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Question Number 130875 by EDWIN88 last updated on 30/Jan/21

Answered by benjo_mathlover last updated on 30/Jan/21

$${\mathrm{y}}^{″}-\left(\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\right){\mathrm{y}}^{\prime }+\left(\frac{2}{1-{\mathrm{x}}^2}\right)\mathrm{y}=0$$$${\mathrm{y}}_2^{\prime }{\mathrm{y}}_1-{\mathrm{y}}_2{\mathrm{y}}_1^{\prime }={\mathrm{Ce}}^{-\int {\mathrm{a}}_1\mathrm{dx}}$$$$\frac{{\mathrm{y}}_2^{\prime }{\mathrm{y}}_1-{\mathrm{y}}_2{\mathrm{y}}_1^{\prime }}{{\mathrm{y}}_1^2}=\frac{1}{{\mathrm{y}}_1^2}{\mathrm{Ce}}^{-\int {\mathrm{a}}_1\mathrm{dx}}$$$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{y}}_2}{{\mathrm{y}}_1}\right)=\frac{1}{{\mathrm{y}}_1^2}{\mathrm{Ce}}^{-\int {\mathrm{a}}_1\mathrm{dx}}$$$$\frac{{\mathrm{y}}_2}{{\mathrm{y}}_1}=\int \frac{{\mathrm{Ce}}^{-\int {\mathrm{a}}_1\mathrm{dx}}}{{\mathrm{y}}_1^2}\mathrm{dx};{\mathrm{y}}_2={\mathrm{y}}_1\int \frac{{\mathrm{Ce}}^{-\int {\mathrm{a}}_1\mathrm{dx}}}{{\mathrm{y}}_1^2}\mathrm{dx}$$$$\mathrm{y}={\mathrm{C}}_1{\mathrm{y}}_1+{\mathrm{C}}_2{\mathrm{y}}_1\int \frac{{\mathrm{e}}^{-\int {\mathrm{a}}_1\mathrm{dx}}}{{\mathrm{y}}_1^2}\mathrm{dx}$$$$\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{this}\mathrm{eq}\mathrm{has}\mathrm{a}\mathrm{particular}\mathrm{solution}$$$${\mathrm{y}}_1=\mathrm{x}\mathrm{then}{\mathrm{a}}_1=\frac{-2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{so}\mathrm{we}\mathrm{have}$$$$\mathrm{y}=\mathrm{x}\int \frac{{\mathrm{e}}^{\int \frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx}}}{{\mathrm{x}}^2}\mathrm{dx}=\mathrm{x}\int \frac{{\mathrm{e}}^{-\ln (1-{\mathrm{x}}^2)}}{{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{y}=\mathrm{x}\int \frac{\mathrm{dx}}{{\mathrm{x}}^2(1-{\mathrm{x}}^2)}=\mathrm{x}\int (\frac{1}{{\mathrm{x}}^2}+\frac{1}{2(1-\mathrm{x})}+\frac{1}{2(1+\mathrm{x})})\mathrm{dx}$$$$\mathrm{y}=\mathrm{x}(-\frac{1}{\mathrm{x}}+\frac{1}{2}\ln \mid \frac{1+\mathrm{x}}{1-\mathrm{x}}\mid )$$$$\mathcal{G}\mathrm{eneral}\mathrm{solution}$$$$\mathrm{y}={\mathrm{C}}_1\mathrm{x}+{\mathrm{C}}_2(\frac{1}{2}\mathrm{xln}\mid \frac{1+\mathrm{x}}{1-\mathrm{x}}\mid -1)$$

Commented by EDWIN88 last updated on 30/Jan/21

$$nice.$$

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