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Question Number 175888 by Mastermind last updated on 08/Sep/22

$$\mathrm{Solve}\mathrm{the}\mathrm{differential}\mathrm{equation}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1+{\mathrm{y}}^2}{\mathrm{y}(1-{\mathrm{x}}^2)}$$

Answered by mahdipoor last updated on 08/Sep/22

$$\Rightarrow \int \frac{y}{1+y^2}dy=\int \frac{1}{1-x^2}dx\Rightarrow$$$$\frac{ln(1+y^2)}{2}=\frac{ln\mid \frac{x+1}{x-1}\mid }{2}+C^{{}^{\prime }}\Rightarrow$$$$1+y^2=C.\mid \frac{x+1}{x-1}\mid$$$$$$

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