L45 - Spring 2003 Math 308/501502 4 Linear Second-Order...

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Spring 2003 Math 308/501–502 4 Linear Second-Order Equations 4.5 Method of Undetermined Coeffs Wed, 01/Oct c ± 2003, Art Belmonte Summary The method of undetermined coefficients is dealt with in full generality in the Section 6.3 lecture, q.v. In this lecture handout, we’ll restrict ourselves to examples of second-order equations. By hand or with MATLAB, you’re a winner! The hand work involved in computing derivatives, substitution, collecting terms, and solving linear systems can be tedious and error-prone. It lends itself quite well, however, to machine power, as you’ll observe in the MATLAB Examples . You may have felt that the formulation for y p given in Section 6.3 is overly complicated. Take solace in the fact that under that stated formulation there is only one case (the general case). What’s more, there is no guesswork as to the correct form of y p . With practice, you’ll get it right every time! Note that in all examples (hand or MATLAB), we carry the work through to a full general solution (or to the unique solution of an initial value problem). Please do this in your homework problems as well. Hand Examples Example A Find a general solution of y 00 + 3 y 0 - 18 y = 18 e 2 t . Solution 1. The homogeneous equation y 00 + 3 y 0 - 18 y = 0has characteristic equation 0 = r 2 + 3 r - 18 = ( r - 3 )( r + 6 ) , with roots r = 3 , - 6. Hence y h = c 1 e 3 t + c 2 e - 6 t .The forcing function is f ( t ) = 18 e 2 t ,whichis not a solution of the homogeneous equation. (“There is no interference.”) 2. So let y p = ae 2 t .Then y 0 p = 2 2 t and y 00 p = 4 2 t . Substituting into the nonhomogeneous equation gives ( 4 2 t ) + 3 ( 2 2 t ) - 18 ( 2 t ) = 18 e 2 t ( - 8 a - 18 ) e 2 t = 0 = 0 e 2 t 3. Equating coefficients of like entities, we have - 8 a - 18 = 0, whence a =- 9 4 . Thus y p 9 4 e 2 t is our particular solution. 4. A general solution (verified by dsolve )is y = y p + y h 9 4 e 2 t + c 1 e 3 t + c 2 e - 6 t Example B Find a general solution of y 00 + 7 y 0 + 10 y 4sin3 t . Solution 1. The homogeneous equation y 00 + 7 y 0 + 10 y = characteristic equation 0 = r 2 + 7 r + 10 = ( r + 2 )( r + 5 ) , with roots r 2 , - 5. Hence y h = c 1 e - 2 t + c 2 e - 5 t forcing function is f ( t ) t not a solution of the homogeneous equation. 2. Let y p = a cos 3 t + b sin 3 t y 0 p 3 a sin 3 t + 3 b cos 3 t y 00 p 9 a cos 3 t - 9 b sin 3 t Thus ( - 9 a cos 3 t - 9 b sin 3 t ) + 7 ( - 3 a sin 3 t + 3 b cos 3 t ) + 10 ( a cos 3 t + b sin 3 t ) t . Accordingly, we have ( a + 21 b ) cos3 t + ( 4 + b - 21 a ) sin3 t = 0 = 0cos3 t + 0sin3 t 3. Equating coefficients of like entities gives a + 21 b = 0and 4 - 21 a + b =
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L45 - Spring 2003 Math 308/501502 4 Linear Second-Order...

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