1
$\begingroup$

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: $$(f*g)(t)=\int_0^tf(w)g(t-w)dw =f(t)*g(t)$$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Amzoti Nov 25 '13 at 2:14
  • $\begingroup$ Thanks, I'll give that a try. Looks like I made an algebra mistake like you said. $\endgroup$ – Bob Shannon Nov 25 '13 at 2:54
2
$\begingroup$

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

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

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.