# Solving a integro-differential equation

I'm trying to solve the following problem, but my solution doesn't seem to be correct. Could I be rewriting the integral incorrectly?

$$\frac{dy}{dt} + 25 \int_0^t{y(t-w)e^{-10w}dw} = L[4]; y(0)=0$$ $$sL[y(t)] - y(0) + 25(L[y(t)]*L[e^{-10t}])=L[4]$$

Let $L[y(t)] = Y(t)$:

$$sY(t) + 25(Y(t)*\frac{1}{s+10})=\frac{4}{s}$$ $$sY(t) + 25(Y(t)*\frac{1}{s+10})=\frac{4}{s}$$ $$sY(t) + 25Y(t)*\frac{25}{s+10}=\frac{4}{s}$$ $$sY(t) + \frac{25*25Y(t)}{s+10}=\frac{4}{s}$$ $$sY(t) + \frac{625Y(t)}{s+10}=\frac{4}{s}$$ $$Y(t) (s + \frac{625}{s+10}) = \frac{4}{s}$$ $$Y(t) =\frac{4(s+10)}{s(s^2+10s+625)}$$

I'm not sure where I am going wrong on this one.

It is worth noting that when rewriting the integral I am using the following definition: $$(f*g)(t)=\int_0^tf(w)g(t-w)dw =f(t)*g(t)$$

• Just looks like you double multiplied the $25$, see my answer. Also, be careful with mixing time and $s$ domain, use one at a time and be consistent. – Amzoti Nov 25 '13 at 2:14
• Thanks, I'll give that a try. Looks like I made an algebra mistake like you said. – Bob Shannon Nov 25 '13 at 2:54

We have:

$$\mathcal{L} (y' + (25 y(t)*e^{-10t}) = 4)$$

This yields:

$$sy(s) - 0 + 25 y(s) \dfrac{1}{s+10} = \dfrac{4}{s}$$

Factoring, we have:

$$y(s) \left(s + \dfrac{25}{s+10}\right) = \dfrac{4}{s}$$

This gives us:

$$y(s) = \dfrac{4(s+10)}{s(s+5)^2} = \dfrac{8}{5 s} -\dfrac{4}{(s+5)^2} - \dfrac{8}{5 (s+5)}$$

Next, find $\mathcal{L}^{-1} (y(s))$

• You did well with acceptances today! +1 – Namaste Nov 26 '13 at 0:04