hw5soln-corrected

# 3 2 1 3 2x y c 0 c 1 x x x e c 2 e 2x 16 4 6 5 the

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Unformatted text preview: y = e 2x s ￿ + 2 s ; ￿ ￿ D 2 y = e 2x s ￿￿ + 4s ￿ + 4s ; ￿ ￿ D 3 y = e 2x s ￿￿￿ + 6s ￿￿ + 12s ￿ + 8s . ￿ ￿￿ ￿ ￿ ￿ ￿ ￿ ￿ D 3 − 2D 2 − 4D + 8 y = e 2x s ￿￿￿ + 6s ￿￿ + 12s ￿ + 8s − 2 s ￿￿ + 4s ￿ + 4s − 4 s ￿ + 2s + 8s ￿ ￿ = s ￿￿￿ − 4s ￿￿ e 2x . Note that s ￿￿ = 6αx + 2β; s ￿￿￿ = 6α; and therefore ￿3 ￿ ￿ ￿ D − 2D 2 − 4D + 8 y = e 2x 6α + 24αx + 8β ￿ ￿ = 24αxe 2x + 6α + 8β e 2x . Substitute this expression into (4.1). ￿ ￿ 24αxe 2x + 6α + 8β e 2x = 6xe 2x 5 Solve for α and β. 24α = 6 α= 1 4 and and Substitute these values into (4.3). y =e 2x ￿ 6 α + 8β = 0 3 + 8β = 0 2 3 β = − 16 13 3 2 x− x 4 16 ￿ The sum of this solution with (4.1) is the general solution. ￿ ￿ 3 2 1 3 2x y = C 0 + C 1 x − x + x e + C 2 e −2x 16 4 6 5 The equation to solve is ￿ ￿ D 2 + 2D + 1 y = e −t ln (t ) . Solve the associated homogeneous equation. ￿ ￿ D 2 + 2D + 1 y ∗ = 0 (D + 1)2 y ∗ = 0 y ∗ = C 1 e −t + C 2 t e −t Now, use variation of parameters. The general solution will be y = u 1 e −t + u 2 t e −t ￿ ￿ with u 1 , u 2 such that ￿ ￿ e −t u 1 + t e −t u 2 = 0; ￿ ￿ −e −t u 1 + (1 − t ) e −t u 2 = e −t ln (t ) . Divide both equations through by e −t : ￿ ￿ u 1 + t u 2 = 0; ￿ ￿ −u 1 + (1 − t ) u 2 = ln (t ) . To isolate u 2 , add (5.1) and (5.2). ￿ u 2 = ln (t ) u2 = ˆ ln (t ) = t (ln (t ) − 1) + C 2 ￿ To get u 1 , substitute u 2 into (5.1). ￿ u 1 + t ln (t ) = 0 ￿ u 1 = −t ln (t ) u1 = − ˆ t ln (t ) ￿ ￿ 1 21 = −t ln (t ) − + C 1 . 2 4 7 (5.1) (5.2) Therefore the general s...
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## This note was uploaded on 01/26/2014 for the course MATH 527 taught by Professor Boucher during the Spring '07 term at New Hampshire.

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