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A5 Solutions

2 u 0 v 4 u 12 x 1 solve as separableor linear de

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Unformatted text preview: u and y2 = (x + 1)u￿￿ + 2u￿. Let 1 second solution: ii) Find 2is a solution of DE on −∞ < x < ∞. Substitute into DE: ￿ ￿￿ (b) Let y2 = u(x)y1 = u(x)(x + 1). Then y2 = (x + 1)u￿ + u and y2 = (x + 1)u￿￿ + 2u￿. x + DE: Substitute (into 1)2 [(x + 1)u￿￿ + 2u￿] + 2(x + 1)[(x + 1)u￿ + u] − 2(x + 1)u = 0 x 1) 1)3 u￿￿ ] − 2( + 1)2 u￿ = 0 (x + 1)2 [(x + 1)u￿￿ + 2u￿] + 2(x + 1)[(x(+ + u￿ + u+ 4(xx + 1)u = 0 3 Let v = u￿ , v ￿ = u￿￿ to get first order DE (x + 1)(x ￿+ 1)3x ￿￿ + 4(v = 0. 2 u￿ = 0 v + 4( u + 1)2 x + 1) Solve as separable￿￿or linear DE. Assuming x > −1, ￿ ￿ 3￿ Let v = u , v = u to get first order DE (x + 1) v + 4(x + 1)2 v = 0. 1 Assuming > Solve as separable or linear DE. dv = − x 4 −1, v dx x+1 ￿ 1 dv ￿4 1 4 =− dv − v dx = x + 1 dx x+1 ￿v ￿ 1 4 ln dv| = −4 ln(x + dx + C0 |v 1) − v x+1 v = −4(ln(x 1)−4 + C = C1 x + + 1) ln |v | 0 −4 u￿ = C 1(x + 1)−4 v = C1 (x + 1) 1 u￿ = C1Cx +x + 4 −3 + C2 u − ( 1 ( 1)− 1) 3 1 u . −C y x (x +−3 + C Take simplest form: u(x) = (x + 1)−3=Thus 1 3 (= + 1) 1)−3 (x2+ 1) = (x + 1)−2 is a solution 2 for x >...
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