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Question Number 27594 by abdo imad last updated on 10/Jan/18

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

Commented by abdo imad last updated on 12/Jan/18

$$e.h\Rightarrow (1-x^2)y^{{}^{\prime }}-xy=0\iff (1-x^2)y^{,}=xy$$$$\iff \frac{y^{{}^{\prime }}}{y}=\frac{x}{1-x^2}\iff ln/y/=\int \frac{x}{1-x^2}dx+k$$$$ln/y/=-\frac{1}{2}ln/1-x^2/+k=ln\left(\frac{1}{\sqrt{/1-x^2}}\right)+k$$$$\Rightarrow y=\frac{\lambda }{\sqrt{/1-x^2/}}letfind\lambda bymvcmethod$$$$if-1<x<1y=\lambda {(1-x^2)}^{-\frac{1}{2}}so$$$$y^{{}^{\prime }}={\lambda }^{,}{(1-x^2)}^{-\frac{1}{2}}-\frac{1}{2}\lambda (-2x){(1-x^2)}^{-\frac{3}{2}}$$$$={\lambda }^{{}^{\prime }}{(1-x^2)}^{-\frac{1}{2}}+\lambda x{(1-x^2)}^{-\frac{3}{2}}$$$$equ.\iff (1-x^2){\lambda }^{{}^{\prime }}{(1-x^2)}^{-\frac{1}{2}}+\lambda x(1-x^2){(1-x^2)}^{-\frac{3}{2}}-\lambda x{(1-x^2)}^{-\frac{1}{2}}=1$$$${\lambda }^{\prime }\sqrt{1-x^2}=1\Rightarrow {\lambda }^{{}^{\prime }}=\frac{1}{\sqrt{1-x^2}}\Rightarrow \lambda =\int \frac{dx}{\sqrt{1-x^2}}+c$$$$\lambda =arcsinx+candy\left(x\right)=\frac{1}{\sqrt{1-x^2}}(arcsinx+c)$$$$y\left(x\right)=\frac{arcsinx}{\sqrt{1-x^2}}+\frac{c}{\sqrt{1-x^2}}.$$$$if1-x^2<0\iff /x/>1y=\frac{\lambda }{\sqrt{x^2-1}}andthesamemethod$$$$givetheresult.$$$$$$

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