Question

Pls help:)

Answer 1

For the first question, here's what I've done so far (with no clear indication of being right or wrong):\[\large\begin{align*}{\bf L}=\mathcal{L}\left\{L_n(t)\right\}(s)&=\mathcal{L}\left\{\frac{e^t}{n!}\frac{d^n}{dt^n}\left[\frac{t^n}{e^t}\right]\right\}(s)\\
&=\frac{1}{n!}\mathcal{L}\left\{\frac{d^n}{dt^n}\left[\frac{t^n}{e^t}\right]\right\}(s-1)
\end{align*}\]
It's a good exercise to derive the \(n\)th derivative for yourself. I got this result:
\[\large\begin{align*}\frac{d^n}{dt^n}\left[\frac{t^n}{e^t}\right]
&=e^{-t}\sum_{k=0}^n\frac{n!}{(n-k)!}\frac{n!}{k!(n-k)!}(-t)^{n-k}
\end{align*}\]
I'm not quite sure how to continue with this, but I'll check back on this question later.

For the second question, is it \(\large f(t)=5te^{2t}\sin^3t\) ? If so, I recommend reducing the power of the sine, then referring to a table for the transform of \(t^n\sin t\) and \(t^n\cos t\).

For the third question, it looks like you can try to rewrite the integral definition of \(\text{erf}(t)\) as a convolution, such that you can apply the rule,
\[\large\mathcal{L}\left\{\int_0^tf(t-\tau)g(\tau)~d\tau\right\}=\mathcal{L}\{f(t)\}\mathcal{L}\{g(t)\}\]
For the fourth question, I think finding the Fourier series representation of \(F(t)\) would be a good start. I can help you with that if you don't know how to find it. Or @sidsiddhartha , he's pretty knowledgeable about this.

Alternatively, you can try looking through this article that discusses the Laplace transform of periodic functions: http://www.math.vt.edu/people/dlr/m2k_opm_lapper1.pdf

Answer 2

@SithsAndGiggles explained it well i think u'll be now able to do first 3 problems
now for the fourth problem u can use this theorem --
let f(t) be piecewise continuous on interval [0,infinity), be of exponential order and periodic with a period T then
\[\Large L[f(t)]=\frac{ 1 }{ 1-e^{-sT} }\int\limits_{0}^{T}e^{-st}f(t)dt~~~~~with ~~~s>0\]
so the problem is-\[F(t)=0~,~~0<t<\pi \\~~~~~~~=t- \pi ~~.~~\pi<t<2 \pi\]
now just apply the above theorem
here T=2 pi
\[\Large L[f(t)]=\frac{ 1 }{ 1-e^{-2 \pi s} }\int\limits_{0}^{2 \pi}e^{-st}f(t)dt \\\\\Large =\frac{ 1 }{ 1-e^{-2 \pi s} }\int\limits_{\pi}^{2 \pi}e^{-st}(t-\pi)dt\]
now just integration :)

Answer 3

Thnx a lot @sidsiddhartha and @SithsAndGiggles