91Ó°ÊÓ

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

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.

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

The function f(x) for interval $(0,\mathrm{Ï€}/4)$is used.

$y\left(x\right)=−\cos \left(x\right)\underset{I_1}{\underset{ÁåŸ}{\left({∫}_0^x\sin \left(x^{\text{'}}\right)x^{\text{'}}d{x}^{\text{'}}\right)}}−\sin \left(x\right)\underset{I_2}{\underset{ÁåŸ}{\left[{∫}_x^{\mathrm{Ï€}/4}\cos \left(x^{\text{'}}\right)x^{\text{'}}d{x}^{\text{'}}+{∫}_{\mathrm{Ï€}/4}^{\mathrm{Ï€}/2}\cos \left(x^{\text{'}}\right)\left(\mathrm{Ï€}/2−x^{\text{'}}\right)d{x}^{\text{'}}\right]}}$

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

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

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

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

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

$$\begin{gathered} y\left(x\right)=−\cos \left(x\right)[\underset{I_{−}3}{\underset{ÁåŸ}{{∫}_0^{\mathrm{Ï€}/4}\sin \left(x^{\text{'}}\right)x^{\text{'}}d{x}^{\text{'}}}}+\underset{I_{−}4}{\underset{ÁåŸ}{{∫}_{\mathrm{Ï€}/4}^x\sin \left(x^{\text{'}}\right)\left(\mathrm{Ï€}/2−x^{\text{'}}\right)d{x}^{\text{'}}}}] \\ −\sin \left(x\right)\left(\underset{I_{−}5}{\underset{ÁåŸ}{{∫}_x^{\mathrm{Ï€}/2}\cos \left(x^{\text{'}}\right)\left(\mathrm{Ï€}/2−x^{\text{'}}\right)d{x}^{\text{'}}}}\right) \end{gathered}$$

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

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

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

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

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

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

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

The solution to the given differential equation, is given by

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