Question

Use the Laplace transform to solve the given integral equation or integrodifferential equation.

$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{-}\frac{\mathbf{8}}{\mathbf{3}}\underset{\mathbf{0}}{\overset{\mathbf{t}}{\int }}{\mathbf{(}\mathbf{\tau }\mathbf{-}\mathbf{t}\mathbf{)}}^{\mathbf{3}}\mathbf{f}\mathbf{\left(}\mathbf{\tau }\mathbf{\right)}\mathbf{d\tau }$

Short answer

The laplace transformation is$f\left(t\right)=\frac{3}{8}{e}^{2t}+\frac{1}{8}{e}^{-2t}+\frac{1}{2}\cos \left(2t\right)+\frac{1}{2}\sin \left(2t\right)$

Step-by-step solution

Using the transformation formulas:

Firstly use the Laplace transformation to solve equation;

$\begin{array}{l}L\left\{1\right\}=\frac{1}{s},L\left\{\sin kt\right\}=\frac{k}{s^2+k^2}\\ \underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau =y\left(t\right)*1\end{array}$

Using Laplace Transformation formulas;

Take Laplace transformation of both the sides;

$\begin{array}{l}f\left(t\right)=1+t-\frac{8}{3}\underset{0}{\overset{t}{\int }}{(\tau -t)}^{3}f\left(\tau \right)d\tau \\ L\left[f\left(t\right)\right]=L\left[1\right]+L\left[t\right]-\frac{8}{3}L\left[\underset{0}{\overset{t}{\int }}{(\tau -t)}^{3}f\left(\tau \right)d\tau \right]\\ L\left[f\left(t\right)\right]=\frac{1}{s}+\frac{1}{s^2}-\frac{8}{3}L\left[\underset{0}{\overset{t}{\int }}{(\tau -t)}^{3}f\left(\tau \right)d\tau \right]\\ \end{array}$

$\begin{array}{l}f*g=\underset{0}{\overset{t}{\int }}f\left(\tau \right)g(t-\tau )d\tau \\ L\left[f*g\right]=L\left[f\left(t\right)\right]\left[g\left(t\right)\right]\\ \\ L\left[\underset{0}{\overset{t}{\int }}{(\tau -t)}^{3}f\left(\tau \right)d\tau \right]=L\left[f*g\right]\end{array}$

$\begin{array}{l}L\left[f\left(t\right)\right]=\frac{1}{s}+\frac{1}{s^2}-\frac{8}{3}L\left[\underset{0}{\overset{t}{\int }}{(\tau -t)}^{3}f\left(\tau \right)d\tau \right]\\ L\left[f\left(t\right)\right]\left(\frac{s^4-16}{s^4}\right)=\frac{s+1}{s^2}\\ L\left[f\left(t\right)\right]=\frac{s^2(s+1)}{s^4-16}\end{array}$

Using the Partial Fraction;

Decompose the fraction using partial fraction technique;

$\begin{array}{l}\frac{s^2(s+1)}{(s-1)\left(s+2\right)\left(s^2+4\right)}=\frac{A}{(s-2)}+\frac{B}{\left(s+2\right)}+\frac{Cs+D}{\left(s^2+4\right)}\\ A+B+C=1\\ A+B-C=0\\ \Rightarrow A=\frac{3}{8},B=\frac{1}{8},C=\frac{1}{2},D=\frac{1}{2}\\ \\ \\ \frac{A}{(s-2)}+\frac{B}{\left(s+2\right)}+\frac{Cs+D}{\left(s^2+4\right)}=\frac{\frac{3}{8}}{(s-2)}+\frac{\frac{1}{8}}{\left(s+2\right)}+\frac{\frac{1}{2}s}{\left(s^2+4\right)}+\frac{\frac{1}{2}}{\left(s^2+4\right)}\\ \\ L\left[f\left(t\right)\right]=\frac{\frac{3}{8}}{(s-2)}+\frac{\frac{1}{8}}{\left(s+2\right)}+\frac{\frac{1}{2}s}{\left(s^2+4\right)}+\frac{\frac{1}{2}}{\left(s^2+4\right)}\end{array}$

$\begin{array}{l}f\left(t\right)=\frac{3}{8}{L}^{-1}\left[\frac{1}{(s-2)}\right]+\frac{1}{8}{L}^{-1}\left[\frac{1}{s+2}\right]+\frac{1}{2}{L}^{-1}\left[\frac{s}{\left(s^2+4\right)}\right]+\frac{1}{2}{L}^{-1}\left[\frac{1}{s^4+4}\right]\\ f\left(t\right)=\frac{3}{8}{e}^{2t}+\frac{1}{8}{e}^{-2t}+\frac{1}{2}\cos \left(2t\right)+\frac{1}{2}\sin \left(2t\right)\end{array}$

$\begin{array}{l}L\left[e^{at}\right]=\frac{1}{s-a}\\ L\left[\sin \left(ax\right)\right]=\frac{a}{s^2+a^2}\\ L\left[\cos \left(ax\right)\right]=\frac{s}{s^2+a^2}\\ \\ f\left(t\right)=\frac{3}{8}{e}^{2t}+\frac{1}{8}{e}^{-2t}+\frac{1}{2}\cos \left(2t\right)+\frac{1}{2}\sin \left(2t\right)\\ \end{array}$

Hence, the final answer is;$f\left(t\right)=\frac{3}{8}{e}^{2t}+\frac{1}{8}{e}^{-2t}+\frac{1}{2}\cos \left(2t\right)+\frac{1}{2}\sin \left(2t\right)$