I am trying to use the Laplace Transformation to find the unknown function $f:[0,\infty)\rightarrow\mathbb{R}$ in the integral equation, $$f(t)=t+\frac{1}{6}\int_{0}^{t} (t-u)^3f(u) \ du.$$
I start by noticing, $$f(t)=t+\frac{1}{6}(f*t^3)(t) \ \ \ \ \text{(where $*$ denotes the convolution)}.$$ Then, \begin{align} \mathcal{L}(f)(z)&=\mathcal{L}(t)(z)+\frac{1}{6}\mathcal{L}(f)(z)\mathcal{L}(t^3)(z) \\ \mathcal{L}(f)(z)&=\frac{1}{z^2}+\frac{\mathcal{L}(f)(z)}{z^4} \\ \mathcal{L}(f)(z)&=\frac{z^2}{z^4-1} \end{align} Using partial fraction, $$\frac{z^2}{z^4-1}=-\frac{1}{4}\left(\frac{1}{z+1}\right)+\frac{1}{4}\left(\frac{1}{z-1}\right)-\frac{1}{4i}\left(\frac{1}{z+i}\right)+\frac{1}{4i}\left(\frac{1}{z-i}\right).$$ So, $$\mathcal{L}(f)(z)=-\frac{1}{4}e^{-t}+\frac{1}{4}e^t-\frac{1}{4i}e^{-ti}+\frac{1}{4i}e^{ti}.$$
Is this solution correct? I was unable to determine this via back substitution using Mathematica.
It's correct, although It's odd for me to see $(f*t^3) (t) $ rather than $f(t) *t^3 $. Also in the last line it should be $f(t)$ instead of $ \mathcal{L}(f)(z)$, after all that is what you want to find, or $\mathcal{L}^{-1}(f)(z)=f(t)$. We can simplify the last line if we use the complex form of the sine function: $ \displaystyle{2\sin z= e^{iz} - e^{-iz}} $, also with $2 \sin h( z) = e^z-e^{-z}$ $$f(t)=\frac12\left(\frac{1}{2} e^t-\frac{1}{4}e^{-t}\right) +\frac12\left(\frac{1}{2i}e^{it}-\frac{1}{2i}e^{-it}\right)=\frac{\sinh t} {2} +\frac{\sin t} {2} $$
Here is another approach to find that inverse laplace trasform. $$\mathcal{L}(f)(z) =\frac{z^2} {z^4 - 1}=\frac12\frac{2 z^2} {(z^2 - 1)(z^2 +1)}$$ $$=\frac12\frac {z^2 +1 +z^2 - 1}{(z^2 - 1)(z^2 +1)}=\frac12\left(\frac{1} {z^2 - 1}+\frac{1} {z^2 +1}\right)$$ Now we can directly see that: $$f(t) =\frac12(\sinh t + \sin t) $$