hw5soln-corrected

Solve for and 4 0 0 and and substitute these values

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Unformatted text preview: , ￿2 ￿ D + 4 y = 4β cos (2x ) − 4α sin (2x ) . Substituting this expression into (2.1) gives 4β cos (2x ) − 4α sin (2x ) = sin (2x ) . Solve for α and β. 4β = 0 β=0 and and Substitute these values into (2.2). −4α = 1 α = −1 4 1 y = − x cos (2x ) 4 The sum of this solution with (1.2) is the general solution. 1 y = − x cos (2x ) + C 1 cos (2x ) + C 2 sin (2x ) 4 2 (2.2) 3 The equation to solve is ￿2 ￿ D − 4D + 10 y = e −t sin (t ) . (3.1) Solve the associated homogeneous equation. ￿2 ￿ D − 4D + 10 y ∗ = 0 Solve the characteristic equation. λ2 − 4λ + 10 = 0 ￿ 4 ± 16 − 40 λ= 2 ￿ = 2±i 6 These are two nonreal roots. So, ￿ ￿￿ ￿ ￿￿ ￿￿ y ∗ = e 2t C 1 cos 6t + C 2 sin 6t . (3.2) Now, make the judicious guess ￿ ￿ y = e −t α cos (t ) + β sin (t ) . For convenience, substitute s = α cos (t ) + β sin (t ). Then y = e −t s ; ￿ ￿ D y = e −t s ￿ − s ; ￿ ￿ D 2 y = e −t s ￿￿ − 2s ￿ + s . Combining these gives ￿2 ￿ ￿￿ ￿ ￿ ￿ ￿ D − 4D + 10 y = e −t s ￿￿ − 2s ￿ + s − 4 s ￿ − s + 10s ￿ ￿ = e −t s ￿￿ − 6s ￿ + 15s . Note that 15s = 15α cos (t ) + 15β sin (t ) ; −6s ￿ = −6β cos (t ) + 6α sin (t ) ; s ￿￿ = −α c...
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