8/13/2019 Fe1007 Lecture 2c

1/22

Part III

-

123

Power Series Method

Methods for solving some special higher-order equations

D-operator method (the most importantmethod we would discuss in this part)

124

Reduction of order by substitution

Power series method

8/13/2019 Fe1007 Lecture 2c

2/22

F(D) e kx = F(k) e kx

Theorem 1:

If k is a constant,

To prove this, we note that

Theorems Involving D-Operator :

125

kx kxd e kedx

=

D e kx = k e kx

.

.

D2e kx = D(De kx) = D(ke kx) = k2e kx

Dne kx = kne kx

Consequently:

F(D) e k = (a 0Dn + a 1Dn1 + + a n1 D + a n) e kx

= (a 0kn + a 1kn1 + + a n1 k + a n) e kx

= F(k) e kx

126

(1) (6D2 + 4 D 2) e 3x = (6 32 + 4 3 2) e 3x

= 64 e 3x

(2) (D2 + 1) e 2x = (22 + 1) e 2x

= 5 e 2x

Examples :

8/13/2019 Fe1007 Lecture 2c

3/22

Theorem 2 ( Shifting Theorem) :

If k is a constant and g(x) is a function of x, then :

F(D) [e kx g(x)] = e kx F (D + k) g(x)

To prove this, we note that :

D [e kx g(x)] = e kx D g(x) + ke kx g(x)

127

= e kx (D + k) g(x)

D2 [ekx g(x)] = D [ e kx (D + k) g(x) ]

= e kx D(D + k) g(x) + ke kx (D + k) g(x)

= e kx (D + k) (D + k) g(x)= e kx (D + k) 2 g(x)

n n n 1 n 1

n

n nD (uv) (D u)v (D u)Dv ... (Du)D v

1 n 1

n

n n! uD v

n k k!(n k)!

= + + +

+

=

Dn [ekx g(x)] = ? Leibniz's Theorem states that :

128

n n 1 n 1

n (D u)v n (D u)Dv ... n (Du)D v

uD v

= + + +

+

3 3 2 2 3D (uv) (D u)v 3 (D u)Dv 3 (Du)D v uD v= + + +

8/13/2019 Fe1007 Lecture 2c

4/22

Dn [g(x) e kx]

= e kx Dn g(x) + n D (e kx) Dn1 g(x) +

+ n D n1 (e kx) D g(x) + (D n e kx) g(x)

Replacing V by e kx and U by g(x) , we have :

129

= e kx Dn g(x) + n ke kx Dn1 g(x) +

+ n k n1 e kx D g(x) + k n e kx g(x)

= e kx ( Dn + nk D n1 + + nk n1 D + k n ) g(x)

= e kx (D + k) n g(x)

2

( 6D 2 + 4D 2) [e3x x2] = e 3x [ 6 (D + 3) 2 + 4(D + 3) 2] x2

= e 3x ( 6D 2 + 40D + 64 ) x2

Example: F(D) [e kx g(x)] = e kx F(D + k) g(x)

130

x2e 6 40 64 xdxdx

= + +

= e 3x (6 2 + 40 2x + 64 x 2)

= e 3x (12 + 80x + 64 x 2)

8/13/2019 Fe1007 Lecture 2c

5/22

Consider the following ODE, (find y as a function of x) :

Solving ODE Using D-Operator :

(D2 + 1) y = e 2x (1)

131

1LHS : F(D) y = y ;

F(D)

=

F(D) 0

Define the inverse D-operator, such that :1

F(D)

,

1RHS : g(x)

F(D)

2x1 eF(D)

F(D) e kx = F(k) e kx (2)

To solve (1), we recall Theorem 1 :

Apply the inverse operator on both sides of eqn (2)and we get :

=kx kx1 = kx1

F k e

132

= kx kx ; F(k)1 1

e eF

0(D)

F(k)

(3)

which gives :

F(D)

F(D)

8/13/2019 Fe1007 Lecture 2c

6/22

8/13/2019 Fe1007 Lecture 2c

7/22

x2x22p e)D(F

1e)3D4D(

1y =++=

(D2 + 4D + 3) y = e 2x

The particular solution is given by :

Example:

135

x2x22p

e151

e)3242(

1y =

++=

3x x 2x1 2

1y c e c e e

15

= + +

,

General solution is

2x 2xp 3 2

1 1y = e = eF(D)(D +4D +3D)

(D3 + 4D 2 + 3D) y = e 2x

The particular solution is given by :

Another Example:

136

2x 2x

p 3 21 1

y = e = e30(2 +4 2 +3 2)

= + + +3x x 2x0 1 21

y c c e c e e30

,

General solution is

8/13/2019 Fe1007 Lecture 2c

8/22

Theorem 3:

If k is a constant,

F(D 2) sin (kx) = F( k2) sin (kx) and

F(D 2) cos (kx) = F( k2) cos (kx)

137

where F(D 2) is a polynomial operator with D replaced byD2.

Example : F(D 2 ) = (D 4+ 3D 2 1)

D2 sin(3x) = D (3cos(3x)) = 3 2sin(3x)

D2 cos(3x) = D (3sin(3x)) = 3 2cos(3x)

(1) (D4 + 3D 2 1) sin (2x) = ((D 2)2 + 3D 2 1) sin (2x)

= (22)2 + 3( 22) 1) sin (2x)

= 3 sin (2x)

Apply to the LHS, we get sin(2x).1

138

)8()x2sin(31

)x2sin()1D3D(

124 =+

Therefore:

+

Apply it to the RHS , We get+ 4 2

1 3sin(2x).

(D 3D 1)

8/13/2019 Fe1007 Lecture 2c

9/22

(2) (D 4 2D 2) cos (2x)

= ((22)2 2(22)) cos (2x)

= 24 cos (2x)

139

Therefore:

)x2cos(241)x2cos(

)D2D(

124

=

xhy Ae

=

(3) (D + 1)y = 10 sin (2x)

+ = =+ +p1 1

(D 1)y y 10 sin(2x)D 1 D 1

140

2(D 1)

10sin(2x)D 1

=

210(D 1)

sin(2x)D 1

=

8/13/2019 Fe1007 Lecture 2c

10/22

8/13/2019 Fe1007 Lecture 2c

11/22

21

sin(4x)(D 5D 6) +

1 sin(4x)

( 10 5D)=

Therefore, y p is given by :

p 100y sin(4x)( 10 5D)=

20 sin(4x)

2 D=+

143

220 (D 2)

sin(4x)D 4

=

20 (D 2)sin(4x)

20=

(D 2)sin(4x)= Dsin(4x) 2sin(4x)=

4cos(4x) 2sin(4x)=

2(D 2D)y 24x ;+ =(5)

x)2D(D

24yp += x

)2D(21

D21 24

+=

2xh 1 2y ec c

= +

21 2r 2r 0 ; r 0 & r 2+ = = =

144

x 2D 1D1 12 +=

)x(fxD1 Let = x Df(x) =

dor x f(x)

dx=

2xIntegrating both sides gives : f(x) xdx

2= =

8/13/2019 Fe1007 Lecture 2c

12/22

=

+ +

1 1 1x x

DD 2 2 12

x ...8

D4

D2D1

21 32

++=

1 1 x 0....2 2 = +

2 31 D D D x x x x ...

2 2 4 8

= + +

145

3x6x6 2 +=

ph yyy +=

2p

x x 1 y 12

2 2 4

= +

43 xy)1D(D =(6)

43p

x)1D(D

1y

=

43

2x

1D1

D

1DD

+=

146

4432

x D11x

D

1

D

1D1

+=

210x

30x

5x 765 = 4432 x ]DDDD1[ +++++

8/13/2019 Fe1007 Lecture 2c

13/22

5 6 7

px x x5 30 210

y = 24x24x12x4x 234

r3 (r 1) = 0 r = 0, 0, 0, 1

43 xy)1D(D =

147

yh = (c 1 + c 2x + c 3x2) e 0 + c 4e x

= (c 1 + c 2x + c 3x2) + c 4e x

h py y y= +

0qppr2r 222 =+++

+ + + =2 2 2 ax(D 2pD p q )y e(7)

where p, q and a are non-zero constants.

q4p4p4p2r222 +=

148

qip +=

( )pxh 1 2y e cos(qx) sin(qx)c c= +

qipr2 =

8/13/2019 Fe1007 Lecture 2c

14/22

Particular solution can be obtained by the undeterminedcoefficient method.

Trial solution : yp = C eax

D-operator Method :

ax1

149

p 2 2 2D 2pD p q=

+ + +

ax222

eqpap2a

1

+++=

ph yyy +=

SummaryD -Operator method

F(D)e kx=F(k)e kx

150

F(D) [e kxg(x)] = e kxF(D+k) g(x)

F(D 2) sinkx = F(-k 2) sinkx

F(D 2) coskx = F(-k 2) coskx

8/13/2019 Fe1007 Lecture 2c

15/22

Methods for solving some special higher-order equations

D-operator method (the most importantmethod we would discuss in this part)

151

Reduction of order by substitution

Power series method

Reduction of Order

dxdPy =

22

dP dP P or dx

dx P = =

2y (y ) =(1)

=Let y P

152

1 x CP

= + 1Px C

= +

1y ln(x C) C= + +

dy 1 1 ; dy dx

dx x C x C= =

+ +

8/13/2019 Fe1007 Lecture 2c

16/22

P2dxdPx =

xdx2

PdP =

2 ln P ln x C = + 21or P C x=

y2yx =(2) = =dP

, yy Pdx

Let :

153

212

d yi.e., C x

dx= 2

31 CxC3

1dxdy +=

324

1 CxCxC121y ++=

324

4 CxCxC ++=

Power Series Method :Solve the differential equation:

y' = x + ysubject to the condition y(1) = 0

The Taylor series for y(x), in powers of x 1, is :

y"(1) = 1 + y'(1) = 2

y(1) = 0 , y' = x + y , y(1)' = 1 + y(1) = 1

(n)(1)y ny(1) + y'(1)(x - 1) + ... + + ...(x-1)n!

8/13/2019 Fe1007 Lecture 2c

17/22

y" = 1 + y' y'" = y"

y'"(1) = y"(1) = 2

(n+1) (n)y (x) = y (x) = 2 for n 2 Recurrence Eqn

155

)1x(!3

12)1x(!2

12)1x(10)x(y 32 ++++=

=

+

2n

n

!n)1(x2)1x(=

xn

n 2

1 (x 1)2 2 1 (x 1)n

e!

=

= + +

n

nxn 2

xSince : 2e 2 1 xn!

=

= + +

156

n

2

x

n

1 (x 1)(x 12e 2 (x ) 2n!

1)

=

= +

x 1 y(x) 2e 2 (x 1) =

n 2

n!x

=+ = +

8/13/2019 Fe1007 Lecture 2c

18/22

Example: Solve the initial value problem

Assume solution of the form :

y = a 0 + a 1x + a 2x2 + ... + a n1 xn1 + a nxn + ... (1)

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

157

y' = a 1 + 2a 2x + 3a 3x2 + ... + na nxn1 + ... (2)

y' y = (a 1 a 0) + (2a 2 a 1)x + (3a 3 a 2) x2

+ ... + (na n a n1)xn1 + ... (3)

Comparing (3) with y' y = x , we get :

2a 2 a 1 = 1

3a 3 a 2 = 0

na n a n1 = 0

a 1 a 0 = 0 a 1 = a 0

From (1) : y(0) = a 0 + 0 + ... = 1

158

a 0 = 1 , a 1 = 1

12a1a 12 =

+=

!32

232

31

3aa 23 ==== 4 n

2 2, a , a

4! n!= =

8/13/2019 Fe1007 Lecture 2c

19/22

2 3 nx x xy 1 x 2 2 2

2! 3! n! = + + + + + +

!nx !3x!2x2x1 n32

++++++=

159

)x1e(2x1 x ++=

x1e2y x =

Summary Reduction of order :

By substitution, the order of the equation may belowered (mostly applies to equations without zero-order derivative item, such asy""-y"= g(x) )

160

Power series method Mostly by using the Taylor power series and building

up a relationship between lower-order derivatives andhigher-order derivative. For example, from y"-y=2, wehave y"'-y'=0,y""-y"=0, etc.)

8/13/2019 Fe1007 Lecture 2c

20/22

Final Review: first-order equations (1)

Variable separable equation (the basic case)

M(x)dx + N(y)dy = 0

Methods for transforming other equations into variable

161

dy = F(ax + by)dx

separable equations:

Let v ax by= +

dy yF

dx x

=

yLet v

x=

Exact Differential Equations:

Final Review: first-order equations (2)

Linear Equations: dy Py Qdx

+ =1PdxLet = e , y = Qdx

Bernoullis Equations: ndy Py Qydx

+ = 1 nLet z y =

162

M(x,y)dx N(x,y)dy 0, where y x+ = =

Solution: F(x,y)=c

where F/ x=M(x,y), and F/ y=N(x,y)

Applications of ODE (Based on the meaning of Derivate:Derivate measures the rate of change)

8/13/2019 Fe1007 Lecture 2c

21/22

Final Review: second-order equations (1)

Linear second-order homogenous ODE (the basic case)a. Two equal/unequal real number roots

1 2r x r xh 1 2 1 2y c e c e , when r r= +

1r xh 1 2 1 2y (c c x)e , when r r= + =

163

. x

h 1 2y e (c cos x c sin x)= +

Linear second-order non-homogenous equations withconstant coefficients

General solution=general solution of homogenous equation+ particular solution

Variation of parameters:

Particular Solution: p 1 1 2 2y u v u v= +where

Final Review: second-order equations (2)Methods for finding particular solutions :

164

2

11 2 1 2v ' u u ' u 'u

=

12

1 2 1 2

u Fv '

u u ' u 'u=

8/13/2019 Fe1007 Lecture 2c

22/22

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

Three different cases, depending on whether 0 is

a root of the characteristic equation.

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

rxF(x) e :=

Final Review: second-order equations (3)

165

F(x) sinkx, coskx :=

ree eren cases e , xe , x e , epen ngon whether r is a root of the characteristic equation.

Two different cases as below, depending on whetherki is a root of the characteristic equation.

Bcoskx Csinkx; Bxcoskx Cxsinkx+ +

D-Operator method

F(D) e kx = F(k)e kx

Shifting method: F(D) [e kxg(x)] = e kxF(D+k) g(x)F(D 2) sinkx = F(-k 2) sinkx;

2 = - 2

Final Review: second- and higher-order equations

166

Reduction of order by substitution

Power series method