Use the convolution theorem to obtain a formula for the solution to the given initial value problem, where g(t)is piecewise continuous on$\mathbf{1}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$and of exponential order.

role="math" localid="1665123516493" $\mathbf{[}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{}\mathbf{+}\mathbf{}\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathit{g}\mathbf{(}\mathit{t}\mathbf{)}\mathbf{}\mathbf{;}\mathbf{}\mathit{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Let f(t) and g(t) be continuous on the interval,$\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$the convolution can be denoted as *. Thus, the convolution for the function f(t) and g(t) be defined as,

role="math" localid="1665123823515" $\mathbf{(}\mathit{f}\mathbf{}\mathbf{*}\mathbf{}\mathit{g}\mathbf{)}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\mathit{f}\mathbf{(}\mathit{t}\mathbf{}\mathbf{-}\mathbf{}\mathit{v}\mathbf{)}\mathit{g}\mathbf{\left(}\mathit{v}\mathbf{\right)}\mathbf{d}\mathit{v}$

Consider the given equation,

$y\text{'}\text{'}+y=g\left(t\right);y\left(0\right)=0,quady\text{'}\left(0\right)=1$

Apply Laplace transform,

$L\{y\text{'}\text{'}+y\}=L\left[g\right(t\left)\right]$

$$\begin{gathered} s^{2}y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)+y\left(s\right)=g\left(s\right) \\ {s}^{2}y\left(s\right)-1+y\left(s\right)=g\left(s\right) \\ \left(s^2+1\right)y\left(s\right)-1=g\left(s\right) \\ y\left(s\right)=\frac{g\left(s\right)}{s^2+1}+\frac{1}{s^2+1} \end{gathered}$$

Take, inverse Laplace transform, we get

$L^{-1}\left[y\right(s\left)\right]=L^{-1}\left[\frac{g\left(s\right)}{\left(s^2+1\right)}\right]+L^{-1}\left[\frac{1}{\left(s^2+1\right)}\right]....\left(1\right)$

Use the convolution formula,

$L^{-1}\left[\int \left(s\right)g\left(s\right)\right]={\int }_0^{t}f(t-v)g\left(v\right)dv$

Since,

$f\left(s\right)=\frac{1}{s^2+1}$

f(t)=sint

Substitute in the formula,

$L^{-1}\left[\frac{g\left(s\right)}{s^2+1}\right]={\int }_0^t[\sin (t-v\left)g\right(v)dv$

Hence, equation becomes,

$y\left(t\right)={\int }_0^t[\sin (t-v\left)g\right(v)dv+\sin t$

Therefore, the obtained solution for the given initial value problem is.

$y\left(t\right)={\int }_0^t[\sin (t-v\left)g\right(v)dv+\sin t$