Tricky inverse Laplace transform

[materdei]

<< Moderator Note -- thread moved to the Homework Help forums >>[/color]

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


However, I feel VERY unsure about this.

Any pointers and feedback are immensly welcome!

 

[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?

 

[HallsofIvy]

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 $(s - 3)^2$

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

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

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

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

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

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


However, I feel VERY unsure about this.

Any pointers and feedback are immensly welcome!

 

[vela]

<< 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:

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

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.