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## Homework Statement

Find the solution to the differential equation using Laplace transforms.

## Homework Equations

[tex]y'' - y' - 2y = 12 \theta{(t-\pi)}\sin{t}[/tex]

[tex]y(0) = 1[/tex]

[tex]y'(0) = -1[/tex]

## The Attempt at a Solution

Using Laplace transforms...

[tex]\mathcal{L}(y'') = s^2F - s + 1[/tex]

[tex]\mathcal{L}(-y') = -sF-1[/tex]

[tex]\mathcal{L}(-2y) = -2F[/tex]

[tex]\mathcal{L}(12\theta{t-pi}\sin{t}) = \mathcal{L}(-12\theta{(t-\pi)}\sin{(t-\pi)}) = \frac{-12e^{-\pi s}}{s^2+1}[/tex]

[tex](s^2-s-2)F = \frac{-12e^{-\pi s}}{s^2+1}[/tex]

I rearrange and end up with partial fractions...

[tex]F = e^{-\pi s}[\frac{-12}{(s^2+1)(s-2)(s+1)}] = e^{-\pi s}(\frac{A}{s-2} \frac{B}{s+1} \frac{Cs+D}{s^2+1})[/tex]

From this point, I can rearrange everything and end up with 4 equations, 4 unknowns. I am thinking, however, that there is an easier way, simply by using one of the common transforms found in a Laplace transform table. Is there a better way to go about this problem without having to resort to partial fractions?

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