Solving a differential equation using Laplace transform?

Question

$$y''+2y'+ 10 = b\,δ(t-T),\,\begin{cases}y(0)=3\\ y'(0) = 0\end{cases}$$

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.

Answer 1

so you want to find the constants $b$ and $T$ so that $$3e^{-t} \cos(3t) - e^{-t}\sin(3t)+\dfrac{1}{3}be^{-(t-T)}\sin(3t-3T)= 0 \text{ for all $t > T.$}$$

\begin{align} 0 &= 9\cos(3t) - 3\sin(3t)+be^T\sin(3t-3T) \\ &= 9\cos(3t) - 3\sin(3t)+be^T(\sin 3t\cos 3T - \cos 3t \sin 3T \\ &= (9 -be^T \sin 3T)\cos 3t + (-3 + be^T \cos 3T)\sin 3t \end{align}

so set $ 9 = be^T \sin 3T, 3 = be^T \cos 3T$ which gives $$T = \dfrac{\arctan(3)}{3} + n\pi,\ be^T = 3\sqrt{10}, \text{ where $n$ is a positive integer.}$$

Answer 2

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: $$ \boxed{T_n=\displaystyle\frac{2\pi n+\arcsin \frac{3}{\sqrt{10}}}{3},\qquad b_n=3\sqrt{10}e^{-T_n}} $$