Solving differential equation with impulse function

Question

Solve the differential equation: $$y''-y'-2y={12 \, \sin t} \, \delta(t-π) $$.

Given:$ y'=dy/dt, y(0) =1, y'(0) =-1$

My attempt:

Applying Laplace transformation on the differential equation, I get:

$$L(y''-y'-2y)=L({12 \, \sin t} \, \delta(t-π)) $$.

Using the fact that:

$f(t) \, \delta(t-a) = f(a) \, \delta(t-a)$

$$L(y''-y'-2y)=L({12 \, \sin π} \, \delta(t-π)) $$.

$y(s^2-s-2) -s+2=0$

Solving this my answer doesn't match the given answer. Can somebody help solve this please.

Answer 1

$$y''-y'-2y={12\sin t}\delta(t-π)$$ Take Laplace Transform gives: $$s^2Y(s)-sy(0)-y'(0)-(sY(s)-y(0))-2Y(s)=12{e^{-\pi s}}\sin \pi$$ $$Y(s)(s^2-s-2)-s+2=0$$ $$Y(s)(s-2)(s+1)=s-2$$ $$Y(s)=\dfrac 1 {(s+1)}$$ $$y(t)=e^{-t}$$

Answer 2

Recall that the Laplace transform of $\delta(t)$ is $\mathcal{L}\{\delta(t)\} = 1$. Also, for a function $f$, the Laplace transform of a shift is $\mathcal{L}\{f(t-a)u(t-a)\} = e^{-at}\mathcal{L}\{f(t)\}$, where $u$ is the unit step function. Using these two rules, you should be able to correctly find the Laplace transform of the right-hand side of your differential equation and solve the resulting equation.