Nothing Special   »   [go: up one dir, main page]
Scribd Logo
Upload

Welcome to Scribd!

  • 33 views

    0perator Methods L#9.2

    Uploaded by

    Kimondo King
    This document discusses using operator methods to solve differential equations where the right hand side is a polynomial. It presents a theorem that expresses the inverse of an operator F(D) as a polynomial in D. It then uses this theorem to determine particular solutions for differential equations of the form F(D)y = Q(x). Examples are worked through, including determining particular solutions and combining them with the general solution to solve initial value problems.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    0perator Methods L#9.2

    Uploaded by

    Kimondo King
    33 views3 pages
    This document discusses using operator methods to solve differential equations where the right hand side is a polynomial. It presents a theorem that expresses the inverse of an operator F(D) as a polynomial in D. It then uses this theorem to determine particular solutions for differential equations of the form F(D)y = Q(x). Examples are worked through, including determining particular solutions and combining them with the general solution to solve initial value problems.

    Original Title

    0PERATOR METHODS L#9.2

    Copyright

    © © All Rights Reserved

    Available Formats

    PDF, TXT or read online from Scribd

    Did you find this document useful?

    Is this content inappropriate?

    This document discusses using operator methods to solve differential equations where the right hand side is a polynomial. It presents a theorem that expresses the inverse of an operator F(D) as a polynomial in D. It then uses this theorem to determine particular solutions for differential equations of the form F(D)y = Q(x). Examples are worked through, including determining particular solutions and combining them with the general solution to solve initial value problems.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    33 views3 pages

    0perator Methods L#9.2

    Uploaded by

    Kimondo King
    This document discusses using operator methods to solve differential equations where the right hand side is a polynomial. It presents a theorem that expresses the inverse of an operator F(D) as a polynomial in D. It then uses this theorem to determine particular solutions for differential equations of the form F(D)y = Q(x). Examples are worked through, including determining particular solutions and combining them with the general solution to solve initial value problems.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 3

    OPERATOR METHODS :

    We wish to consider the determination of when Q(x) is a polynomial using operator method.

    The following theorem will be handy in finding .

    Theorem: If F(D) = aD2+bD+c then +

    Proof : Since ]-1 = [aD2+bD+c ]-1 then this can be expanded in ascending powers of D (
    say, by binomial theorem or by long division)

    REM:
    2
    y3
    1 . . . for -1
    2! 3!

    2 If +

    1
    Then y p  x n  [a0  a1 D  a 2 D 2  ...]x n a0  0
    F ( D)

    Now since D n (xm )  0 for n  m ,

    then y p  +

    Ex#1: Determine:

    1 2 1 1
    (a) (x ) (b) 2
    (x3 ) c) (12 x)
    D D D3

    SOLN (c)
    1
    D 3
    1 1  1
    (12 x ) = 2  (12 x)  2
    D D  D
    12xdx 
    1 1 1 
    = (6x 2 ) =  (6 x 2 ) 
    D 2
    D D 
    1 1
     6x (2 x 3 )
    2
    = dx =
    D D

     2 x dx
    3
    =

    1 4
    = x
    2

    Exercise to the learner....? (a) and (b).

    x3 1 5
    [ ANS: (a) (b) x ]
    3 20

    Page 1 of 3
    Ex#2: Determine :

    1 1 1 1 1
    (a) ( x 2 ) (b) 2 ( x 3 ) (c) 2 ( x 2  1) (d) 2 ( x 2  2 x) (e) ( x 2  2 x  1)
    D 1 D 1 D 1 D  D 1 2D  2D  3
    2

    SOLN: ( Would you like to try (a) on your own first, before looking at the solution ....? )

    1
    (a) Now ( x 2 ) = (1+D) 1 (x 2 )
    D 1
    2
    y3
    But  . . . (#)
    2! 3!

    D2
    So
    2

    = (1–D+ + . . .)

    = -D( ) +D 2 (

    = - 2x +2

    1
    REM: Series expansion of may be obtained by long division...Recall ?
    y 1

    1 y  y2
    1 y 1
    1 y
    0 y
    0  y  y2
    0  0  y2
    0  0  y2  y3

    1
    So = 1- y +y2 +. . .
    y 1

    1
    Therefore ( x 2 ) = ( 1 - D +D2 +. . .) (x2)
    D 1

    1
    (c) ( x 2  1)  ( D 2  1) 1 ( x 2  1)
    D 1
    2

    Now (-1 + y) 1 =(-1)-1 (1-y)-1

    = -1[1+(-1)y +(-1)(-1-1) +(-1)(-1-1)(-1-2) +. . .

    = -{ 1 – y +y2-y3+. . . }

    =- 1 +y+y2+y3+ . . .

    Page 2 of 3
    1
    ( x 2  1)  (1  D 2  D 4  D 6 ....)( x 2  1)
    D 1
    2

    = (1  D 2  D 4  D 6 ....) ( x 2 )  (1  D 2  D 4  D 6 ....) 1

    = (x  2)  1
    2

    Exercise to the learner? . . . try (b) (d) and (e)

    [ ANS: (b) ( 1  D  D 2  ... )( x3 )  x3  6 x (d) (1  D  D 3  ...)( x 2  2 x)  x 2  4 x  2

    1 2 2 25
    (e) x  x
    3 9 27

    REM: We are now equipped to solve the ODE a2 y   a1 y   a0 y  Q( x) where Q(x) :

    Ex#2: Determine the particular solution for each of the following equations by a suitable operator
    method and hence solve the equation.

    (a) ( D 2  1) y  x 2 (b) ( D 2  2D  1) y  x  1

    SOLN:

    (a) For y H : :

    ( D 2  1) y  0 and the auxiliary equation is m 2  1  0

    Then m1  1 , m2  1

    So that y H  C1e m1x  C2 e m2 x

     y H  C1e  x  C2 e x

    For y p :

    1
    yp  (x 2 )
    D 1
    2

     (1  D 2  D 4  ...)( x 2 )

    = - x2  2

    Hence y  y h  y p

    => y  C1e  x  C2 e x - x 2  2

    Exercise: (b)

    { ANS: y  (C1 x  C2 )e x  x  1 }

    Page 3 of 3

    You might also like