So I am given the $ODE$:

$$y'(t) + 2e^{-2t}\int_{0}^{t} e^{2u} y(u) du=e^{-t}\sin(t), y(0)=0$$

and I'm supposed to find $y(t)$

So if I move the exponential inside, I get:

$$y'(t) + 2\int_{0}^{t}e^{-2t} e^{2u} y(u) du=e^{-t}\sin(t), \quad y(0)=0$$

$$y'(t) + 2\int_{0}^{t}e^{2(u-t)}y(u) du=e^{-t}\sin(t), \quad y(0)=0$$

Now I know that the integral inside in a convolution integral. I'm just having trouble writing out the Laplace transform. Would the integral inside just be:

$$y'(t) + 2(e^{2t}\ast y(t)) =e^{-t}\sin(t), \quad y(0)=0\;?$$

I am not really sure how to do the convolution integral part...

If someone can clarify this small doubt, that would be awesome thanks!


The convolution theorem states that


where $f*g=\int_{-\infty}^\infty f(t-u)g(u)\,du$, and $F(s)=\int_0^\infty f(t)e^{-st}\,dt$, and $G(s)=\int_0^\infty g(t)e^{-st}\,dt$ are the Laplace Transforms of $f$ and $g$, respectively.

If $f$ and $g$ are causal functions, then $f*g=\int_{0}^t f(t-u)g(u)\,du$

There was a slight error in the OP. Note that for the integro-differential equation given by

$$y'(t)+2\int_0^t e^{-2(t-u)}y(u)\,du=e^{-t}\sin(t)\tag 1$$

the convolution term is $e^{-2t}*y(t)$ and not $e^{2t}y(t)$.

Taking the Laplace Transform of the integro-differential equation in $(1)$ yields

$$Y(s)-sy(0)+2\left(\frac{1}{s+2}\right)Y(s)=\frac{1}{s^2+2s+2} \tag 2$$

Using $y(0)=0$ and solving $(2)$ for $Y(s)$ reveals


Can you finish by using partial fraction expansion?

  • $\begingroup$ Yes I can do it myself now. Thanks a lot! I just needed to be sure. $\endgroup$ – Future Math person Mar 19 '17 at 5:29
  • $\begingroup$ You're welcome! My pleasure. -Mark $\endgroup$ – Mark Viola Mar 19 '17 at 5:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.