# Abdullah's question via email about solving an integral equation using a Laplace Transform

#### Prove It

##### Well-known member
MHB Math Helper

We would need to recognise that the integral in the equation is a convolution integral, which has Laplace Transform: $\displaystyle \mathcal{L}\,\left\{ \int_0^t{ f\left( u \right) \,g\left( t - u \right) \,\mathrm{d}u } \right\} = F\left( s \right) \,G\left( s \right)$.

In this case, $\displaystyle g\left( t - u \right) = \sin{ \left[ 4\left( t - u \right) \right] } \implies g\left( t \right) = \sin{ \left( 4\,t \right) } \implies G\left( s \right) = \frac{4}{s^2 + 16}$.

Taking the Laplace Transform of the equation gives

\displaystyle \begin{align*} F\left( s \right) &= 5 \left( \frac{1}{s} \right) + 12 \left( \frac{2}{s^3} \right) + 4\,F\left( s \right) \left( \frac{4}{s^2 + 16} \right) \\ F\left( s \right) &= \frac{5}{s} + \frac{24}{s^3} + \left( \frac{16}{s^2 + 16} \right) F\left( s \right) \\ F\left( s \right) - \left( \frac{16}{s^2 + 16} \right) F\left( s \right) &= \frac{5}{s} + \frac{24}{s^3} \\ \left( 1 - \frac{16}{s^2 + 16} \right) F\left( s \right) &= \frac{5}{s} + \frac{24}{s^3} \\ \left( \frac{s^2}{s^2 + 16} \right) F\left( s \right) &= \frac{5\,s^2 + 24}{s^3} \\ F\left( s \right) &= \frac{\left( s^2 + 16 \right) \left( 5\,s^2 + 24 \right) }{s^5} \\ F\left( s \right) &= \frac{5\,s^4 + 104\,s^2 + 384}{s^5} \\ F\left( s \right) &= \frac{5}{s} + \frac{104}{s^3} + \frac{384}{s^5} \\ F\left( s \right) &= 5\left( \frac{1}{s} \right) + 52 \left( \frac{2}{s^3} \right) + 16 \left( \frac{4!}{s^5} \right) \\ \\ f\left( t \right) &= 5 + 52\,t^2 + 16\,t^4 \end{align*}