# Solving differential equation with impulse function

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:

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

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

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

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