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

Solve the differential equation  (dy/dx)=((1+y^2 )/(y(1−x^2 )))

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

Answered by mahdipoor last updated on 08/Sep/22

⇒∫(y/(1+y^2 ))dy=∫(1/(1−x^2 ))dx  ⇒  ((ln(1+y^2 ))/2)=((ln∣((x+1)/(x−1))∣)/2)+C^′   ⇒   1+y^2 =C.∣((x+1)/(x−1))∣

$$\Rightarrow\int\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} }{dy}=\int\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:\:\Rightarrow \\ $$$$\frac{{ln}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}{\mathrm{2}}=\frac{{ln}\mid\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\mid}{\mathrm{2}}+{C}^{'} \:\:\Rightarrow\: \\ $$$$\mathrm{1}+{y}^{\mathrm{2}} ={C}.\mid\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\mid \\ $$$$ \\ $$

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