How to solve differential equation problem involving Dirac delta function?

Question

$$ y''+2y'+ 10y=b\,δ\left(\, t - T\,\right)\,\qquad y\left(\, 0\,\right)=3\,,\quad y'\left(\, 0\,\right)=0 $$

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.

Answer 1

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 $T$...

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\,{\rm y}''\pars{t} + 2\,{\rm y}'\pars{t} + 10\,{\rm y}\pars{t} =b\,δ\pars{t - T}\,,\qquad \,{\rm y}\pars{0}=3\,,\quad \,{\rm y}'\pars{0}=0 \:\ {\large ?}}$

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 $\ds{A\ \mbox{and}\ B}$.