Hey so this question came up on a differential equations exam with regards to Laplace transforms and convolution integrals $$ y''(t)+9y(t)=\sin^{2018}(t) $$ $$ y(0)=y'(0)=0 $$
But I wasn't sure what the sine Laplace transform was. The professor said that the transform was just the sine transform, but raised to the power of 2018, however I am pretty sure that that is not the case.
How would someone go about doing this problem?
Taking a look at this table, we have the recurring relation
$$ \int e^{-st}\sin^{n}(t)\ dt = -\frac{e^{-st}\sin^{n-1}(t)}{s^2+n^2}(s\sin t + n\cos t) + \frac{n(n-1)}{s^2+n^2}\int e^{-st}\sin^{n-2}(t)\ dt $$
Integrating from $0$ to $\infty$, we obtain $$ L_n = \frac{n(n-1)}{s^2+n^2} L_{n-2} $$
where $L_n$ is the Laplace transform of $\sin^n(t)$. The base case is $L_0 = \dfrac{1}{s}$
Putting it all together $$ L_{2018} = \frac{2018\cdot 2017}{s^2+2018^2}\cdot \frac{2016\cdot2015}{s^2+2016^2}\cdots \frac{1}{s} = \frac{2018!}{s} \prod\limits_{k=1}^{1009} \frac{1}{\big(s^2 + (2k)^2\big)} $$
This leads to the solution $$ Y(s) = \frac{2018!}{s(s^2+9)} \prod\limits_{k=1}^{1009} \frac{1}{\big(s^2 + (2k)^2\big)} $$
which you can reverse transform by taking the convolution of $\sin^{2018}(t)$ and $\mathcal L \{\frac{1}{s^2+9}\} = \frac13 \sin 3t$
$$ y(t) = \frac13 \int_0^t \sin^{2018}(\tau)\sin \big(3(t-\tau)\big) d\tau $$
and using more recurrence relations