Question

Use the Laplace transform to solve the given integral equation or integraodifferential equation._{$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\underset{\mathbf{0}}{\overset{\mathbf{t}}{\int }}\mathbf{f}\mathbf{\left(}\mathbf{\tau }\mathbf{\right)}\mathbf{d\tau }\mathbf{=}\mathbf{1}$}

Short answer

Using the Laplace transformation we get$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}$

Step-by-step solution

We will use the Convolution Theorem;

_{Take the Laplace transform both side of equation,}

_{$L\left\{\underset{0}{\overset{t}{\int }}f\left(\tau \right)g\left(t-\tau \right)d\tau \right\}=L\left\{f\left(t\right)\right\}L\left\{g\left(t\right)\right\}$}

Using the linearity of the Laplace transform we have:

$\begin{array}{l}f*g=\underset{0}{\overset{t}{\int }}f\left(\tau \right)g(t-\tau )d\tau \\ L\left[f*g\right]=L\left[f\left(t\right)\right]L\left[g\left(t\right)\right]\\ \\ g(t-\tau )=1\to g\left(t\right)=1\\ \end{array}$

Evaluation further:

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

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

Solving further ;

_{counting on step 1 we have, in the process above we have used the formula foe Laplace transform:}

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

$f\left(t\right)=e^{-t}$

Hence, the answer is:$f\left(t\right)=e^{-t}$