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Question Number 146260 by mathmax by abdo last updated on 12/Jul/21

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

Answered by qaz last updated on 12/Jul/21

$${\mathrm{y}}_{\mathrm{p}}=\frac{1}{{\mathrm{D}}^2-2\mathrm{D}+1}{\mathrm{e}}^{-\mathrm{x}}\sin \mathrm{x}$$$$={\mathrm{e}}^{\mathrm{x}}\frac{1}{{\mathrm{D}}^2}{\mathrm{e}}^{-2\mathrm{x}}\sin \mathrm{x}$$$$={\mathrm{e}}^{\mathrm{x}}\int \int {\mathrm{e}}^{-2\mathrm{x}}\sin \mathrm{xdxdx}$$$$......$$$$=\frac{1}{25}{\mathrm{e}}^{-\mathrm{x}}(4\cos \mathrm{x}+3\sin \mathrm{x})$$$$----------$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{-\mathrm{x}}\frac{1}{{(\mathrm{D}-2)}^2}\sin \mathrm{x}$$$$={\mathrm{e}}^{-\mathrm{x}}\frac{{(\mathrm{D}+2)}^2}{{({\mathrm{D}}^2-4)}^2}\sin \mathrm{x}$$$$={\mathrm{e}}^{-\mathrm{x}}\frac{-1^2+4\mathrm{D}+4}{{(-1^2-4)}^2}\sin \mathrm{x}$$$$=\frac{1}{25}{\mathrm{e}}^{-\mathrm{x}}(4\cos \mathrm{x}+3\sin \mathrm{x})$$$$\Rightarrow \mathrm{y}={\mathrm{e}}^{\mathrm{x}}({\mathrm{C}}_1+{\mathrm{C}}_2\mathrm{x})+\frac{1}{25}{\mathrm{e}}^{-\mathrm{x}}(4\cos \mathrm{x}+3\sin \mathrm{x})$$

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