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Sol-HW2 - Math Methods Fall 2010 Solutions for Assignment 2...

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Math Methods Solutions for Assignment 2 Sep. 01, 2010 Fall 2010 Differential Equations Due Sep. 08, 2010 1 (6). Find the general solutions of the following equations: a) (4 y - 3 x ) dx + ( y - 2 x ) dy = 0. Since the coefficients are homogenous functions of order one, we use u = y/x . 4 y x - 3 dx + y x - 2 dy = 0 (4 u - 3) dx + ( u - 2)( udx + xdu ) = 0 ( u 2 + 2 u - 3) dx + x ( u - 2) du = 0 dx x + u - 2 ( u + 3)( u - 1) du = 0 dx x + 5 / 4 u + 3 - 1 / 4 u - 1 du = 0 ln( x ) + 5 4 ln( u + 3) - 1 4 ln( u - 1) = c x 4 ( u + 3) 5 u - 1 = c x 4 ( y/x + 3) 5 y/x - 1 = c ( y + 3 x ) 5 y - x = c b) y (4) + 2 y 00 + y = 0. The substitution y = e λx yields λ 4 + 2 λ 2 + 1 = ( λ - i ) 2 ( λ + i ) 2 = 0 . Linearly independent solutions are e ix , xe ix , e - ix and xe - ix . In terms of functions of real variables these are sin x, x sin x, cos x and x cos x . y = c 1 cos x + c 2 sin x + c 3 x cos x + c 4 x sin x c) y 00 + 1 x y 0 + 1 x 2 y = x . This is an Euler equation. First, we search for solution of homogenous equation as y = x λ . λ ( λ - 1) + λ + 1 = 0; λ 2 + 1 = 0; λ = ± i.

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Linearly independent solutions are x i = e i ln x and x - i = e - i ln x , and the general solution of homogenous equation is y = c 1 cos(ln x ) + c 2 sin(ln x ). Since the inhomogenous part of the equation is a polynomial, it is easier to search for the particular solution as a polynomial. We can search for it as y = cx 3 , substituting we get 6 cx + 3 cx + cx = x , i.e. c = 1 / 10. Hence y = c 1 cos(ln x ) + c 2 sin(ln x ) + 1 10 x 3 2 (4). Find the general solutions of the equation (1 - 2 x ) y 00 + 4 xy 0 - 4 y = 0. ( Hint: y = x is a solution) If we know one solution, we can find the other one by “reduction of order”.
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Sol-HW2 - Math Methods Fall 2010 Solutions for Assignment 2...

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