u1 + c2u2 + v1u1 + v2u2

The general solution of (5) is then given by :

98

31

xx2u e u, e

==

dxdx

21 1

12

d u du2 3u 0

dxdx+ =

22 2

22

d u du2 3u 0

dxdx+ =

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Undetermined coefficients (3 special F(x) functions):

rxF(x) e :=

F x sinkx coskx :=

Three different cases, depending on whether r is aroot of the characteristic equation.

101

Three different cases, depending on whether 0 isa root of the characteristic equation.

2F(x) ax bx c := + +

Two different cases, depending on whether ki is aroot of the characteristic equation.

Undetermined Coefficient Method:

Find a particular solution of

x2

2ey3

dx

yd=+

Guess yp = A ex

102

xp e

4

1 y =

xp

x2p2

e3A=3y,eAdx

yd=

4

1=A

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Assume: yp = A sin x

" = A sin x ' = A cos x

Example : Find a solution of

y" 3y' 4y = 2 sin x

by the undetermined coefficient method.

103

yp" =A sin x B cos x

A sin x 3 A cos x 4 A sin x = 2 sin x

No solution. Wrong Guess!!

Try: yp = A sin x + B cos x yp' = A cos x B sin x

17

5A =

17

3B=

(A + 3B 4A) sin x + (B 3A 4B) cos x = 2 sin x

-5A + 3B = 2 ; -3A 5B = 0

104

2

2

d y9y 9x cos x

dx+ =

using the undetermined coefficient method.

Example : Find a particular solution of

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Table 15.1The method of undetermined coefficients for selected equationsof the form : 2

2

d y dya by

dxdx

xF( )+ + =

If F(x) hasa term thatis aconstantmulti le of

And if yp

107

e rx r not a root of the characteristic eqn

r a single root of the characteristiceqn

r a double root of the characteristiceqn

A erx

A x erx

A x2 erx

sin kx,

cos kx

ki not a root of the characteristic

eqn

ki a root of the characteristic eqn

B cos kx + C sin kx

B x cos kx +C x sin kx

Table 15.1

The method of undetermined coefficients for selected equations ofthe form : 2

2

d y dya by

dxdx

xF( )+ + =

If F(x) hasa term thatis aconstantmultiple of

And if yp

108

ax2 + bx

+ c

0 not a root of the characteristic eqn

0 a single root of the characteristiceqn

0 a double root of the characteristiceqn

Dx + Ex + F

(chosen to matchthe degree of ax2 +bx + c)

Dx3 + Ex2 + Fx(degree 1 higherthan the degree ofax2 + bx + c)

Dx4 + Ex3 + Fx2

(degree 2 higherthan the degree ofax2 + bx + c)

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2x

23d y dy 2y e

dxdx+ =(1) (r2 + r 2) = 0

r1 = 2 , r2 = 12x x

h 1 2; y c e c e= +

xp

3 y Ae=

109

2x

22d y dy 2y e

dxdx

+ =(2)

r1 = 2 , r2 = 1

2xpy A ex

=

2x xh 1 2; y c e c e

= +

Trial function : yp = A x2 ex

2

2xd y dy2 y e

dxdx+ + =(3) (r2 + 2r +1) = 0

r1 = 1 , r2 = 1 ; yh = ( c1+ c2x) ex

110

22

d y y sin(x)dx

+ =(4) (r2 + 1) = 0

r = i , = 1

yh = c1 sin(x) + c2 cos(x) (1)

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)x3sin(8

1yp =

(2) + (3) = 8C1sin(3x) 8C2cos(3x) = sin(3x)

8

1C,0C 12 ==

113

1 21

y sin(x) cos(x) sin(3x)8

a a = +

0dx

yd2

2=

2

223

d yx

dx

x4 5+ +=

F(x)

(6)

r2 = 0 r1 = 0, r2 = 0

114

Trial function:

yp = Ax4 + Bx3 + Cx2

yh

= ( c1

+ c2x ) e x = c

1+ c

2x

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6x5x4dx

dy4

dx

yd 22

2++=+(7)

r1 = 0, r2 = 4 ,r2 + 4r = 0

Trial function:yp = Ax

3 + Bx2 + Cx

115


0y16dx

dy8

dx

yd2

2=++

for which y(2) = 3e8 and y'(2) = 10e8

The general solution (i.e. homogeneous) is

r2 + 8r + 16 = 0 r = 4, 4

4x 4xx1 2y c e c e = +

= = +8 8 8e e1 2y(2) 3e c c

116

= = +8 8 8 81 2 2y '(2) 10e 4c e 4c 2e c e

= +1 2

c c

= +1 210 4c 7c

1 2 c 1, c 2= =4x 4xy e 2xe = +

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,)x(ty)x(rdx

dy)x(r

dx

yd1212

2=++

=2d y dy

and y2(x) is a solution of

(9) If y1(x) is a solution of

117

2,

dxdx

+ + =2

1 2 1 22

d y dyr (x) r (x)y t (x) +t (x) .

dxdx

prove that y3(x) = y1(x) + y2(x) is a solution of

)x(ty)x(rdx

dy)x(r

dx

yd112

112

12

=++

dd2

y2(x) is a solution

Solution : y1(x) is a solution

118

xyxr

dx

xr

dx2221

2

=++

If y3(x) = y2(x) + y1(x)

dx

dy

dx

dy

dx

dy 213 +=then:22

2

21

2

23

2

dx

yd

dx

yd

dx

yd+=and

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323

123

2y)x(r

dx

dy)x(r

dx

yd ++

= + + + +

21 1

1 2 1

2

22

2 2

2

d y dd y dy

y

r (x) r (x)+ (ydx dxdxx yd )

119

1 2

y3 = y1 + y2 is a solution of

)x(t)x(ty)x(rdx

dy)x(r

dx

yd

21212

2+=++ (proved).

2

2

d y dy2 3y 6 7sinx

dxdx+ = + (i)

2d d

is given by the solution of

(10) The solution of:

120

2

ydxdx

+ =

2

2

d y dy2 3y 7sinx

dxdx+ =

plus the solution of

(iii)

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3x x3 4

7 7y c e c e sin x cos x= +

The solution of (iii) is

The solution of (ii) is

3x x1 2y c e c e 2

= +

121

3x x5 6

7 7y c e c e sin x cos x

5 102= +

The solution of (i) is therefore


(basic problem: three different cases incharacteristic eqns. )

Linear 2-nd order non-homogenous equations(general solution for the corresponding homogenous

Summary of Part II:

equat on p us part cu ar so ut on :

Method for calculating particular solution:

Variation of parameters


D-operator method (special cases, to bedi d l )

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2bFE1007F2-200809