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$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$.$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{cc}x,& 0<x<\pi /4\\ \pi /2-x,& \pi /4<x<\pi /2.\end{array}$

The given expressions are $f\left(x\right)=\left\{\begin{array}{ll}x,& 0<x<\pi /4\\ \pi /2& \pi /4<x<\pi /2\end{array}\right.$-

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.

The solution of differential equation

$y^{\text{'}\text{'}}+y=f\left(x\right)$

The boundary conditions applied.

$y\left(x\right)=-\cos x{\int }_0^x\left(\sin x^{\text{'}}\right)f\left(x^{\text{'}}\right)d{x}^{\text{'}}-\sin x{\int }_x^{\sin }\cos \left(x^{\text{'}}\right)f\left(x^{\text{'}}\right)d{x}^3$

The function f(x) for interval $(0,\pi /4)$is used.

$y\left(x\right)=-\cos \left(x\right)\underset{I_1}{\underbrace{\left({\int }_0^x\sin \left(x^{\text{'}}\right)x^{\text{'}}d{x}^{\text{'}}\right)}}-\sin \left(x\right)\underset{I_2}{\underbrace{\left[{\int }_x^{\pi /4}\cos \left(x^{\text{'}}\right)x^{\text{'}}d{x}^{\text{'}}+{\int }_{\pi /4}^{\pi /2}\cos \left(x^{\text{'}}\right)\left(\pi /2-x^{\text{'}}\right)d{x}^{\text{'}}\right]}}$

$\begin{array}{c}{I}_1=\left(\sin \left(x^{\text{'}}\right)-{\left.x^{\text{'}}\cos \left(x^{\text{'}}\right)\right|}_x^0\right.\\ =\left(\sin \right(x)-x\cos (x)-0-0)\end{array}$

And, evaluating the integral $I_{2}$, we get

$\begin{array}{c}{I}_2=\left(x^{\text{'}}\sin \left(x^{\text{'}}\right)+{\left.\cos \left(x^{\text{'}}\right)\right|}_{\pi /4}^x+\left(\pi /{\left.2\sin \left(x^{\text{'}}\right)\right|}_{\pi /2}^{\pi /4}-\left(x^{\text{'}}\sin \left(x^{\text{'}}\right)+{\left.\cos \left(x^{\text{'}}\right)\right|}_{\pi /2}^{\pi /4}\right.\right.\right.\\ =\left(\pi /4\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-x\sin \left(x\right)-\cos \left(x\right)\right)+\left(\pi /2-\pi /2\frac{1}{\sqrt{2}}\right)-\left(\pi /2-\pi /4\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\\ =-x\sin \left(x\right)-\cos \left(x\right)+\frac{1}{\sqrt{2}}(\pi /4+1-\pi /2+\pi /4+1)+(\pi /2-\pi /2)\\ =-x\sin \left(x\right)-\cos \left(x\right)+\frac{1}{\sqrt{2}}(2+\pi /2-\pi /2)+\left(0\right)\\ \begin{array}{l}=-x\sin \left(x\right)-\cos \left(x\right)+\frac{1}{\sqrt{2}}\left(2\right)\\ =-x\sin \left(x\right)-\cos \left(x\right)+\sqrt{2}\end{array}\\ \end{array}$

$\begin{array}{c}y\left(x\right)=-\cos \left(x\right)I_1-\sin \left(x\right)I_2\\ =-\cos \left(x\right)\left[\sin \right(x)-x\cos (x\left)\right]-\sin \left(x\right)[-x\sin (x)-\cos (x)+\sqrt{2}]\\ =-\cos \left(x\right)\sin \left(x\right)+x\cos {\left(x\right)}^2+x\sin {\left(x\right)}^2+\sin \left(x\right)\cos \left(x\right)-\sqrt{2}\sin \left(x\right)\\ =x-\sqrt{2}\sin \left(x\right)\end{array}$

And, for the second case where $x>\pi / 4$, we get

$$\begin{gathered} y\left(x\right)=-\cos \left(x\right)[\underset{I_{-}3}{\underbrace{{\int }_0^{\pi /4}\sin \left(x^{\text{'}}\right)x^{\text{'}}d{x}^{\text{'}}}}+\underset{I_{-}4}{\underbrace{{\int }_{\pi /4}^x\sin \left(x^{\text{'}}\right)\left(\pi /2-x^{\text{'}}\right)d{x}^{\text{'}}}}] \\ -\sin \left(x\right)\left(\underset{I_{-}5}{\underbrace{{\int }_x^{\pi /2}\cos \left(x^{\text{'}}\right)\left(\pi /2-x^{\text{'}}\right)d{x}^{\text{'}}}}\right) \end{gathered}$$

And, hence we have

$y\left(x\right)=-\cos \left(x\right)\left[I_3+I_4\right]-\sin \left(x\right)\left(I_5\right)$

And, hence evaluating the integral $I_{3}$, we get

$\begin{array}{c}{I}_3={\int }_0^{\pi /4}{x}^{\text{'}}\sin \left(x^{\text{'}}\right)d{x}^{\text{'}}=\\ \left(\sin \left(x^{\text{'}}\right)-{\left.x^{\text{'}}\cos \left(x^{\text{'}}\right)\right|}_{\pi /4}^0\right.\\ =\sin (\pi /4)-\pi /4\cos (\pi /4)-0-0\\ =\frac{1}{\sqrt{2}}(1-\pi /4)\end{array}$

$\begin{array}{c}{I}_4={\int }_{\pi /4}^x\sin \left(x^{\text{'}}\right)\left(\pi /2-x^{\text{'}}\right)d{x}^{\text{'}}\\ =\left(-\pi /2\cos \left(x^{\text{'}}\right)-{\left.\left(\sin \left(x^{\text{'}}\right)-x^{\text{'}}\cos \left(x^{\text{'}}\right)\right)\right|}_x^{\pi /4}\right.\\ =\left(-\pi /2\cos \left(x^{\text{'}}\right)-\sin \left(x^{\text{'}}\right)+{\left.x^{\text{'}}\cos \left(x^{\text{'}}\right)\right|}_x^{\pi /4}\right.\\ =(-\pi /2\cos (x)-\sin (x)+x\cos (x)+\pi /2\cos (\pi /4)+\sin (\pi /4)-\pi /4\cos (\pi /4\left)\right)\\ =x\cos \left(x\right)-\sin \left(x\right)+\frac{1}{\sqrt{2}}(\pi /2+1-\pi /4)-\pi /2\cos \left(x\right)\end{array}$

$$\begin{gathered} I_3+I_4=x\cos \left(x\right)-\sin \left(x\right)+\frac{1}{\sqrt{2}}(\pi /2+1-\pi /4+1-\pi /4)-\pi /2\cos \left(x\right) \\ \mathrm{Thus},\mathrm{we}\mathrm{get} \\ {\mathrm{I}}_3+{\mathrm{I}}_4=\mathrm{xcos}\left(\mathrm{x}\right)-\sin \left(\mathrm{x}\right)+\frac{1}{\sqrt{2}}\left(2\right)-\mathrm{\pi }/2\cos \left(\mathrm{x}\right) \\ \begin{array}{l}{\mathrm{I}}_5={\int }_{\mathrm{x}}^{\mathrm{\pi }/2}\cos \left({\mathrm{x}}^{\text{'}}\right)\left(\mathrm{\pi }/2-{\mathrm{x}}^{\text{'}}\right){\mathrm{dx}}^{\text{'}}\\ =\left(\mathrm{\pi }/2\sin \left({\mathrm{x}}^{\text{'}}\right)-{\left.\left({\mathrm{x}}^{\text{'}}\sin \left({\mathrm{x}}^{\text{'}}\right)+\cos \left({\mathrm{x}}^{\text{'}}\right)\right)\right|}_{\mathrm{\pi }/2}^{\mathrm{x}}\right.\\ =\left(\mathrm{\pi }/2\sin \left({\mathrm{x}}^{\text{'}}\right)-{\mathrm{x}}^{\text{'}}\sin \left({\mathrm{x}}^{\text{'}}\right)-{\left.\cos \left({\mathrm{x}}^{\text{'}}\right)\right|}_{\mathrm{\pi }/2}^{\mathrm{x}}\right.\\ =(\mathrm{\pi }/2\cdot 1-\mathrm{\pi }/2\cdot 1-0-\mathrm{\pi }/2\sin (\mathrm{x})+\mathrm{xsin}(\mathrm{x})+\cos (\mathrm{x}\left)\right)\\ =(-\mathrm{\pi }/2\sin (\mathrm{x})+\mathrm{xsin}(\mathrm{x})+\cos (\mathrm{x}\left)\right)\end{array} \\ \mathrm{y}\left(\mathrm{x}\right)=-\cos \left(\mathrm{x}\right)\left[{\mathrm{I}}_3+{\mathrm{I}}_4\right]-\sin \left(\mathrm{x}\right)\left[{\mathrm{I}}_5\right] \end{gathered}$$

And, hence the solution to the differential equation in the stated case, is

$\begin{array}{l}y\left(x\right)=-\cos \left(x\right)\left[x\cos \left(x\right)-\sin \left(x\right)+\frac{1}{\sqrt{2}}\left(2\right)-\pi /2\cos \left(x\right)\right]-\sin \left(x\right)[-\pi /2\sin (x)+x\sin (x)+\cos (x\left)\right]y\left(x\right)\\ =-x\cos {\left(x\right)}^2+\cos \left(x\right)\sin \left(x\right)-\sqrt{2}\cos \left(x\right)+\pi /2\cos {\left(x\right)}^2\\ \begin{array}{l}+\pi /2\sin {\left(x\right)}^2-x\sin {\left(x\right)}^2-\cos \left(x\right)\sin \left(x\right)y\left(x\right)\\ =-x\left(\cos {\left(x\right)}^2+\sin {\left(x\right)}^2\right)-\sqrt{2}\cos \left(x\right)+\pi /2\left(\sin {\left(x\right)}^2+\cos {\left(x\right)}^2\right)+\left(\cos \right(x\left)\sin \right(x)-\cos (x\left)\sin \right(x\left)\right)y\left(x\right)\\ =-x-\sqrt{2}\cos \left(x\right)+\pi /2+0y\left(x\right)\\ =\pi /2-x-\sqrt{2}\cos \left(x\right)\end{array}\end{array}$

and, hence the general solution to the given differential equation, is thus

The solution to the given differential equation, is given by

$y\left(x\right)=\left\{\begin{array}{ll}\pi /2-x-\sqrt{2}\cos \left(x\right),& x>\pi /4\\ x-\sqrt{2}\sin \left(x\right),& x<\pi /4\end{array}\right.$