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Unformatted text preview: B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 18 U. Mu˘gan July 16, 2008 1) a) First we find the solution of the corresponding homogenous D.E. y 00 2 y 3 y = 0 , which has the C.E. r 2 2 r 3 = ( r 3)( r + 1) = 0 . Hence y h ( x ) = c 1 e 3 x + c 2 e x , c 1 ,c 2 = constant , and we assume y p = Ae 2 x for the particular solution. Thus y = 2 Ae 2 x and y 00 = 4 Ae 2 x and substituting into the D.E. yields 4 Ae 2 x + 2(2 Ae 2 x ) 3( Ae 2 x ) = 3 e 2 x Thus 3 A = 3 and A = 1, yielding y ( x ) = c 1 e 3 x + c 2 e x e 2 x . b) Homogenous solution y h ( x ) = c 1 + c 2 e 2 x , c 1 ,c 2 = constant . Initially we assume y p = A + B 1 sin2 x + B 2 cos2 x. However, since a constant is a homogenous solution of the D.E. we must modify y p by multiplying the constant A by x and thus the correct form of y p is y p = Ax + B 1 sin2 x + B 2 cos2 x. If we take the derivatives if y p and substitute into D.E. the coefficients of the similar terms give A = 3 / 2, B 1 = B 2 = 1 / 2. c) Homogenous solution y h ( x ) = c 1 e x + c 2 xe x , c 1 ,c 2 = constant . Assume y p = Ax 2 e x , 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. so that y p = 2 Axe x Ax 2 e x and y 00 p = 2 Ae x 4 Axe x + Ax 2 e x . Substituting in the d.E. gives ( Ax 2 4 Ax + 2 A ) e x + 2( Ax 2 + 2 Ax ) e x + Ax 2 e x = 2 e x . Note that all terms on the left involving x 2 add to zero and we left with 2 A = 2 or A = 1. Hence y = c 1 e x + c 2 xe x + x 2 e x . d) Assume that y p = ( Ax + B )sin2 x + ( Cx + D )cos2 x which is appropriate for both terms appearing in R ( x ). Since none of the term in y p is the homogenous solution, we do not need to modify y p . Ans.: y = c 1 cos x + c 2 sin x 1 3 x cos2 x 5 9 sin2 x, c 1 ,c 2 = constant . e) C.E. of the corresponding homogenous equation is r 2 + r + 4 = 0. Hence, y h = e x/ 2 [ c 1 cos( √ 15 x/ 2) + c 2 sin( √ 15 x/ 2)] , c 1 ,c 2 = constant ....
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This note was uploaded on 06/18/2010 for the course COMPUTER S Math 225 taught by Professor Yosumhoca during the Spring '10 term at Bilkent University.
 Spring '10
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