I am asked to show that $x'' + w^2x = f\sin(wt)$
has a solution given by $x = \frac{f}{2w^2}(\sin(wt) – wt\cos(wt))$
where $w$ and $f$ are constants, by means of Laplace transforms.
By taking a Laplace transform of each side, I get that $ L(x)(s) = \frac{wf}{(s^2 + w^2)^2}$
Here's where I'm stymied. My problem lies in taking the inverse Laplace to find the solution; I have found a couple inverse Laplace functions that appear in http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx, but I'm uncertain how to manipulate my result into one of those forms.
What's more, the textbook for this class (curse it) has a remarkably limited table. The closest pairs it has would be the pairs for $\sin(t)$ and $\cos(t)$.
if I leave $w$ in, then I can get $\frac{w}{s^2 + w^2}\frac{1}{s^2 + w^2}$ where the first term resembles the $\sin(t)$ transform, but I don't know if I'm just chasing rainbows with that…
I thought of adding and subtracting a value but I always end up with $1 + something$ (I've pulled the $wf$ out as a constant before the transform) as a numerator, and I'm not sure what to do with that.
So … how do I do this inverse Laplace transform?
Best Answer
Here, you have to use the result $\mathcal L[t\, p(t)] = -\dfrac{d}{ds} \mathcal L[p(t)]$.
Using this:
$\mathcal L[t \cos \omega t] = -\dfrac{d}{ds}\mathcal L[\cos \omega t] = -\dfrac{d}{ds} \dfrac{s}{s^2 + \omega^2} = \dfrac{s^2 - \omega^2}{(s^2 + \omega^2)^2}$
This can be split up as:
$\dfrac{s^2 + \omega^2 - 2\omega^2}{(s^2 + \omega^2)^2} = \dfrac{1}{s^2 + \omega^2} - \dfrac{2\omega^2}{(s^2 + \omega^2)^2} = \dfrac{1}{\omega}\mathcal L[\sin \omega t] -\dfrac{2\omega^2}{(s^2 + \omega^2)^2}$
Thus:
$ \mathcal L[t \cos \omega t]= \dfrac{1}{\omega}\mathcal L[\sin \omega t] -\dfrac{2\omega^2}{(s^2 + \omega^2)^2} \Rightarrow\\ \mathcal L\left[\dfrac{1}{\omega}\sin \omega t - t \cos \omega t\right] = \dfrac{2\omega^2}{(s^2 + \omega^2)^2}$. $
Multiply by $\dfrac{f}{2\omega}$ and you get $\dfrac{\omega f}{(s^2 + \omega^2)^2}$.