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Question Number 99989 by bemath last updated on 24/Jun/20

$$({\mathrm{D}}^2-4\mathrm{D}+4)\mathrm{y}={xe}^{2x}$$

Answered by bobhans last updated on 24/Jun/20

$$\mathrm{homogenous}\mathrm{solution}$$$${\beta }^2-4\beta +4=0,\mathrm{which}\mathrm{has}\mathrm{root}\beta =2$$$$\mathrm{hence}\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{solution}$$$$\mathrm{Now}\mathrm{we}\mathrm{aplly}\mathrm{reduction}\mathrm{of}\mathrm{order}\mathrm{with}\mathrm{this}$$$$\mathrm{solution}.\mathrm{set}\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}\mathrm{z},\mathrm{substituting}\mathrm{this}$$$$\mathrm{into}\mathrm{the}\mathrm{original}\mathrm{diff}\mathrm{eq}\mathrm{yields}$$$${\mathrm{e}}^{2\mathrm{x}}({\mathrm{z}}^{″}+{4\mathrm{z}}^{\prime }+4\mathrm{z})-{4\mathrm{e}}^{2\mathrm{x}}({\mathrm{z}}^{\prime }+2\mathrm{z})+{4\mathrm{e}}^{2\mathrm{x}}\mathrm{z}={xe}^{2x}$$$$\mathrm{which}\mathrm{simplifies}\mathrm{to}{\mathrm{z}}^{″}=x$$$$integratingthistwiceinsuccessionyields$$$$\mathrm{z}=\frac{1}{6}{x}^3+C_{1}x+C_2$$$$\mathrm{generall}\mathrm{solution}\mathrm{y}=({\mathrm{C}}_{1}x+C_2){\mathrm{e}}^{2\mathrm{x}}+\frac{1}{6}{x}^3{e}^{2x}$$

Commented by bemath last updated on 24/Jun/20

$$\mathrm{great}...\mathrm{thank}\mathrm{much}$$

Answered by mathmax by abdo last updated on 24/Jun/20

$${\mathrm{y}}^{{}^{″}}-{4\mathrm{y}}^{{}^{\prime }}+4\mathrm{y}={\mathrm{xe}}^{2\mathrm{x}}$$$$\left(\mathrm{he}\right)\to {\mathrm{r}}^2-4\mathrm{r}+4=0\Rightarrow {(\mathrm{r}-2)}^2=0\Rightarrow \mathrm{r}=2\Rightarrow {\mathrm{y}}_{\mathrm{h}}=(\mathrm{ax}+\mathrm{b}){\mathrm{e}}^{2\mathrm{x}}$$$$={\mathrm{axe}}^{2\mathrm{x}}+{\mathrm{be}}^{2\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{xe}}^{2\mathrm{x}}{\mathrm{e}}^{2\mathrm{x}}\\ (2\mathrm{x}+1){\mathrm{e}}^{2\mathrm{x}}{2\mathrm{e}}^{2\mathrm{x}}\end{array}\right|=2\mathrm{x}{\mathrm{e}}^{4\mathrm{x}}-(2\mathrm{x}+1){\mathrm{e}}^{4\mathrm{x}}=-{\mathrm{e}}^{4\mathrm{x}}$$$${\mathrm{W}}_1=\left|\begin{array}{c}0{\mathrm{e}}^{2\mathrm{x}}\\ {\mathrm{xe}}^{2\mathrm{x}}{2\mathrm{e}}^{2\mathrm{x}}\end{array}\right|=-{\mathrm{xe}}^{4\mathrm{x}}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{xe}}^{2\mathrm{x}}0\\ (2\mathrm{x}+1){\mathrm{e}}^{2\mathrm{x}}{\mathrm{xe}}^{2\mathrm{x}}\end{array}\right|={\mathrm{x}}^2{\mathrm{e}}^{4\mathrm{x}}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=\int \frac{-{\mathrm{xe}}^{4\mathrm{x}}}{-{\mathrm{e}}^{4\mathrm{x}}}\mathrm{dx}=\int \mathrm{xdx}=\frac{{\mathrm{x}}^2}{2}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=\int \frac{{\mathrm{x}}^2{\mathrm{e}}^{4\mathrm{x}}}{-{\mathrm{e}}^{4\mathrm{x}}}\mathrm{dx}=-\int {\mathrm{x}}^2\mathrm{dx}=-\frac{{\mathrm{x}}^3}{3}$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2={\mathrm{xe}}^{2\mathrm{x}}\left(\frac{{\mathrm{x}}^2}{2}\right)+{\mathrm{e}}^{2\mathrm{x}}(-\frac{{\mathrm{x}}^3}{3})=(\frac{{\mathrm{x}}^3}{2}-\frac{{\mathrm{x}}^3}{3}){\mathrm{e}}^{2\mathrm{x}}=\frac{{\mathrm{x}}^3}{6}{\mathrm{e}}^{2\mathrm{x}}\Rightarrow$$$$\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}=(\mathrm{ax}+\mathrm{b}){\mathrm{e}}^{2\mathrm{x}}+\frac{{\mathrm{x}}^3}{6}{\mathrm{e}}^{2\mathrm{x}}$$

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