s ds t 1 Z t s 2 3 s 2 3 ds t 2 Z t s 1 3 s 2 3 ds t 1 t 3 3 t 3 t 2 ln t 1 6 t

# S ds t 1 z t s 2 3 s 2 3 ds t 2 z t s 1 3 s 2 3 ds t

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]( s ) ds, = - t - 1 Z t s 2 (3 - s - 2 ) 3 ds + t 2 Z t s - 1 (3 - s - 2 ) 3 ds = - t - 1 t 3 3 - t 3 + t 2 ln( t ) + 1 6 t - 2 = - t 2 3 + 1 3 + t 2 ln( t ) + 1 6 Since the first term is part of the homogeneous solution, we write the general solution as y ( t ) = c 1 t - 1 + c 2 t 2 + 1 2 + t 2 ln( t ) , t > 0 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations Part 2 - Nonhomogeneous — (27/27)
• Fall '08
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