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Chem Differential Eq HW Solutions Fall 2011 183

# Chem Differential Eq HW Solutions Fall 2011 183 - Section...

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Section A.3 Linear Ordinary Differential Equations with Nonconstant Coefficients A183 17. Put the equation in standard form: xy + 2 (1 - x ) y + ( x - 2) y = 0 , y 1 = e x ; y + 2(1 - x ) x y + x - 2 x y = 0 , p ( x ) = 2 x - 2; p ( x ) dx = 2 ln x - 2 x e - p ( x ) dx = e - 2 ln x +2 x = e 2 x x 2 ; y 2 = e x e 2 x x 2 e 2 x dx = e x 1 x 2 dx = - e x x . Hence the general solution y = c 1 e x + c 2 e x x . 21. y - 4 y + 3 y = e - x . λ 2 - 4 λ + 3 = 0 ( λ - 1)( λ - 3) = 0 λ = 1 or λ = 3 . Linearly independent solutions of the homogeneous equation: y 1 = e x and y 2 = e 3 x . Wronskian: W ( x ) = e x e 3 x e x 3 e 3 x = 3 e 4 x - e 4 x = 2 e 4 x . We now apply the variation of parameters formula with g ( x ) = e - x ; y p = y 1 - y 2 g ( x
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