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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 # 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 + 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∞ &lt; x &lt; 1. Since then initial point is at x = 2 which is contained by the interval. b) p ( x ) = cos x, q ( x ) = 3ln  x  , g ( x ) = 0 . So, the largest interval is 0 &lt; x &lt; ∞ which contains the initial point x = 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 &lt; x &lt; 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....
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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
 YosumHoca

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