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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!

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The convolution theorem states that

$$\mathscr{L}\{f*g\}(s)=F(s)G(s)$$

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

$$Y(s)=\frac{s+2}{(s+4)(s^2+s+2)}$$


Can you finish by using partial fraction expansion?

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  • $\begingroup$ 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

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