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Unformatted text preview: M 427K Variation of Parameters – PS October 10, 2011 Problem 1 Find the general solution to the following differential equation t 2 y 00 + 4 t 2 y + 4 t 2 y = e 2 t (1) Solution : The standard form of the differential equation is y 00 + 4 y + 4 y = e 2 t t 2 . Corresponding Homogeneous solution , y h : Solve y 00 + 4 y + 4 y = 0 . The characteristic equation is r 2 + 4 r + 4 = 0 , and the roots are r 1 = 2 , r 2 = 2 . Therefore, we have y 1 = e 2 t , y 2 = te 2 t . Particular solution, y p : Recall that y p has the following form: y p = u 1 y 1 + u 2 y 2 , where u 1 and u 2 can be evaluated using the following equations: u 1 ( t ) = Z y 2 f ( t ) W ( y 1 ,y 2 ) dt u 2 ( t ) = Z y 1 f ( t ) W ( y 1 ,y 2 ) dt We have, W ( y 1 ,y 2 ) := det y 1 y 2 y 1 y 2 = det e 2 t te 2 t 2 e 2 t e 2 t 2 te 2 t = e 4 t . [email protected] 1 c hf, 2011 M 427K Variation of Parameters – PS October 10, 2011 Therefore, we have u 1 ( t ) = Z y 2 f ( t ) W ( y 1 ,y 2 ) dt = Z te 2 t e...
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This note was uploaded on 02/03/2012 for the course M 427K taught by Professor Fonken during the Fall '08 term at University of Texas.
 Fall '08
 Fonken
 Differential Equations, Equations

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