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Question Number 94903 by mathmax by abdo last updated on 21/May/20

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

Answered by niroj last updated on 21/May/20

$${\mathrm{y}}^{{}^{″}}+{2\mathrm{y}}^{{}^{\prime }}+\mathrm{y}=\mathrm{x}{\mathrm{e}}^{-\mathrm{x}}$$$$({\mathrm{D}}^2+2\mathrm{D}+1)\mathrm{y}={\mathrm{xe}}^{-\mathrm{x}}$$$$\mathrm{A}.\mathrm{E}.,({\mathrm{m}}^2+2\mathrm{m}+1)=0$$$${(\mathrm{m}+1)}^2=0,\mathrm{m}=-1,-1$$$$\mathrm{CF}=({\mathrm{C}}_1+{\mathrm{C}}_2\mathrm{x}){\mathrm{e}}^{-\mathrm{x}}$$$$\mathrm{PI}=\frac{{\mathrm{xe}}^{-\mathrm{x}}}{{(\mathrm{D}+1)}^2}$$$$={\mathrm{e}}^{-\mathrm{x}}\frac{\mathrm{x}}{{(\mathrm{D}-1+1)}^2}$$$$={\mathrm{e}}^{-\mathrm{x}}\frac{\mathrm{x}}{{\mathrm{D}}^2}$$$$={\mathrm{e}}^{-\mathrm{x}}\int (\int \mathrm{xdx})\mathrm{dx}$$$$={\mathrm{e}}^{-\mathrm{x}}\int \frac{{\mathrm{x}}^2}{2}\mathrm{dx}=\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}\times \frac{{\mathrm{x}}^3}{3}$$$$=\frac{1}{6}{\mathrm{e}}^{-\mathrm{x}}{\mathrm{x}}^3$$$$\mathrm{y}=\mathrm{CF}+\mathrm{PI}$$$$\mathrm{y}=({\mathrm{C}}_1+{\mathrm{C}}_2\mathrm{x})+\frac{1}{6}{\mathrm{e}}^{-\mathrm{x}}{\mathrm{x}}^3//.$$$$$$$$$$$$$$

Commented by mathmax by abdo last updated on 22/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

Commented by niroj last updated on 22/May/20

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Answered by ElOuafi last updated on 22/May/20

$$let\left(E\right):y^{″}+2y^{\prime }+y={xe}^{-x}$$$$and\left(E_0\right):y^{″}+2y^{\prime }+y=0beitsequationwithout$$$$secondmember.\Rightarrow \left(E_r\right):r^2+2r+1=0\Rightarrow r=-1$$$$so{y}_1=e^{-x}\wedge y_2={xe}^{-x}aresolutionsof\left(E_0\right)$$$$thegeneralsolutionof\left(E_0\right)is:y_{g_0}={\lambda }_1{e}^{-x}+{\lambda }_2{xe}^{-x}$$$$with{\lambda }_1,{\lambda }_2\in \mathbb{R},letsearchnowaparticularsolution$$$$of\left(E\right)undertheforme:y_p={\lambda }_1\left(x\right)y_1+{\lambda }_2\left(x\right)y_2$$$$witch{\lambda }_1\left(x\right),{\lambda }_2\left(x\right)satisfying$$$${\lambda }_1^{{}^{\prime }}\left(x\right)e^{-x}+{\lambda }_2^{{}^{\prime }}\left(x\right).{xe}^{-x}=0\wedge {\lambda }_1^{{}^{\prime }}\left(x\right).{\left(e^{-x}\right)}^{{}^{\prime }}+{\lambda }_2^{{}^{\prime }}\left(x\right).{\left({xe}^{-x}\right)}^{\prime }={xe}^{-x}$$$$aftersolvingsystemwefind:{\lambda }_1^{{}^{\prime }}\left(x\right)=-x^2\wedge {\lambda }_2^{{}^{\prime }}\left(x\right)=x$$$$\Rightarrow {\lambda }_1=-\frac{x^3}{3}\wedge {\lambda }_2=\frac{x^2}{2}$$$$\Rightarrow y_p=-\frac{x^3}{3}{y}_1+\frac{x^2}{2}{y}_2\iff y_p=-\frac{x^3}{3}{e}^{-x}+\frac{x^3}{2}{e}^{-x}=\frac{x^3}{6}{e}^{-x}$$$$thegeneralsolutionof\left(E\right)is:$$$$y_g=y_p+y_{g_0}$$$$y\left(x\right)=\frac{x^3}{6}{e}^{-x}+{\lambda }_1{e}^{-x}+{\lambda }_2{xe}^{-x};({\lambda }_1;{\lambda }_2)\in {\mathbb{R}}^2$$

Commented by mathmax by abdo last updated on 22/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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