Tricky inverse Laplace transform

1. Oct 9, 2014

materdei

<< Moderator Note -- thread moved to the Homework Help forums >>

I'm stuck on a problem, and I'm in serious need of help.

I) Problem:

Find the solution to $f (t) = 2 \int^t_0 f'(u) sin 3 (t-u) \ du + 2 cos (3t)$.
Also find $f (0)$.

II) Solution, so far:

$F(s) = 2 (s F(s) - f(0)) * \frac{3}{(s^2 + 3^2)} + \frac{2}{(s^2 + 3^2)}$

$F(s) (1 - \frac{(6 s)}{(s^2 + 3^2)} = \frac{(2 s - 6 f(0)}{(s^2 + 3^2)}$

$F(s) = \frac{(2 s - 6 f(0)}{(s^2 + 3^2 - 6s)}$

And this is basically where I'm stuck.

III) Proposed final solution:

Rewrites $(s^2 + 3^2 - 6s)$ to $(s - 3)^2$

$F(s) = 2 s \frac{(1)}{(s - 3)^2)} - 6 f(0) \frac{(1)}{((s - 3)^2)}$

$f(t) = 2 \mathcal {L^{-1}} (s) \mathcal {L^{-1}} \frac{(1)}{(s - 3)^2} - 6 f(0) \mathcal {L^{-1}} \frac{(1)}{(s - 3)^2}$

$f(t) = 2 t * t e^{3t} - 6 f(0) t e^{3t}$

$f(t) = 2 t^2 e^{3t} - 6 f(0) t e^{3t}$

Insert $t = 0$

$f(0) = 0* 2 e^{3t} - 0 * 6 f(0) e^{3t} = 0$

Hence, $f(0) = 0$ and therefore $- 6 f(0) t e^{3t} = 0$

which gives $f(t) = 2 t^2 e^{3t}$

Any pointers and feedback are immensly welcome!

Last edited by a moderator: Oct 24, 2014
2. Oct 15, 2014

Greg Bernhardt

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Oct 24, 2014

HallsofIvy

Staff Emeritus
Well, this is an error. The inverse Laplace transform of $\frac{s}{(s- 3)^2}$ is NOT the inverse Laplace transform of s times the inverse Laplace transform of $\frac{1}{(x- 3)^2}$

Try writing $\frac{s}{(s- 3)^2}$ as $$\frac{s- 3}{(s- 3)^2}+ \frac{3}{(s- 3)^2}= \frac{1}{s- 3}+ 3\frac{1}{(s- 3)^2}$$

4. Oct 24, 2014

vela

Staff Emeritus
You're not convolving the Laplace transforms, and you didn't transform cos 3t correctly. You should have
$$F(s) = 2\left[(s F(s)-f(0)) \frac{3}{s^2+3^2}\right] + \frac{2s}{s^2+3^2}.$$

Also, you can find f(0) by simply setting t=0 in the original equation.