Question Number 87944 by jagoll last updated on 07/Apr/20
$$\mathrm{y}\:'\:=\:\frac{\mathrm{2xy}}{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} } \\ $$
Commented by john santu last updated on 07/Apr/20
![(dy/dx) = ((2xy)/(y^2 −x^2 )) [ let x = vy , 1 = v(dy/dx) + y(dv/dx) ] (dy/dx) = ((1−y(dv/dx))/v) ⇒((1−y(dv/dx))/v) = ((2vy^2 )/(y^2 −v^2 y^2 )) = ((2v)/(1−v^2 )) 1−y(dv/dx) = ((2v^2 )/(1−v^2 )) ⇒ y(dv/dx) = ((1+v^2 )/(1−v^2 )) ((1−v^2 )/(1+v^2 )) dv = (dx/(((x/v)))) ⇒∫ (((1−v^2 )/(v+v^3 ))) dv = ln∣x∣ +c ∫ ((1−v^2 )/(v(1+v^2 ))) dv = ln ∣x∣ + c ∫ (dv/v)−∫ (2/(1+v^2 )) dv = ln∣x∣ + c ln ∣v∣ − 2 tan^(−1) (v) = ln∣x∣ +c ln ∣(x/y)∣ − 2tan^(−1) ((x/y)) = ln ∣x∣ + c](https://www.tinkutara.com/question/Q87945.png)
$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2xy}}{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} } \\ $$$$\left[\:\mathrm{let}\:\mathrm{x}\:=\:\mathrm{vy}\:,\:\mathrm{1}\:=\:\mathrm{v}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{y}\frac{\mathrm{dv}}{\mathrm{dx}}\:\right]\:\: \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}−\mathrm{y}\frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}} \\ $$$$\Rightarrow\frac{\mathrm{1}−\mathrm{y}\frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}}\:=\:\frac{\mathrm{2vy}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{2v}}{\mathrm{1}−\mathrm{v}^{\mathrm{2}} } \\ $$$$\mathrm{1}−\mathrm{y}\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:\frac{\mathrm{2v}^{\mathrm{2}} }{\mathrm{1}−\mathrm{v}^{\mathrm{2}} }\:\Rightarrow\:\mathrm{y}\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}+\mathrm{v}^{\mathrm{2}} }{\mathrm{1}−\mathrm{v}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}−\mathrm{v}^{\mathrm{2}} }{\mathrm{1}+\mathrm{v}^{\mathrm{2}} }\:\mathrm{dv}\:=\:\frac{\mathrm{dx}}{\left(\frac{\mathrm{x}}{\mathrm{v}}\right)} \\ $$$$\Rightarrow\int\:\left(\frac{\mathrm{1}−\mathrm{v}^{\mathrm{2}} }{\mathrm{v}+\mathrm{v}^{\mathrm{3}} }\right)\:\mathrm{dv}\:=\:\mathrm{ln}\mid\mathrm{x}\mid\:+\mathrm{c}\: \\ $$$$\int\:\frac{\mathrm{1}−\mathrm{v}^{\mathrm{2}} }{\mathrm{v}\left(\mathrm{1}+\mathrm{v}^{\mathrm{2}} \right)}\:\mathrm{dv}\:=\:\mathrm{ln}\:\mid\mathrm{x}\mid\:+\:\mathrm{c} \\ $$$$\int\:\frac{\mathrm{dv}}{\mathrm{v}}−\int\:\frac{\mathrm{2}}{\mathrm{1}+\mathrm{v}^{\mathrm{2}} }\:\mathrm{dv}\:=\:\mathrm{ln}\mid\mathrm{x}\mid\:+\:\mathrm{c} \\ $$$$\mathrm{ln}\:\mid\mathrm{v}\mid\:−\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{v}\right)\:=\:\:\mathrm{ln}\mid\mathrm{x}\mid\:+\mathrm{c} \\ $$$$\mathrm{ln}\:\mid\frac{\mathrm{x}}{\mathrm{y}}\mid\:−\:\mathrm{2tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)\:=\:\mathrm{ln}\:\mid\mathrm{x}\mid\:+\:\mathrm{c} \\ $$