Question

In Problems 35–42 use the Laplace transform to solve the given

equation.

$\mathbf{42}\mathbf{.}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\mathbf{f}\mathbf{\left(}\mathbf{\tau }\mathbf{\right)}\mathbf{f}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{\tau }\mathbf{)}\mathbf{d}\mathbf{\tau }\mathbf{=}\mathbf{6}{\mathbf{t}}^{\mathbf{3}}$

Short answer

The equation is $f\left(t\right)=6t$

Step-by-step solution

To Find the laplace transform equation.

Solution of Integrodifferential equation of using Laplace equation is

$\underset{0}{\overset{t}{\int }}f\left(\tau \right)f(t-\tau )d\tau =6t^3$

Apply Laplace operator on both side of expression and bu using Laplace linearity property we get

$L\left(\underset{0}{\overset{t}{\int }}f\left(\tau \right)f(t-\tau )d\tau \right)=L\left(6t^3\right)$

Since, according to Convolution Theorem

$L{\int }_0^{t}f\left(\tau \right)g(t-\tau )d\tau =L\left\{f\right(t\left)\right\}L\left\{g\right(t\left)\right\}$

Thus, Laplace Transform of given expression become

$\begin{array}{rcl}F\left(s\right)F\left(s\right)& =& 6\frac{3!}{s^4}\\ {\left(F\left(s\right)\right)}^2& =& \frac{36}{s^4}\\ F\left(s\right)& =& \frac{6}{s^2}\end{array}$

Final proof

Now, find function value $f\left(t\right)$ by using Inverse Laplace Transform

$$\begin{gathered} L^{-1}\left\{F\right(s\left)\right\}=6t \\ f\left(t\right)=6t \end{gathered}$$