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Question Number 28529 by abdo imad last updated on 26/Jan/18

$$solvethed.e.(x^2-1)y^{{}^{\prime }}+xy=x^2-e^x.$$

Commented by abdo imad last updated on 28/Jan/18

$$h.e.(x^2-1)y^{{}^{\prime }}=-xy\Rightarrow \frac{y^{{}^{\prime }}}{y}=\frac{-x}{x^2-1}$$$$\Rightarrow ln\mid y\mid =\int \frac{x}{1-x^2}dx=\frac{-1}{2}ln\mid 1-x^2\mid +c$$$$=ln\left(\frac{1}{\sqrt{\mid 1-x^2\mid }}\right)+c\Rightarrow y=\frac{\lambda }{\sqrt{\mid 1-x^2\mid }}letusemvcmethod$$$$case1\mid x\mid <1\Rightarrow y=\lambda {(1-x^2)}^{-\frac{1}{2}}$$$$y^{{}^{\prime }}={\lambda }^{{}^{\prime }}{(1-x^2)}^{\frac{-1}{2}}+\lambda x{(1-x^2)}^{\frac{-3}{2}}\left(e\right)\Rightarrow$$$$-{\lambda }^{{}^{\prime }}{(1-x^2)}^{\frac{1}{2}}-\lambda x{(1-x^2)}^{\frac{-1}{2}}+\lambda x{(1-x^2)}^{\frac{-1}{2}}=x^2-e^x$$$$\Rightarrow {\lambda }^{{}^{\prime }}{(1-x^2)}^{\frac{1}{2}}=e^x-x^2$$$$\Rightarrow {\lambda }^{{}^{\prime }}=(e^x-x^2){(1-x^2)}^{\frac{-1}{2}}\Rightarrow$$$$\lambda \left(x\right)={\int }_{.}^x\frac{e^t-t^2}{\sqrt{1-t^2}}dt+kand$$$$y\left(x\right)=\frac{1}{\sqrt{1-x^2}}({\int }_{.}^x\frac{e^t-t^2}{\sqrt{1-t^2}}dt+k).$$

Commented by abdo imad last updated on 28/Jan/18

$$forcase2\mid x\mid >1wefolowthesamemethod.$$

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