Question: Obtain (12.6) by using the convolution integral to solve (12.1).

The given expressions are $y_0^{\text{'}}=y_0=0$.

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandantiderivative.

Use the transform.

$\begin{array}{c}L\left(y^{\text{'}\text{'}}\right)+\omega \cdot L\left(y\right)\\ =L\left(f\right(t\left)\right)\left(p^{2}Y-p{y}_0-y_0^{\text{'}}\right)+{\omega }^2\cdot y\\ =L\left(f\right(t\left)\right)\end{array}$

Substitute$y_0^{\text{'}}=y_0=0$.

role="math" localid="1664352491593" $\begin{array}{c}{p}^{2}Y+{\omega }^2\cdot Y=L\left(f\right(t\left)\right)\left(p^2+{\omega }^2\right)\\ Y=L\left(f\right(t\left)\right)\\ Y=\frac{1}{\left(p^2+{\omega }^2\right)}\cdot L\left(f\right(t\left)\right)\\ L\left(Y\right)=\frac{1}{\left(p^2+{\omega }^2\right)}\cdot L\left(f\right(t\left)\right)n\end{array}$

The value of H ( p )

$\begin{array}{c}H\left(p\right)=\frac{1}{\left(p^2+{\omega }^2\right)}\\ L\left[h\right(t\left)\right]=\frac{1}{\left(p^2+{\omega }^2\right)}\left[h\right(t\left)\right]\\ =L^{-1}\left(\frac{1}{\left(p^2+{\omega }^2\right)}\right)Y\\ =L^{-1}\left(\frac{1}{\left(p^2+{\omega }^2\right)}.L\left(f\right(t\left)\right)\right)h\left(t\right)\\ =\frac{\sin \omega t}{\omega }\dots \mathrm{..}\left(2\right)\end{array}$

The value of G (p)

$\begin{array}{rcl}& & \begin{array}{c}G\left(p\right)=L\left(f\right(t\left)\right)\\ L\left(g\right(t\left)\right)=L\left(f\right(t\left)\right)\\ g\left(t\right)=L^{-1}\left(f\right(t\left)\right)\\ g\left(t\right)=f\left(t\right)g\left(\tau \right)\\ =f\left(\tau \right)\dots \left(3\right)\end{array}\end{array}$

Substitute equation (2),(3) in (1)

$\begin{array}{rcl}& & \begin{array}{c}y=L^{-1}\left[H\right(p)\cdot G(p\left)\right]\\ =g\cdot h\left[L_{34}:[g\times h]=H\left(p\right)\cdot G\left(p\right)\right]\\ ={\int }_0^{l}g\left(\tau \right)\cdot h(t-\tau )d\tau \\ y=g*f={\int }_0^t\frac{1}{w}\sin \left(w\left(t-t^{\text{'}}\right)\right)f\left(t^{\text{'}}\right)d{t}^{\text{'}}\end{array}\end{array}$

Therefore, taking the Laplace inverse of the equation and substituting with the boundaries values, we get

$\begin{array}{rcl}Y& =& \frac{1}{p^2+w^2}L\left\{f\right(t\left)\right\}\end{array}$

And, taking the Laplace inverse of Y in order to find the solution to the differential equation, which can be done using the convolution integral, we get

$\begin{array}{rcl}y& =& g*f={\int }_0^t\frac{1}{w}\sin \left(w\left(t-t^{\text{'}}\right)\right)f\left(t^{\text{'}}\right)d{t}^{\text{'}}\end{array}$