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homework set 19-Solutions - BILKENT UNIVERSITY Department...

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B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 19 U. Mu˘gan July 16, 2008 1) a) Corresponding homogenous equation is y 00 - y 0 - 2 y = 0 Since, it is a constant coefficient D.E. we look for the solution of the form y = e rx , then its C.E. is r 2 - r - 2 = 0 with the roots r 1 = 2 and r 2 = - 1. Hence the homogenous solution is y h ( x ) = c 1 e 2 x + c 2 e - x , c 1 , c 2 = constant . Assume y p = c 1 ( x ) e 2 x + c 2 ( x ) e - x then y 0 p = [2 c 1 ( x ) e 2 x - c 2 ( x ) e - x ] + [ c 0 1 ( x ) e 2 x + c 0 2 ( x ) e - x ] . We set c 0 1 ( x ) e 2 x + c 0 2 ( x ) e - x . Computing y 00 p and substituting in the D.E. gives 2 c 0 1 ( x ) e 2 x - c 0 2 ( x ) e - x = 2 e - x . Thus we have two algebraic equations for c 0 1 ( x ) and c 0 2 ( x ) with the solution c 0 1 ( x ) = (2 / 3) e - 3 x , c 0 2 ( x ) = - 2 / 3 Hence c 1 ( x ) = - (2 / 9) e - 3 x and c 1 ( x ) = - (2 x/ 3). Substituting c 1 ( x ) and c 2 ( x ) in y p gives y p ( x ) = - (2 x/ 3) e - x . b) Similarly, first find the homogenous solution y h ( x ) = c 1 e - x + c 2 xe - x c 1 , c 2 = constant . Assume y p = c 1 ( x ) e - x + c 2 ( x ) xe - x . c 0 1 ( x ) and c 0 2 ( x ) satisfy c 0 1 ( x ) e - x + c 0 2 ( x ) xe - x = 0 , and - c 0 1 ( x ) e - x + c 0 2 ( x )[ e - x - xe - x ] = 3 e - x with the solutions c 0 1 ( x ) = - 3 x, and c 0 2 ( x ) = 3 . 1 I made every effort to avoid the calculation errors and/or typos while I prepared the solution set. You are responsible to check all the solutions and correct the errors if there is any. If you find any errors and/or misprints, please notify me.
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Therefore y p ( x ) = (3 / 2) x 2 e - x .
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