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2y-x-y-xy-x-2-1-




Question Number 85834 by sahnaz last updated on 25/Mar/20
2y^′ −(x/y)=((xy)/(x^2 −1))
$$\mathrm{2y}^{'} −\frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{xy}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}} \\ $$
Answered by mind is power last updated on 27/Mar/20
⇔2y′y−x=((xy^2 )/(x^2 −1))  u=y^2 ⇒2y′y=u′  ⇒u′−x=((xu)/(x^2 −1))  ⇒xu−(x^2 −1)u′−x(x^2 −1)=0 easy now
$$\Leftrightarrow\mathrm{2}{y}'{y}−{x}=\frac{{xy}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${u}={y}^{\mathrm{2}} \Rightarrow\mathrm{2}{y}'{y}={u}' \\ $$$$\Rightarrow{u}'−{x}=\frac{{xu}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{xu}−\left({x}^{\mathrm{2}} −\mathrm{1}\right){u}'−{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0}\:{easy}\:{now} \\ $$

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