Question

Use the Laplace transform to solve the given integral equation or integraodifferential equation.: $t-2f\left(t\right)=\underset{0}{\overset{t}{\int }}(e^{\tau }-e^{-\tau })f(t-\tau )d\tau$

Short answer

Using the Laplace transformation we get_{$f\left(t\right)=\frac{1}{2}\left[t-\frac{1}{6}{t}^3\right]$}

Step-by-step solution

We will use the Convolution Theorem;

Here we will use the convolution theorem to evaluate the sum further,

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

Solving further;

Evaluating the integrodifferential equation using Laplace equation. Applying Laplace linearity property we get

$t-2f\left(t\right)=\underset{0}{\overset{t}{\int }}(e^{\tau }-e^{-\tau })f(t-\tau )$

$L\left\{t-2f\left(t\right)\right\}=L\left\{\underset{0}{\overset{t}{\int }}(e^{\tau }-e^{-\tau })f(t-\tau )d\tau \right\}$

$$\begin{gathered} L\left\{t-2f\left(t\right)\right\}=L\left\{(e^{\tau }-e^{-\tau })\right\}L\left\{f\left(t\right)\right\} \\ \frac{1}{s}-2F\left(s\right)=\left[\frac{1}{s-1}-\frac{1}{s+1}\right]F\left(s\right) \end{gathered}$$

find function value of f(t) by using Inverse Laplace Theorem;

Laplace transformation of the given function becomes,

$$\begin{gathered} F\left(s\right)=\frac{s^2-1}{2s^4} \\ F\left(s\right)=\frac{1}{2s^2}-\frac{1}{2s^4} \end{gathered}$$

$$\begin{gathered} L\left\{F\left(s\right)\right\}=L\left\{\frac{1}{2}\left[\frac{1}{s^2}-\frac{1}{s^4}\right]\right\} \\ f\left(t\right)=\frac{1}{2}\left[t-\frac{1}{6}{t}^3\right] \end{gathered}$$

Hence, the answer is$f\left(t\right)=\frac{1}{2}\left[t-\frac{1}{6}{t}^3\right]$