Solving a differential equation using Laplace transform? $$y''+2y'+ 10 = b\,δ(t-T),\,\begin{cases}y(0)=3\\ y'(0) = 0\end{cases}$$
I managed to solve this equation. My answer is $$y(t) = 3e^{-t} \cos(3t) - e^{-t}\sin(3t)+\dfrac{1}{3}be^{-(t-T)}\sin(3t-3T)u(t-T)$$
I am asked to find values for $b$ and $T$ such that $y(t) = 0$ for all $t>T$. Answer at the back of the book is $$\begin{cases}b_n=3\sqrt{10} e^{-T_n}\\\\T_n=\dfrac{1}{3} \arcsin\dfrac{3}{\sqrt10}+\dfrac{2}{3}n\pi&n=0,1,...\end{cases}$$ I have no idea how they got that. I would greatly appreciated help. 

Cancelling the $e^{-t}$ in $y(t)=0$ for $t>T$ implies that
$$\begin{align*}
3\cos(3t)-\sin(3t)+ \frac{b}{3}e^T\sin 3(t-T)\equiv 0\qquad (1)
\end{align*}$$
Recall the formula (found in most Diff. Eq. books):
$$
A\sin\theta - B\cos\theta = R\sin(\theta-\alpha),
$$
where $R=\sqrt{A^2+B^2}$ and $\alpha=\arcsin\frac{B}{R}$. We choose $A=1,$ $B=3,$ and $\theta=3t$. Therefore
$$
\sin3t-3\cos 3t=\sqrt{10}\sin\left(3t-\arcsin \frac{3}{\sqrt{10}}\right).
$$
Plugging into equation $(1)$ yields:
$$\begin{align*}
\frac{b}{3}e^T\sin 3(t-T)&\equiv \sqrt{10}\sin\left(3t-\arcsin \frac{3}{\sqrt{10}}\right)\\
\implies b&=3\sqrt{10}e^{-T},\ T=\frac{\arcsin \frac{3}{\sqrt{10}}}{3}.
\end{align*}$$
To be completely correct, when we determined the angle $\alpha$, any integer multiple of $2\pi$ could have been added without affecting the trig functions. Therefore $\alpha$ could take on any of the values
$$
\alpha_n=2\pi n+\arcsin \frac{3}{\sqrt{10}},\qquad n=\cdots,-1,0,1,\cdots
$$
Modifying $T$ and $b$ accordingly yields the answer:
$$
\boxed{T_n=\displaystyle\frac{2\pi n+\arcsin \frac{3}{\sqrt{10}}}{3},\qquad b_n=3\sqrt{10}e^{-T_n}}
$$
