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