# Laplace Transform

1. Homework Statement

y''+y = f(t)

y(0) = 0; y'(0)=1

f(t) = 1, 0<=t<pi/2
0, pi/2<=t

3. The Attempt at a Solution
so far, i have

(s^2+1)*L{y} = $$\frac{s-e^(-pi/2s)}{s}$$ +1

what is next ?

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#### gabbagabbahey

Homework Helper
Gold Member
so far, i have

(s^2+1)*L{y} = $$\frac{s-e^(-pi/2s)}{s}$$ +1

what is next ?
That doesn't look quite right; the Laplace Transform of $f(t)$ is not $$\frac{s-e^{-\pi s/2}}{s}$$ (although it's close). Double check that calculation.

Once you correct your expression, solve for $\mathcal{L}[y(t)]$ and then take the inverse Laplace transform to get $y(t)$

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