# Solving a differential equation using the laplace transform involving convolution

The problem is the following

The thing that puzzles me here is the integral on the right hand side, so:

How to take the laplace transform on the right hand side?

Any help to get me going would be greatly appreciated.

• Thanks for accepting my answer. However, I went a bit too fast and there was a mistake. Please see my updated answer for the correct solution. May 11, 2014 at 19:20

The integral on the right-hand side can be rewritten as

$$\int_{0}^t(\delta(t-\tau)-\delta(t-2-\tau))f(\tau)d\tau=f(t)-f(t-2)=f(t),\quad 0\le t<2\quad\textrm{(and zero otherwise)}$$

So its Laplace transform is simply

$$\int_0^2f(t)e^{-st}dt$$

Also note that $h(t-\pi)\sin(2t)=h(t-\pi)\sin(2(t-\pi))$, so you can simply use the transform of $h(t)\sin(2t)$ and use the shifting property, i.e. multiply by $e^{-s\pi}$.

• Not sure I got why that's 'allowed' to do, but it was the rewriting of the integral which I needed. Thanks a lot. Also, in taking the Laplace transform of f(t) notice that it's periodic, so it should be $$\frac{\int_0^2 f(t) e^{-s t} \, dt}{1-e^{-s T}}$$ May 11, 2014 at 19:29
• @Julian: The integral is just the convolution $(\delta*f)(t)-(\delta*f)(t-2)$, which is just $f(t)-f(t-2)$, because any function convolved with the delta impulse is simply the function itself. If the delta impulse is shifted, the function is shifted accordingly. May 11, 2014 at 19:33
• @Julian: The formula in your comment is correct for a periodic function, but since you have $f(t)-f(t-2)$ (and the period is 2) everything outside the interval $[0,2]$ is canceled, so only the first period is left, and the resulting function is not periodic anymore. May 11, 2014 at 19:34

The convolution theorem is \begin{align} F(s)*G(s) = \int_{0}^{t} g(t-\tau) f(\tau) d\tau \end{align} such that \begin{align} L\{ f*g\} = L\{ \int_{0}^{t} g(t-\tau) f(\tau) d\tau \} = F(s)G(s). \end{align} The differential equation then becomes \begin{align} L\{ y^{''} + 3 y^{'} + 2y \} = L\{ H(t-\pi) \sin(2t) \} + \delta(s) f(s) - \delta(s-2) f(s). \end{align} Completing the transform the equation becomes \begin{align} (s+1)(s+2) y(s) - (s+1) y(0) - y^{'}(0) = \frac{2 e^{-\pi s}}{s^{2} + 4} + \delta(s) f(s) - \delta(s-2) f(s). \end{align} Since $y^{'}(0) = 0$ and $y(0) = 1$, and $f(t+2) = f(t)$ then \begin{align} y(s) &= \frac{1}{s+2} + \frac{1}{(s+1)(s+2)} \left[ \frac{2 e^{-\pi s}}{s^{2} + 4} + \delta(s) f(s) - \delta(s-2) f(s) \right] \\ &= \frac{1}{s+2} + \frac{1}{(s+1)(s+2)} \left[ \frac{2 e^{-\pi s}}{s^{2} + 4} + f(0) - f(2) \right] \\ &= \frac{1}{s+2} + \frac{1}{(s+1)(s+2)} \left[ \frac{2 e^{-\pi s}}{s^{2} + 4} \right] \\ \end{align} Inversion leads to \begin{align} y(t) &= e^{-2t} + L^{-1}\{ \frac{1}{(s+1)(s+2)} \cdot \frac{2 e^{-\pi s}}{s^{2} + 4} \} \\ &= e^{-2t} + \frac{1}{20} H(t-\pi) \left[ 8 e^{\pi - t} - 5 e^{2(\pi -t)} - \sin(2t) - 3 \cos(2t) \right]. \end{align}

note: the convolution theorem is in most standard tables such as this one http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

• You should use $\mathcal{L}\{(f*g)(t)\}=F(s)G(s)$, where $F(s)$ and $G(s)$ are the Laplace transforms of $f(t)$ and $g(t)$, respectively. You just used $f(s)$ and $g(s)$, which is not the way to go. May 11, 2014 at 18:44
• @Leucippus this is great thanks a lot May 11, 2014 at 21:48