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Question Number 93815 by i jagooll last updated on 15/May/20

$$({\mathrm{D}}^2+6\mathrm{D}+9)\mathrm{y}=\frac{{\mathrm{e}}^{-3\mathrm{x}}}{{\mathrm{x}}^3}$$

Answered by john santu last updated on 15/May/20

$$\mathrm{homogenous}\mathrm{solution}$$$${\lambda }^2+6\lambda +9=0$$$$\lambda =-3,-3$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{Ae}}^{-3\mathrm{x}}+{\mathrm{Bxe}}^{-3\mathrm{x}}$$$$\mathrm{particular}\mathrm{solution}$$$$(\mathrm{D}+3)\left[(\mathrm{D}+3)\left({\mathrm{e}}^{3\mathrm{x}}\right)\right]=\frac{1}{{\mathrm{x}}^3}$$$$(\mathrm{D}+3)\left[\mathrm{D}\left({\mathrm{e}}^{3\mathrm{x}}\mathrm{y}\right)\right]=\frac{1}{{\mathrm{x}}^3}$$$$(\mathrm{D}+3)\left({\mathrm{e}}^{3\mathrm{x}}\mathrm{y}\right)=\int \frac{1}{{\mathrm{x}}^3}\mathrm{dx}=-\frac{1}{{2\mathrm{x}}^2}$$$${\mathrm{e}}^{3\mathrm{x}}\mathrm{y}=\int -\frac{1}{{2\mathrm{x}}^2}\mathrm{dx}=$$$${\mathrm{y}}_{\mathrm{p}}=\frac{1}{2\mathrm{x}.{\mathrm{e}}^{3\mathrm{x}}}=\frac{{\mathrm{e}}^{-3\mathrm{x}}}{2\mathrm{x}}$$$$\mathrm{generall}\mathrm{solution}$$$$\mathrm{y}={\mathrm{Ae}}^{-3\mathrm{x}}+{\mathrm{Bxe}}^{-3\mathrm{x}}+\frac{{\mathrm{e}}^{-3\mathrm{x}}}{2\mathrm{x}}$$

Commented by i jagooll last updated on 15/May/20

$$\mathrm{thAnk}\mathrm{you}$$

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