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Question Number 145578 by rexford last updated on 06/Jul/21

$$y^{″}\mathrm{\_}y=xsin2x$$$$solvethedifferentialeqn..$$

Answered by mathmax by abdo last updated on 06/Jul/21

$${\mathrm{y}}^{{}^{″}}-\mathrm{y}=\mathrm{xsin}\left(2\mathrm{x}\right)$$$$\mathrm{e}\to {\mathrm{r}}^2-1=0\Rightarrow \mathrm{r}=\stackrel{-}{+}1\Rightarrow \mathrm{y}={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{-\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{w}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\\ {\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right|=-2\ne 0$$$${\mathrm{w}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{xsin}\left(2\mathrm{x}\right)-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right|=-{\mathrm{xe}}^{-\mathrm{x}}\sin \left(2\mathrm{x}\right)$$$${\mathrm{w}}_2=\left|\begin{array}{c}{\mathrm{e}}^{\mathrm{x}}0\\ {\mathrm{e}}^{\mathrm{x}}\mathrm{xsin}\left(2\mathrm{x}\right)\end{array}\right|={\mathrm{xe}}^{\mathrm{x}}\sin \left(2\mathrm{x}\right)$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=\frac{1}{2}\int {\mathrm{xe}}^{-\mathrm{x}}\sin \left(2\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}\mathrm{Im}(\int {\mathrm{xe}}^{-\mathrm{x}+2\mathrm{ix}}\mathrm{dx})$$$$\int {\mathrm{xe}}^{(-1+2\mathrm{i})\mathrm{x}}\mathrm{dx}=\frac{\mathrm{x}}{-1+2\mathrm{i}}{\mathrm{e}}^{(-1+2\mathrm{i})\mathrm{x}}-\int \frac{1}{-1+2\mathrm{i}}{\mathrm{e}}^{(-1+2\mathrm{i})\mathrm{x}}\mathrm{dx}$$$$=\frac{-\mathrm{x}}{1-2\mathrm{i}}{\mathrm{e}}^{(-1+2\mathrm{i})\mathrm{x}}+\frac{1}{1-2\mathrm{i}}\times \frac{1}{-1+2\mathrm{i}}{\mathrm{e}}^{(-1+2\mathrm{i})\mathrm{x}}$$$$=\frac{-\mathrm{x}(1+2\mathrm{i})}{5}{\mathrm{e}}^{-\mathrm{x}}(\cos \left(2\mathrm{x}\right)+\mathrm{isin}\left(2\mathrm{x}\right))-\frac{1}{{(1-2\mathrm{i})}^2}{\mathrm{e}}^{-\mathrm{x}}(\cos \left(2\mathrm{x}\right)+\mathrm{isin}\left(2\mathrm{x}\right))$$$$$$$$=\frac{-{\mathrm{xe}}^{-\mathrm{x}}}{5}(\cos \left(2\mathrm{x}\right)+\mathrm{isin}\left(2\mathrm{x}\right)+2\mathrm{icos}\left(2\mathrm{x}\right)-2\sin \left(2\mathrm{x}\right))$$$$-\frac{{(1+2\mathrm{i})}^2}{25}{\mathrm{e}}^{-\mathrm{x}}(\cos \left(2\mathrm{x}\right)+\mathrm{isin}\left(2\mathrm{x}\right))=....$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=-\frac{1}{2}\int {\mathrm{xe}}^{\mathrm{x}}\sin \left(2\mathrm{x}\right)\mathrm{dx}$$$$=-\frac{1}{2}\mathrm{Im}(\int {\mathrm{xe}}^{\mathrm{x}+2\mathrm{ix}}\mathrm{dx})$$$$\int \mathrm{x}{\mathrm{e}}^{(1+2\mathrm{i})\mathrm{x}}\mathrm{dx}=\frac{\mathrm{x}}{1+2\mathrm{i}}{\mathrm{e}}^{(1+2\mathrm{i})\mathrm{x}}-\int \frac{1}{1+2\mathrm{i}}{\mathrm{e}}^{(1+2\mathrm{i})\mathrm{x}}\mathrm{dx}=....$$$$\Rightarrow {\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2\mathrm{and}\mathrm{general}\mathrm{solution}\mathrm{is}$$$$\mathrm{y}={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{-\mathrm{x}}+{\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2$$

Commented by mathmax by abdo last updated on 06/Jul/21

$$\mathrm{you}\mathrm{can}\mathrm{use}\mathrm{mvc}\mathrm{method}\mathrm{but}\mathrm{contains}\mathrm{alos}\mathrm{of}\mathrm{calculus}$$$$\mathrm{the}\mathrm{best}\mathrm{is}\mathrm{wronskien}...$$

Commented by rexford last updated on 06/Jul/21

$$Please,isthereadifferentwayofsolvingthisasinintroducingconstantsascoefficientsfortheparticularsolution$$

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