How to solve differential equation problem involving Dirac delta function? $$
y''+2y'+ 10y=b\,δ\left(\, t - T\,\right)\,\qquad y\left(\, 0\,\right)=3\,,\quad y'\left(\, 0\,\right)=0
$$
Can you choose values for $b$ and $T$ ( $b$ and $T$ positive numbers) such that $y\left(\, t\,\right) = 0\,,\  \forall\ t > T$ ?.
I am working on this problem. I managed to solve the IVP. The answer is
$$
y\left(\, t\,\right)
=3{\rm e}^{-t}\cos\left(\, 3t\,\right) - {\rm e}^{-t}\sin\left(\, 3t\,\right)
+ 1/3b{\rm e}^{-\left(\, t - T\,\right)}\sin\left(\, 3t - 3T\,\right)
u\left(\, t - T\,\right)
$$. But I am stuck at figuring the values for b and T. Answer given to question above is bn=3Sqrt(10)e^(-Tn) and Tn=1/3 arcsin(3/ sqrt(10)) + 2/3n pi, n=0,1,2...
But I have no idea how to figure out that solution. I would be really grateful for any help.
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\,{\rm y}''\pars{t} + 2\,{\rm y}'\pars{t} + 10\,{\rm y}\pars{t}
     =b\,δ\pars{t - T}\,,\qquad \,{\rm y}\pars{0}=3\,,\quad \,{\rm y}'\pars{0}=0
     \:\ {\large ?}}$

The solution is given by:
  \begin{align}
\,{\rm y}\pars{t}=
\left\{\begin{array}{lcl}
\expo{-t}\bracks{3\cos\pars{3t} + \sin\pars{3t}} & \mbox{if} & t < T
\\[2mm]
\expo{-t}\bracks{A\cos\pars{3t} + B\sin\pars{3t}} & \mbox{if} & t > T
\end{array}\right.
\end{align}

Now, you have two equations to satisfy:
$$
\,{\rm y}\pars{T^{-}} = \,{\rm y}\pars{T^{+}}\,,
\qquad
\,{\rm y}'\pars{T^{+}}  - \,{\rm y}'\pars{T^{-}} = b
$$

Then, you have two equations which determine $\ds{A\ \mbox{and}\ B}$.

A: what do you mean with $u(t)$ on the answer? Is the function $u$ defined anywhere?
Anyway, I haven't tried the math myself, but if you have the analitic solution, shouldn't the rest come by the initial conditions? Set $y(t)=0$, since you know that $e^0=1$, $sin(0)=0$, $cos(0)=1$ you will have a equation on $b$ and $T$. Do the same for the derivative, you will have a system of two equations and two variables. Then solve it for $b$ and $T$...
