Question

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

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

Short answer

The laplace transformation is f (t) = sint

Step-by-step solution

Applying Laplace transformation property;

Take Laplace transform of both side of the given equation

Using linearity of the Laplace transform we have;

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

Evaluating further;

To evaluate $L\left[\underset{0}{\overset{t}{\int }}(t-\tau )f\left(\tau \right)d\tau \right]$use formulas

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

Solving further;

Continuing on step 1 we have:

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

In the process above we used formula:

$L\left[\sin \left(at\right)\right]=\frac{a}{s^2+a^2}$

Hence the final answer is $f\left(t\right)=sint$