Question: Use the given solutions of the homogeneous equation to find a particular solution of the given equation. You can do this either by the Green function formulas in the text or by the method of variation of parameters

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}\left(cs{c}^{2}x\right)\mathit{y}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x}\mathbf{;}\mathit{c}\mathit{o}\mathit{t}\mathit{x}\mathbf{,}\mathbf{1}\mathbf{-}\mathit{x}\mathit{c}\mathit{o}\mathit{t}\mathit{x}$

The given parameters are cot $x,1-x\cot x$

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.

$y^{\text{'}\text{'}}-2\left(\csc {\left(x\right)}^2\right)y=\sin {\left(x\right)}^2$

And, given that the homogeneous solution to such differential equation, is given by

$y_1=\cot \left(x\right)y_2=1-x\cot \left(x\right)$

Hence, we can find the particular solution to such differential equation, using the following equation,

$y_p=y_2\left(x\right)\int \frac{y_1\left(x\right)f\left(x\right)}{W\left(x\right)}dx-y_1\left(x\right)\int \frac{y_2\left(x\right)f\left(x\right)}{W\left(x\right)}dx$

the Wronskian is given by

$W\left(x\right)=\left|\begin{array}{ll}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right|$And, the differentiation of the cosec function is given by,

$\frac{d}{dx}\cot \left(x\right)=-\csc {\left(x\right)}^2$And, hence the Wronskian of the homogeneous equation, is thus

$\begin{array}{l}W\left(x\right)=\left|\begin{array}{cc}\cot \left(x\right)& 1-x\cot \left(x\right)\\ -\csc {\left(x\right)}^2& -\cot \left(x\right)-x\left(-\csc {\left(x\right)}^2\right)\end{array}\right|\\ =\left|\begin{array}{cc}\cot \left(x\right)& 1-x\cot \left(x\right)\\ -\csc {\left(x\right)}^2& -\cot \left(x\right)+x\csc {\left(x\right)}^2\end{array}\right|\\ =-\cot {\left(x\right)}^2+x\csc {\left(x\right)}^2\cot \left(x\right)-\left(-\csc {\left(x\right)}^2+x\cot \left(x\right)\csc {\left(x\right)}^2\right)\\ =-\cot {\left(x\right)}^2+x\csc {\left(x\right)}^2\cot \left(x\right)+\csc {\left(x\right)}^2-x\cot \left(x\right)\csc {\left(x\right)}^2\\ =-\cot {\left(x\right)}^2+\csc {\left(x\right)}^2\\ \begin{array}{l}=-{\left(\frac{\cos \left(x\right)}{\sin \left(x\right)}\right)}^2+\frac{1}{\sin {\left(x\right)}^2}\\ =\frac{-\cos {\left(x\right)}^2}{\sin {\left(x\right)}^2}+\frac{1}{\sin {\left(x\right)}^2}\\ =\frac{-\cos {\left(x\right)}^2+1}{\sin {\left(x\right)}^2}\end{array}\end{array}$

And, knowing the Wronskian of the solutions to the homogeneous equation, and comparing the given differential equation, we find that

$f\left(x\right)=\sin {\left(x\right)}^2$

find the particular solution to the given differential equation,

role="math" localid="1664285934259" $$\begin{gathered} y_p=(1-x\cot (x\left)\right)\int \frac{\cot \left(x\right)\sin {\left(x\right)}^2}{1}dx-\cot \left(x\right)\int \frac{(1-x\cot (x\left)\right)\sin {\left(x\right)}^2}{1}dx \\ \mathrm{Simplifying},\mathrm{the}\mathrm{integration},\mathrm{we}\mathrm{get} \\ \begin{array}{l}{\mathrm{y}}_{\mathrm{p}}=(1-\mathrm{xcot}(\mathrm{x}\left)\right)\int \cot \left(\mathrm{x}\right)\sin {\left(\mathrm{x}\right)}^2\mathrm{dx}-\cot \left(\mathrm{x}\right)\int (1-\mathrm{xcot}(\mathrm{x}\left)\right)\sin {\left(\mathrm{x}\right)}^2\mathrm{dx}\\ {\mathrm{y}}_{\mathrm{p}}=(1-\mathrm{xcot}(\mathrm{x}\left)\right)\underset{{\mathrm{I}}_1}{\underbrace{\int \cot \left(\mathrm{x}\right)\sin {\left(\mathrm{x}\right)}^2\mathrm{dx}}}-\cot \left(\mathrm{x}\right)\int (1-\mathrm{xcot}(\mathrm{x}\left)\right)\sin {\left(\mathrm{x}\right)}^2\mathrm{dx}\end{array} \\ \mathrm{And},\mathrm{hence}\mathrm{evaluating}\mathrm{the}\mathrm{integral},\mathrm{where} \\ {\mathrm{y}}_{\mathrm{p}}=(1-\mathrm{xcot}(\mathrm{x}\left)\right){\mathrm{I}}_1-\cot \left(\mathrm{x}\right){\mathrm{I}}_2 \\ \mathrm{Thus},\mathrm{evaluating}\mathrm{the}\mathrm{integral}\$\mathrm{I\_}\left\{1\right\}\$,\mathrm{we}\mathrm{get} \\ {\mathrm{I}}_1=\int \cot \left(\mathrm{x}\right)\sin {\left(\mathrm{x}\right)}^2\mathrm{dx} \\ \mathrm{Simplifying}\mathrm{the}\mathrm{integral}, \\ \int \frac{\cos \left(\mathrm{x}\right)}{\sin \left(\mathrm{x}\right)}\sin {\left(\mathrm{x}\right)}^2\mathrm{dx}=\int \cos \left(\mathrm{x}\right)\sin \left(\mathrm{x}\right)\mathrm{dx} \\ \mathrm{hence}\mathrm{the}\mathrm{integration}\mathrm{is}\mathrm{thus} \\ {\mathrm{I}}_1=-\frac{\cos {\left(\mathrm{x}\right)}^2}{2}+\text{const} \\ {\mathrm{I}}_2=\int (1-\mathrm{xcot}(\mathrm{x}\left)\right)\sin {\left(\mathrm{x}\right)}^2\mathrm{dx} \\ \mathrm{Separating}\mathrm{the}\mathrm{integral},\mathrm{we}\mathrm{get} \\ \begin{array}{l}=\int \sin {\left(\mathrm{x}\right)}^2\mathrm{dx}-\int \mathrm{x}\cot \left(\mathrm{x}\right)\sin {\left(\mathrm{x}\right)}^2\mathrm{dx}\\ =\int \sin {\left(\mathrm{x}\right)}^2\mathrm{dx}-\int \mathrm{x}\frac{\cos \left(\mathrm{x}\right)}{\sin \left(\mathrm{x}\right)}\sin {\left(\mathrm{x}\right)}^2\mathrm{dx}\\ =\int \sin {\left(\mathrm{x}\right)}^2\mathrm{dx}-\int \mathrm{x}\cos \left(\mathrm{x}\right)\sin \left(\mathrm{x}\right)\mathrm{dx}\end{array} \\ \mathrm{Thus},\mathrm{we}\mathrm{have} \\ \begin{array}{l}=\int \frac{1}{2}(1-\cos (2\mathrm{x}\left)\right)\mathrm{dx}-\int \mathrm{x}\frac{\sin \left(2\mathrm{x}\right)}{2}\mathrm{dx}\\ =\frac{1}{2}\int \mathrm{d}\mathrm{x}-\frac{1}{2}\int \cos \left(2\mathrm{x}\right)\mathrm{dx}-\int \mathrm{x}\frac{\sin \left(2\mathrm{x}\right)}{2}\mathrm{dx}\end{array} \\ \mathrm{Evaluating}\mathrm{the}\mathrm{integral},\mathrm{we}\mathrm{get} \\ \begin{array}{l}{\mathrm{I}}_2=\frac{1}{2}\mathrm{x}-\frac{\sin \left(2\mathrm{x}\right)}{4}-\int \mathrm{x}\frac{\sin \left(2\mathrm{x}\right)}{2}\mathrm{dx}\\ =\frac{\mathrm{x}}{2}-\frac{\sin \left(2\mathrm{x}\right)}{4}-\underset{{\mathrm{I}}_3}{\underbrace{\int \mathrm{x}\frac{\sin \left(2\mathrm{x}\right)}{2}\mathrm{dx}}}\end{array} \\ \mathrm{Where},\mathrm{the}\mathrm{integral} \\ {\mathrm{I}}_3=\int \mathrm{x}\frac{\sin \left(2\mathrm{x}\right)}{2}\mathrm{dx} \end{gathered}$$

Which can be evaluated by integration by parts, where

Hence, the integration is

$$\begin{gathered} \begin{array}{l}{I}_3=uv-\int vdu\\ =-\frac{x\cos \left(2x\right)}{4}-\int -\frac{\cos \left(2x\right)}{4}dx\\ =-\frac{x\cos \left(2x\right)}{4}+\int \frac{\cos \left(2x\right)}{4}dx\\ =-\frac{x\cos \left(2x\right)}{4}+\frac{\sin \left(2x\right)}{8}\end{array} \\ \begin{array}{l}{I}_2=\frac{1}{2}x-\frac{\sin \left(2x\right)}{4}-\left[-\frac{x\cos \left(2x\right)}{4}+\frac{\sin \left(2x\right)}{8}\right]+\text{const.}\\ =\frac{x}{2}-\frac{\sin \left(2x\right)}{4}+\frac{x\cos \left(2x\right)}{4}-\frac{\sin \left(2x\right)}{8}+\text{const.}\end{array} \\ =\frac{x}{2}-\frac{3\sin \left(2x\right)}{8}+\frac{x\cos \left(2x\right)}{4}+\text{const.} \\ \mathrm{And},\mathrm{from}\left(\mathrm{a}\right)\mathrm{the}\mathrm{particular}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ {\mathrm{y}}_{\mathrm{p}}=(1-\mathrm{xcot}(\mathrm{x}\left)\right){\mathrm{I}}_1-\cot \left(\mathrm{x}\right){\mathrm{I}}_2 \\ \mathrm{Hence}, \\ \begin{array}{l}={\mathrm{I}}_1-\mathrm{xcot}\left(\mathrm{x}\right){\mathrm{I}}_1-\cot \left(\mathrm{x}\right)\\ {\mathrm{I}}_2=-\frac{\cos {\left(\mathrm{x}\right)}^2}{2}-\mathrm{xcot}\left(\mathrm{x}\right)\left[-\frac{\cos {\left(\mathrm{x}\right)}^2}{2}\right]\end{array} \\ -\cot \left(\mathrm{x}\right)\left[\frac{\mathrm{x}}{2}-\frac{3\sin \left(2\mathrm{x}\right)}{8}+\frac{\mathrm{xcos}\left(2\mathrm{x}\right)}{4}\right] \\ \mathrm{Now},\mathrm{we}\mathrm{try}\mathrm{to}\mathrm{find}\mathrm{a}\mathrm{simpler}\mathrm{form}\mathrm{for}\mathrm{this}\mathrm{particular}\mathrm{solution}, \\ =-\frac{\cos {\left(\mathrm{x}\right)}^2}{2}+\mathrm{xcot}\left(\mathrm{x}\right)\frac{\cos {\left(\mathrm{x}\right)}^2}{2}\mathrm{xcot}\left(\mathrm{x}\right),3\cot \left(\mathrm{x}\right)\sin \left(2\mathrm{x}\right)\mathrm{xcot}\left(\mathrm{x}\right)\cos \left(2\mathrm{x}\right) \\ \begin{array}{l}-\frac{\mathrm{xcot}\left(\mathrm{x}\right)}{2}+\frac{3\cot \left(\mathrm{x}\right)\sin \left(2\mathrm{x}\right)}{8}-\frac{\mathrm{xcot}\left(\mathrm{x}\right)\cos \left(2\mathrm{x}\right)}{4}\\ =-\frac{\cos {\left(\mathrm{x}\right)}^2}{2}+\mathrm{xcot}\left(\mathrm{x}\right)\left[\frac{\cot {\left(\mathrm{x}\right)}^2}{2}-\frac{\cos \left(2\mathrm{x}\right)}{4}\right]-\frac{\mathrm{xcot}\left(\mathrm{x}\right)}{2}+\frac{3}{8}\cot \left(\mathrm{x}\right)\left(2\sin \right(\mathrm{x}\left)\cos \right(\mathrm{x}\left)\right)\\ =-\frac{\cos {\left(\mathrm{x}\right)}^2}{2}+\mathrm{xcot}\left(\mathrm{x}\right)\left[\frac{\cot {\left(\mathrm{x}\right)}^2}{2}-\frac{\cos \left({\mathrm{x}}^2\right)}{4}+\frac{\sin {\left(\mathrm{x}\right)}^2}{4}\right]-\frac{\mathrm{xcot}\left(\mathrm{x}\right)}{2}+\frac{3}{84}\frac{\cos \left(\mathrm{x}\right)}{\sin \left(\mathrm{x}\right)}\left(\cancel{2}\sin \right(\mathrm{x}\left)\cos \right(\mathrm{x}\left)\right)\end{array} \\ \begin{array}{l}=-\frac{\cos {\left(\mathrm{x}\right)}^2}{2}+\mathrm{xcot}\left(\mathrm{x}\right)\left[\frac{\cos \left({\mathrm{x}}^2\right)}{4}+\frac{\sin {\left(\mathrm{x}\right)}^2}{4}\right]-\frac{\mathrm{xcot}\left(\mathrm{x}\right)}{2}+\frac{3}{4}\cos \\ =\cos {\left(\mathrm{x}\right)}^2\left[-\frac{1}{2}+\frac{3}{4}\right]+\frac{\mathrm{xcot}\left(\mathrm{x}\right)}{4}\left[\cos \left({\mathrm{x}}^2\right)+\sin {\left(\mathrm{x}\right)}^2\right]-\frac{\mathrm{xcot}{\left(\mathrm{x}\right)}^2}{2}\\ =\cos {\left(\mathrm{x}\right)}^2\left[\frac{1}{4}\right]+\frac{\mathrm{xcot}\left(\mathrm{x}\right)}{4}\left[1\right]-\frac{\mathrm{xcot}\left(\mathrm{x}\right)}{2}\\ =\frac{\cos {\left(\mathrm{x}\right)}^2}{4}+\mathrm{xcot}\left(\mathrm{x}\right)\left[\frac{1}{4}-\frac{1}{2}\right]\\ =\frac{\cos {\left(\mathrm{x}\right)}^2}{4}-\frac{\mathrm{xcot}\left(\mathrm{x}\right)}{4}\\ =\frac{1}{4}\left(\cos {\left(\mathrm{x}\right)}^2-\mathrm{xcot}\left(\mathrm{x}\right)\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{particular}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation},\mathrm{is}\mathrm{hence}\\ {\mathrm{y}}_{\mathrm{p}}=\frac{1}{4}\left(\cos {\left(\mathrm{x}\right)}^2-\mathrm{xcot}\left(\mathrm{x}\right)\right)\end{array} \end{gathered}$$uncaught exception: Invalid chunk

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