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Question Number 46609 by maxmathsup by imad last updated on 29/Oct/18

$$solvex{y}^{{}^{″}}-e^{-x}{y}^{{}^{\prime }}=xsinx$$

Commented by maxmathsup by imad last updated on 11/Nov/18

$$hangement{y}^{{}^{\prime }}=zgive{xz}^{{}^{\prime }}-e^{-x}z=xsinx$$$$\left(he\right)\Rightarrow {xz}^{{}^{\prime }}-e^{-x}z=0\Rightarrow {xz}^{{}^{\prime }}=e^{-x}z\Rightarrow \frac{z^{{}^{\prime }}}{z}=\frac{e^{-x}}{x}\Rightarrow ln\mid z\mid =\int \frac{e^{-x}}{x}dx+\alpha \Rightarrow$$$$z=K{e}^{\int \frac{e^{-x}}{x}dx}mvcmethodgve{z}^{{}^{\prime }}=K^{{}^{\prime }}{e}^{\int \frac{e^{-x}}{x}dx}+K\frac{e^{-x}}{x}{e}^{\int \frac{e^{-x}}{x}dx}$$$$\left(e\right)\Rightarrow {xK}^{{}^{\prime }}{e}^{\int \frac{e^{-x}}{x}dx}+{Ke}^{-x}{e}^{\int \frac{e^{-x}}{x}dx}-e^{-x}K{e}^{\int \frac{e^{-x}}{x}dx}=xsinx\Rightarrow$$$${xK}^{{}^{\prime }}{e}^{\int \frac{e^{-x}}{x}dx}=xsinx\Rightarrow K^{{}^{\prime }}=sinx{e}^{-\int \frac{e^{-x}}{x}dx}\Rightarrow$$$$K\left(x\right)={\int }_{.}^x\left(sint{e}^{-\int \frac{e^{-t}}{t}dt}\right)dt+\lambda \Rightarrow z=({\int }_{.}^x\left(sint{e}^{-\int \frac{e^{-t}}{t}dt}\right)dt+\lambda )e^{\int \frac{e^{-x}}{x}}dx$$$$but{y}^{{}^{\prime }}\left(x\right)=z\left(x\right)\Rightarrow y\left(x\right)=\int z\left(x\right)dx+\beta andthefunctionzisdetermined.$$

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