Use laplace transform to solve the ODE $y''(t)+4y(t)=4u(\pi-t)cos(t)$,,,,,,, $y(0)=y'(0)=0$

u is the unit step function (heaviside function)

I use: $u(\pi-t)=1-u(t-pi)$ By inserting this and transforming i get:

$Y(s)(s^2+4)=4 \frac{s}{s^2+1}- 4\frac{s}{s^2+1} \frac{e^{-\pi s}}{s}$

But the answer to the exercise is $Y(s)(s^2+4)=4 \frac{s}{s^2+1}+ 4\frac{s}{s^2+1} e^{-\pi s}$ What am i doing wrong? Or is the answer wrong? I know i am not done yet but i first need to find the correct transformation.

  • $\begingroup$ You seem to be claiming that $\mathcal{L}\left\{4\,\mathcal{U}(\pi-t)\cos t\right\}=4\mathcal{L}\left\{\mathcal{U}(\pi-t)\right\}\times\mathcal{L}\left\{ \cos t\right\}$ but that's not true. Are you using a table of transforms or computing the integral? $\endgroup$
    – user170231
    Commented May 13, 2016 at 18:18

1 Answer 1


Since the problem with your work is something to do with the RHS of the ODE, I'll focus on computing that side's transform only. $$\begin{align*} \mathcal{L}\left\{4\,\mathcal{U}(\pi-t)\cos t\right\}&=4\int_0^\infty \mathcal{U}(\pi-t)\cos t\,e^{-st}\,\mathrm{d}t\\[1ex] &=4\int_0^\pi \cos t\,e^{-st}\,\mathrm{d}t\\[1ex] &=\frac{4}{1+s^2}\bigg[e^{-st}(\sin t-s\cos t)\bigg]_{t=0}^{t=\pi}\\[1ex] &=\frac{4(se^{-\pi s}+s)}{1+s^2} \end{align*}$$ which agrees with the proposed solution.

Let's check again using the table here. $\#28$ and the relation you refer to tell us that $$\begin{align*}\mathcal{L}\left\{4\mathcal{U}(t-\pi)\cos t\right\}&=4\mathcal{L}\left\{(1-\mathcal{U}(\pi-t)\cos t\right\}\\[1ex] &=4\mathcal{L}\{\cos t\}-\mathcal{L}\left\{4\,\mathcal{U}(\pi-t)\cos t\right\}\\[1ex] \mathcal{L}\left\{4\,\mathcal{U}(\pi-t)\cos t\right\}&=\frac{4s}{s^2+1}-e^{-\pi s}\mathcal{L}\left\{4\cos(t+\pi)\right\}\\[1ex] &=\frac{4s}{s^2+1}+4e^{-\pi s}\mathcal{L}\left\{\cos t\right\}\\[1ex] &=\frac{4(s+se^{-\pi s})}{1+s^2} \end{align*}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .