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Question Number 121517 by liberty last updated on 09/Nov/20

$${\mathrm{y}}^{″}+{4\mathrm{y}}^{\prime }+4\mathrm{y}=(\mathrm{x}+3){\mathrm{e}}^{-2\mathrm{x}}$$$$\mathrm{y}\left(0\right)=2\wedge {\mathrm{y}}^{\prime }\left(0\right)=5$$

Answered by benjo_mathlover last updated on 09/Nov/20

Answered by TANMAY PANACEA last updated on 09/Nov/20

$$C.F$$$$y=e^{mx}$$$$(m^2+4m+4)e^{mx}=0\to m=-2,-2$$$$C.F=C_1{e}^{-2x}+C_2{xe}^{-2x}$$$$P.I=\frac{(x+3)e^{-2x}}{(D^2+4D+4)}=\frac{1}{{(D+2)}^2}{e}^{-2x}\times (x+3)$$$$=e^{-2x}\times \frac{1}{{(D-2+2)}^2}\times (x+3)$$$$=e^{-2x}\times \frac{1}{D^2}(x+3)$$$$e^{-2x}(\frac{x^3}{6}+\frac{3x^2}{2})$$$$y=C_1{e}^{-2x}+C_2{xe}^{-2x}+e^{-2x}(\frac{x^3}{6}+\frac{3x^2}{2})$$$$$$

Answered by mathmax by abdo last updated on 09/Nov/20

$${\mathrm{y}}^{{}^{″}}+{4\mathrm{y}}^{{}^{\prime }}+4\mathrm{y}=(\mathrm{x}+3){\mathrm{e}}^{-2\mathrm{x}}$$$$\mathrm{h}\to {\mathrm{r}}^2+4\mathrm{r}+4=0\Rightarrow {(\mathrm{r}+2)}^2=0\Rightarrow \mathrm{r}=-2\left(\mathrm{double}\right)\Rightarrow$$$${\mathrm{y}}_{\mathrm{h}}=(\mathrm{ax}+\mathrm{b}){\mathrm{e}}^{-2\mathrm{x}}={\mathrm{axe}}^{-2\mathrm{x}}+{\mathrm{be}}^{-2\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{xe}}^{-2\mathrm{x}}{\mathrm{e}}^{-2\mathrm{x}}\\ (1-2\mathrm{x}){\mathrm{e}}^{-2\mathrm{x}}-{2\mathrm{e}}^{-2\mathrm{x}}\end{array}\right|=-2\mathrm{x}{\mathrm{e}}^{-4\mathrm{x}}-(1-2\mathrm{x}){\mathrm{e}}^{-4\mathrm{x}}$$$$=-{\mathrm{e}}^{-4\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}0{\mathrm{e}}^{-2\mathrm{x}}\\ (\mathrm{x}+3){\mathrm{e}}^{-2\mathrm{x}}-{2\mathrm{e}}^{-2\mathrm{x}}\end{array}\right|=-(\mathrm{x}+3){\mathrm{e}}^{-4\mathrm{x}}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{xe}}^{-2\mathrm{x}}0\\ (1-2\mathrm{x}){\mathrm{e}}^{-2\mathrm{x}}(\mathrm{x}+3){\mathrm{e}}^{-2\mathrm{x}}\end{array}\right|=({\mathrm{x}}^2+3\mathrm{x}){\mathrm{e}}^{-4\mathrm{x}}$$$${\mathrm{V}}_1=\int \frac{{\mathrm{W}}_1}{\mathrm{W}}\mathrm{dx}=-\int \frac{(\mathrm{x}+3){\mathrm{e}}^{-4\mathrm{x}}}{-{\mathrm{e}}^{-4\mathrm{x}}}=\int (\mathrm{x}+3)\mathrm{dx}=\frac{{\mathrm{x}}^2}{2}+3\mathrm{x}$$$${\mathrm{V}}_2=\int \frac{{\mathrm{W}}_2}{\mathrm{W}}\mathrm{dx}=\int \frac{({\mathrm{x}}^2+3\mathrm{x}){\mathrm{e}}^{-4\mathrm{x}}}{-{\mathrm{e}}^{-4\mathrm{x}}}\mathrm{dx}=-\int ({\mathrm{x}}^2+3\mathrm{x})\mathrm{dx}$$$$=-\frac{{\mathrm{x}}^3}{3}-\frac{3}{2}{\mathrm{x}}^2\Rightarrow {\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2$$$$={\mathrm{xe}}^{-2\mathrm{x}}(\frac{{\mathrm{x}}^2}{2}+3\mathrm{x})+{\mathrm{e}}^{-2\mathrm{x}}(-\frac{{\mathrm{x}}^3}{3}-\frac{3}{2}{\mathrm{x}}^2)$$$$=(\frac{{\mathrm{x}}^3}{2}+{3\mathrm{x}}^2-\frac{{\mathrm{x}}^3}{3}-\frac{3}{2}{\mathrm{x}}^2){\mathrm{e}}^{-2\mathrm{x}}=(\frac{{\mathrm{x}}^3}{6}+\frac{3}{2}{\mathrm{x}}^2){\mathrm{e}}^{-2\mathrm{x}}$$$$\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}$$$$\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}={\mathrm{axe}}^{-2\mathrm{x}}+{\mathrm{be}}^{-2\mathrm{x}}+(\frac{{\mathrm{x}}^3}{6}+\frac{3}{2}{\mathrm{x}}^2){\mathrm{e}}^{-2\mathrm{x}}$$$$=(\mathrm{ax}+\mathrm{b}+\frac{{\mathrm{x}}^3}{6}+\frac{3}{2}{\mathrm{x}}^2){\mathrm{e}}^{-2\mathrm{x}}$$$$\mathrm{y}\left(0\right)=2\Rightarrow \mathrm{b}=2$$$${\mathrm{y}}^{{}^{\prime }}=(\mathrm{a}+\frac{1}{2}{\mathrm{x}}^2+3\mathrm{x}){\mathrm{e}}^{-2\mathrm{x}}-{2\mathrm{e}}^{-2\mathrm{x}}(\mathrm{ax}+\mathrm{b}+\frac{{\mathrm{x}}^3}{6}+\frac{3}{2}{\mathrm{x}}^2)$$$${\mathrm{y}}^{{}^{\prime }}\left(0\right)=5\Rightarrow \mathrm{a}-2\left(\mathrm{b}\right)=5\Rightarrow \mathrm{a}=5+2\mathrm{b}=5+4=9\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=(9\mathrm{x}+2+\frac{{\mathrm{x}}^3}{6}+\frac{3}{2}{\mathrm{x}}^2){\mathrm{e}}^{-2\mathrm{x}}$$

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