Unformatted text preview: A = 0 and A 1 = 1 / 6. Therefore x p ( t ) = t 3 e 2 t / 6 is a particular solution to the given nonhomogeneous equation. 23. The righthand side of this equation suggests that y p ( θ ) = θ s ( A 2 θ 2 + A 1 θ + A ). We choose s = 1 because r = 0 is a simple root of the auxiliary equation, r 2 − 7 r = 0. Therefore, y p ( θ ) = θ ( A 2 θ 2 + A 1 θ + A ) = A 2 θ 3 + A 1 θ 2 + A θ ⇒ y p ( θ ) = 3 A 2 θ 2 + 2 A 1 θ + A ⇒ y p ( θ ) = 6 A 2 θ + 2 A 1 . 193...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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