hw5soln-corrected

# 3 y e t 3 7 cos t sin t 116 116 the sum of

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Unformatted text preview: os (t ) + −β sin (t ) ; and therefore ￿2 ￿ ￿￿ ￿ ￿ ￿ ￿ D − 4D + 10 y = e −t 14α − 6β cos (t ) + 6α + 14β sin (t ) . 3 (3.3) Substitute this expression into (3.1). e −t Solve for α and β. ￿￿ ￿ ￿ ￿ ￿ 14α − 6β cos (t ) + 6α + 14β sin (t ) = e −t sin (t ) 14α − 6β = 0 6 α = 14 β ... ... ... 6 7 α = 14 · 116 3 = 116 and and . . . and 6 ￿ 6 6α + 14β = 1 ￿ β + 14β = 1 14 116 7β =1 7 β = 116 Substitute these values into (3.3). y =e −t ￿ 3 7 cos (t ) + sin (t ) 116 116 ￿ The sum of this solution with (3.2) is the general solution. ￿ ￿ ￿ ￿￿ ￿ ￿￿ ￿￿ 7 3 y = e −t cos (t ) + sin (t ) + e 2t C 1 cos 6t + C 2 sin 6t 116 116 4 4 The equation to solve is ￿3 ￿ D − 2D 2 − 4D + 8 y = 6xe 2x . (4.1) Solve the associated homogeneous equation. ￿ ￿ D 3 − 2D 2 − 4D + 8 y ∗ = 0 (D − 2)2 (D + 2) y ∗ = 0 y ∗ = (C 0 + C 1 x ) e 2x + C 2 e −2x (4.2) ￿ ￿ The obvious judicious guess, y = e 2x αx + β , won’t work since it’s already a homogeneous solution. So instead use ￿ ￿ y = e 2x α x 3 + β x 2 . (4.3) For convenience, substitute s = αx 3 + βx 2 . Then Combining these gives ￿ y = e 2x s ; ￿ ￿ D...
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