Question

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

Short answer

Using the Laplace transformation we get_{$\mathrm{f}\left(\mathrm{t}\right)=\mathrm{cost}+\mathrm{sint}$}

Step-by-step solution

Solution of integral equation of laplace equation is;

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

Applying Laplace operator;

Apply Laplace operator on both side of expression and by using Lapace linearity property we get,

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

Solving further;

Thus, laplace transformation of given expression become

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

Hence, the answer is:$f\left(t\right)=\cos t+\sin t$