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homework set 16-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 # 16 U. Mu˘gan July 16, 2008 1) a) W ( e 2 x , e - 3 x/ 2 ) = fl fl fl fl fl e 2 x e - 3 x/ 2 2 e 2 x - 3 2 e - 3 x/ 2 fl fl fl fl fl = - 3 2 e x/ 2 - 2 e x/ 2 = - 7 2 e x/ 2 (1) b) W ( x, xe x ) = fl fl fl fl fl x xe x 1 e x + xe x fl fl fl fl fl = xe x + x 2 e x - xe x = x 2 e x (2) c) W ( e x sin x, e x cos x ) = fl fl fl fl fl e x sin x e x cos x e x sin x + e x cos x e x cos x - e x sin x fl fl fl fl fl = - e 2 x (3) 2) a) If y 00 + p ( x ) y 0 + q ( x ) y = g ( x ), then for the given equation p ( x ) = - 3 x x - 1 , q ( x ) = 4 x - 1 , g ( x ) = sin x x - 1 So, only point of discontinuity of the given differential equation (D.E.) is x = 1. By the existence and uniqueness theorem the largest interval is -∞ < x < 1. Since then initial point is at x 0 = - 2 which is contained by the interval. b) p ( x ) = cos x, q ( x ) = 3 ln | x | , g ( x ) = 0 . So, the largest interval is 0 < x < which contains the initial point x 0 = 2. c) p ( x ) = 1 x - 2 , q ( x ) = tan x, g ( x ) = 0 So, point of discontinuity of p ( x ) is x = 2 and the points of the discontinuities of tan x are x = ± (2 n + 1) π 2 , n = integer. Therefore, the largest interval which contains the initial point is 2 < x < 3 π/ 2. 3) If you take the derivatives of y 1 ( x ) = 1 and substitute into given D.E. equation is identically satisfied. Similarly, for y 2 ( x ) = x 1 / 2 . If y = c 1 + c 2 x 1 / 2 is substituted in the given D.E., we get - 1 4 c 1 c 2 x - 3 / 2 = 0 which is zero only if c 1 = 0 or c 2 = 0. Thus, the superposition of two L.I. solutions is not, in general a solution. Note that the D.E. is nonlinear. 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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4) W ( e 2 x , e - 3 x/ 2 ) = fl fl fl fl fl x g 1 g 0 fl fl fl fl fl = xg 0 -
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