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Question Number 103788 by bemath last updated on 17/Jul/20

$$y^{″}-y=\cot x$$

Answered by mathmax by abdo last updated on 18/Jul/20

$${\mathrm{y}}^{{}^{″}}-\mathrm{y}=\frac{\mathrm{cosx}}{\mathrm{sinx}}$$$$\mathrm{h}\to {\mathrm{r}}^2-1=0\Rightarrow \mathrm{r}=\stackrel{-}{+}1\Rightarrow {\mathrm{y}}_{\mathrm{h}}=\mathrm{a}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{be}}^{-\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\\ {\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right|=-2\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{-\mathrm{x}}\\ \frac{\mathrm{cosx}}{\mathrm{sinx}}-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right|=-\frac{{\mathrm{e}}^{-\mathrm{x}}\mathrm{cosx}}{\mathrm{sinx}}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}0\\ {\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{cosx}}{\mathrm{sinx}}\end{array}\right|=\frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}}{\mathrm{sinx}}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=\int \frac{{\mathrm{e}}^{-\mathrm{x}}\mathrm{cosx}}{2\mathrm{sinx}}\mathrm{dx}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=-\frac{1}{2}\int \frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}}{\mathrm{sinx}}\mathrm{dx}$$$$\Rightarrow {\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2={\mathrm{e}}^{\mathrm{x}}\int \frac{{\mathrm{e}}^{-\mathrm{x}}\mathrm{cosx}}{2\mathrm{sinx}}\mathrm{dx}-{\mathrm{e}}^{-\mathrm{x}}\int \frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}}{2\mathrm{sinx}}\mathrm{dx}$$$$\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{y}={\mathrm{y}}_{\mathrm{p}}+{\mathrm{y}}_{\mathrm{h}}$$

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