Solve $\left(12.3\right)$if $\mathit{G}\mathbf{=}\mathbf{0}$and $\mathit{d}\mathit{G}\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathbf{0}$at $\mathit{t}\mathbf{=}\mathbf{0}$ to obtain (12.5). Hint: Use L28 and L3 to find the inverse transform.

The given expressions are $G\left(t,t\text{'}\right)=\left\{\begin{array}{l}00<t<t\\ \frac{1}{\omega t}\sin \omega \left(t-t\right)0<t<t\end{array}\right.$

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t)

Use the equation.

$G"+{\omega }^{2}G=\delta \left(t\text{'}-t\right)$

Use Laplace transform.

$$\begin{gathered} L\left(G"\right)+{\omega }^2·L\left(G\right)=L\left[\delta \left(t\text{'}-t\right)\right] \\ =L\left[\delta \left(t\text{'}-t\right)\right]\left(p^{2}Y-p{y}_0-Y_0^{\text{'}}\right)+{\omega }^2·Y \\ =e^{-pt} \end{gathered}$$

Substitute $Y=L\left[G\right],y_{}=0,y_0^{\text{'}}=G_0^{\text{'}}=0$

$$\begin{gathered} p^2+{\omega }^2·e^{-pt} \\ \left(p^2+{\omega }^2\right)Y=e^{-pt} \\ Y=\frac{e^{-pt}}{\left(p^2+{\omega }^2\right)} \\ L\left(G\right)=\frac{1}{p^2+{\omega }^2}·e^{-pt} \end{gathered}$$

Taking inverse;

$$\begin{gathered} G=L^{-1}\left[\frac{1}{p^2+{\omega }^2}·e^{-pt}\right] \\ G=L^{-1}\left[G\left(p\right)·e^{-pt}\right] \\ G=f\left(t\right) \\ G=\left\{\begin{array}{l}g\left(t-t\text{'}\right)t>t>\text{'}>0\\ 0t<t\text{'}\end{array}\right. \end{gathered}$$

Substitute $a=t\text{'}$

$$\begin{gathered} G\left(p\right)=\frac{1}{p^2+{\omega }^2} \\ L\left[g\left(t\right)\right]=\frac{1}{p^2+{\omega }^2} \end{gathered}$$

Taking inverse;

$$\begin{gathered} g\left(t\right)=L^{-1}\left[\frac{1}{p^2+{\omega }^2}\right] \\ g\left(t\right)=\frac{\sin \omega t}{\omega } \\ g\left(t-t\text{'}\right)=\frac{\sin \omega \left(t-t\right)}{\omega }t>t\text{'}>0 \end{gathered}$$

Using equation (2) and (1)

role="math" localid="1654498243592" $$\begin{gathered} G=\left\{\begin{array}{l}\frac{\sin \omega \left(I-I\text{'}\right)}{\omega }t>t\text{'}>0\\ 0t<t\text{'}\end{array}\right. \\ G\left(t,t\text{'}\right)=\left\{\begin{array}{l}\frac{\sin \omega \left(t-t\text{'}\right)}{\omega }\\ \end{array}\right.t>t\text{'}0,0<t\text{'}<t \end{gathered}$$

Thus, the value of value of $\frac{d^2}{d^2}\left[G\left(t,t\text{'}\right)\right]+{\omega }^{2}G\left(t,t\text{'}\right)=\delta \left(t,t\text{'}\right)$is$G\left(t,t\text{'}\right)=\left\{\begin{array}{l}0t<t\text{'}<t\text{'}\\ \frac{\sin \omega \left(t-t\text{'}\right)}{\omega t}t>t\text{'}>0,0<t\text{'}<t\end{array}\right.$.