integral equation by Laplace transforms and contour integration

i have to solve $\displaystyle \int^t_0 f(\tau)sin(t - \tau) d\tau = \cos t - \cos 3t$ by using Laplace transforms and then inverting it using contour integration (which i'm incredibly vague on)

we were given the rule that if $\displaystyle f(t) = \int^t_0 g(t - \tau)h(\tau) d\tau$ then $\displaystyle F(s) = G(s)H(s)$, and also that the Laplace transform of an integral is $\displaystyle L[\int^t_0 f(\tau)d\tau] = \frac{1}{s}L[f(\tau)]$

i found the Laplace transform of $\displaystyle f(t) = \cos t$ is $\displaystyle F(s) = \frac{s}{s^2 + 1}$ and of $\displaystyle f(t) = \cos 3t$ is $\displaystyle F(s) = \frac{s}{s^2 + 9}$

do i have to Laplace transform the integral on the left as well? or do I multiply both Laplace transforms together and then integrate by contour integration?

i'm a little confused as to what to do next!

Re: integral equation by Laplace transforms and contour integration

Quote:

Originally Posted by

**wik_chick88** i have to solve $\displaystyle \int^t_0 f(\tau)sin(t - \tau) d\tau = \cos t - \cos 3t$ by using Laplace transforms and then inverting it using contour integration (which i'm incredibly vague on)

we were given the rule that if $\displaystyle f(t) = \int^t_0 g(t - \tau)h(\tau) d\tau$ then $\displaystyle F(s) = G(s)H(s)$, and also that the Laplace transform of an integral is $\displaystyle L[\int^t_0 f(\tau)d\tau] = \frac{1}{s}L[f(\tau)]$

i found the Laplace transform of $\displaystyle f(t) = \cos t$ is $\displaystyle F(s) = \frac{s}{s^2 + 1}$ and of $\displaystyle f(t) = \cos 3t$ is $\displaystyle F(s) = \frac{s}{s^2 + 9}$

do i have to Laplace transform the integral on the left as well? or do I multiply both Laplace transforms together and then integrate by contour integration?

i'm a little confused as to what to do next!

Take the LT of your integal equaltion:

$\displaystyle \mathcal{L}\left[ \int^t_0 f(\tau)sin(t - \tau) d\tau = \cos t - \cos 3t\right]$

$\displaystyle \left[\mathcal{L} f (s)\right] \left[ \mathcal{L} \sin(s)\right] = \left[ \mathcal{L} \cos(s)\right]- \left[ \mathcal{L} h(s)\right]$

where $\displaystyle h(t)=\cos(3t)$

CB

Re: integral equation by Laplace transforms and contour integration

so am I solving for $\displaystyle f(t)$? ie. i find the Laplace transform of $\displaystyle sin(t - \tau)$ and sub that and also my Laplace transforms of $\displaystyle \cos t$ and $\displaystyle \cos 3t$ into the equation, simplify and then inverse Laplace transform to get $\displaystyle f(t)$?

Yes.

CB