Question: find the solution of ( 12.7 )with$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathit{y}\mathbf{(}\mathit{\pi }\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{0}$when the forcing function is given f ( x ).

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{x}$

The given expressions are $f\left(x\right)=\sin 2x$.

In mathematics, a Green's function is the impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions.

$f\left(x\right)=\sin \left(2x\right)$

Hence, we have

$y\left(x\right)=-\cos \left(x\right){\int }_0^x\left(\sin \left(x^{\text{'}}\right)\right)\sin \left(2x^{\text{'}}\right)d{x}^{\text{'}}-\sin \left(x\right){\int }_x^{\pi /2}\left(\cos \left(x^{\text{'}}\right)\right)\sin \left(2x^{\text{'}}\right)d{x}^{\text{'}}$

using double angle identity, to simplify the integration, we get

$\begin{array}{l}=-2\cos \left(x\right){\int }_0^x\left(\sin \left(x^{\text{'}}\right)\right)\sin \left(x^{\text{'}}\right)\cos \left(x^{\text{'}}\right)d{x}^{\text{'}}-2\sin \left(x\right){\int }_x^{\pi /2}\left(\cos \left(x^{\text{'}}\right)\right)\sin \left(x^{\text{'}}\right)\cos \left(x^{\text{'}}\right)d{x}^{\text{'}}\\ =-2\cos \left(x\right){\int }_0^x\sin {\left(x^{\text{'}}\right)}^2\cos \left(x^{\text{'}}\right)d{x}^{\text{'}}-2\sin \left(x\right){\int }_x^{\pi /2}\sin \left(x^{\text{'}}\right)\cos {\left(x^{\text{'}}\right)}^{2}d{x}^{\text{'}}\end{array}$

Now, using the substitution

$$\begin{gathered} d\cos \left(x\right)=-\sin \left(x\right) \\ d\sin \left(x\right)=\cos \left(x\right) \end{gathered}$$

Hence, we have

$\begin{array}{l}=-2\cos \left(x\right){\int }_0^x\sin {\left(x^{\text{'}}\right)}^{2}d\sin \left(x^{\text{'}}\right)+2\sin \left(x\right){\int }_x^{\pi /2}\cos {\left(x^{\text{'}}\right)}^{2}d\cos \left(x^{\text{'}}\right)\\ =2\cos \left(x\right)\left({\left.\frac{\sin {\left(x^{\text{'}}\right)}^3}{3}\right|}_x^0+2\sin \left(x\right)\left({\left.\frac{\cos {\left(x^{\text{'}}\right)}^3}{3}\right|}_{\pi /2}^x\right.\right.\end{array}$

Hence, we have

$\begin{array}{l}A\left(x^{\text{'}}\right)=\frac{\left|\begin{array}{cc}0& -\cos x^{\text{'}}\\ -1& \sin x^{\text{'}}\end{array}\right|}{\left|\begin{array}{cc}\sin x^{\text{'}}& -\cos x^{\text{'}}\\ \cos x^{\text{'}}& \sin x^{\text{'}}\end{array}\right|}\\ =2\cos \left(x\right)\left(\frac{\sin {\left(x\right)}^3}{3}-0\right)+2\sin \left(x\right)\left(\frac{0}{3}-\frac{\cos {\left(x\right)}^3}{3}\right)\end{array}$

Taking,$-2\sin \left(x\right)\cos \left(x\right)/3$as a common factor we get

Taking,$-2\sin \left(x\right)\cos \left(x\right)/3$as a common factor we get

$=\frac{-2\cos \left(x\right)\sin \left(x\right)}{3}\left(\sin {\left(x\right)}^2+\cos {\left(x\right)}^2\right)$

And, knowing that$\sin {\left(x\right)}^2+\cos {\left(x\right)}^2=1$, hence we have

$=\frac{-2\cos \left(x\right)\sin \left(x\right)}{3}$

And, using the double angle identity, where$\cos \left(x\right)\sin \left(x\right)=\sin \left(2x\right)/2$, hence we have

$=\frac{-\sin \left(2x\right)}{3}$

Which, is the solution to the given differential equation and with the stated boundary conditions.