Solve the integral equation
$$f(t)=1+t-\frac{8}{3}\int ^t_0 (t-\tau)^3f(\tau)d\tau.$$
My attempt: Re-write the given integral equation,
$$f(t)=1+t-\frac{8}{3}{t^3*f(t)}$$
Apply Laplace transform
$$\mathcal{L}\{f(t)\}=\mathcal{L}\left\{1+t-\frac{8}{3}{t^3*f(t)}\right\}$$
I got $f(t)=\mathcal{L}^{-1}\left\{\frac{p^2+p^3}{p^4+16}\right\}$. How to continue from here?
Best Answer
Here is a way of tackling some inverse Laplace transform problems assuming you are good at manipulating summations $$ \mathcal{L}^{-1}\left[\frac{s^2+s^3}{16+s^4}\right]=\mathcal{L}^{-1}\left[\frac{s^2}{16+s^4}\right]+\mathcal{L}^{-1}\left[\frac{s^3}{16+s^4}\right] $$ expanding about $s$ equals infinity we have $$ \frac{s^2}{16+s^4} = \sum_{n=0}^\infty \frac{(-1)^n 16^n}{s^{4n+2}}\\ \frac{s^3}{16+s^4} = \sum_{n=0}^\infty \frac{(-1)^n 16^n}{s^{4n+1}} $$ using that $\mathcal{L}^{-1}(1/s^n)=t^{n-1}/(n-1)!$ $$ \mathcal{L}^{-1}\left[\frac{s^2}{16+s^4}\right] = \sum_{n=0}^\infty \frac{(-1)^n 16^n t^{4n+1}}{(4n+1)!}=\frac{ch(x)s(x)+c(x)sh(x)}{2\sqrt{2}}\\ \mathcal{L}^{-1}\left[\frac{s^3}{16+s^4}\right] = \sum_{n=0}^\infty \frac{(-1)^n 16^n t^{4n}}{(4n)!}=c(x)ch(x) $$ where $ch=\cosh,sh=\sinh,s=\sin,c=\cos,x=\sqrt{2}t$.