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Question Number 123593 by Bird last updated on 26/Nov/20

$$solve{y}^{{}^{″}}+2y^{{}^{\prime }}-y={xe}^{-x^2}$$

Answered by mathmax by abdo last updated on 27/Nov/20

$$\mathrm{h}\to {\mathrm{r}}^2+2\mathrm{r}-1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1+1=2\Rightarrow {\mathrm{r}}_1=-1+\sqrt{2}\mathrm{and}{\mathrm{r}}_2=-1-\sqrt{2}$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{ae}}^{(-1+\sqrt{2})\mathrm{x}}+\mathrm{b}{\mathrm{e}}^{(-1-\sqrt{2})\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{e}}^{(-1+\sqrt{2})\mathrm{x}}{\mathrm{e}}^{(-1-\sqrt{2})\mathrm{x}}\\ (-1+\sqrt{2}){\mathrm{e}}^{(-1+\sqrt{2})}(-1-\sqrt{2}){\mathrm{e}}^{(-1-\sqrt{2})}\end{array}\right|$$$$=(-1-\sqrt{2}){\mathrm{e}}^{-2\mathrm{x}}+(1-\sqrt{2}){\mathrm{e}}^{-2\mathrm{x}}=-2\sqrt{2}{\mathrm{e}}^{-2\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{(-1-\sqrt{2})\mathrm{x}}\\ {\mathrm{xe}}^{-{\mathrm{x}}^2}(-1-\sqrt{2}){\mathrm{e}}^{(-1-\sqrt{2})\mathrm{x}}\end{array}\right|={\mathrm{xe}}^{-{\mathrm{x}}^2}{\mathrm{e}}^{(-1-\sqrt{2})\mathrm{x}}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{e}}^{(-1+\sqrt{2})\mathrm{x}}0\\ (-1+\sqrt{2}){\mathrm{e}}^{(-1+\sqrt{2})\mathrm{x}}{\mathrm{xe}}^{-{\mathrm{x}}^2}\end{array}\right|={\mathrm{xe}}^{-{\mathrm{x}}^2}{\mathrm{e}}^{(-1+\sqrt{2})\mathrm{x}}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=\int \frac{{\mathrm{xe}}^{-{\mathrm{x}}^2}{\mathrm{e}}^{(-1-\sqrt{2})\mathrm{x}}}{-2\sqrt{2}{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}=-\frac{1}{2\sqrt{2}}\int {\mathrm{xe}}^{-{\mathrm{x}}^2}{\mathrm{e}}^{(1-\sqrt{2})\mathrm{x}}\mathrm{dx}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=\int \frac{{\mathrm{xe}}^{-{\mathrm{x}}^2}{\mathrm{e}}^{(-1+\sqrt{2})\mathrm{x}}}{-2\sqrt{2}{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}=-\frac{1}{2\sqrt{2}}\int {\mathrm{xe}}^{-{\mathrm{x}}^2}{\mathrm{e}}^{(1+\sqrt{2})\mathrm{x}}\mathrm{dx}$$$$\Rightarrow {\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2$$$$\mathrm{and}\mathrm{general}\mathrm{soution}\mathrm{is}\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}$$

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