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Question Number 93003 by i jagooll last updated on 10/May/20

(x^2 −2xy−y^2 )dx−(x−y)^2 dy=0

$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}−\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx}−\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} \:\mathrm{dy}=\mathrm{0} \\ $$

Answered by TANMAY PANACEA ... last updated on 10/May/20

(dy/dx)=((x^2 −2xy−y^2 )/(x^2 −2xy+y^2 ))=((1−2(y/x)−(y^2 /x^2 ))/(1−2(y/x)+(y^2 /x^2 ))) (y/x)=v→ (dy/dx)=x.(dv/dx)+v v+x(dv/dx)=((1−2v−v^2 )/(1−2v+v^2 )) x(dv/dx)=((1−2v−v^2 )/(1−2v+v^2 ))−v x(dv/dx)=((1−2v−v^2 −v+2v^2 −v^3 )/(1−2v+v^2 )) (((1−2v+v^2 )dv)/(1−3v+v^2 −v^3 ))=(dx/x) wait

$$\frac{{dy}}{{dx}}=\frac{{x}^{\mathrm{2}} −\mathrm{2}{xy}−{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{2}{xy}+{y}^{\mathrm{2}} }=\frac{\mathrm{1}−\mathrm{2}\frac{{y}}{{x}}−\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{2}\frac{{y}}{{x}}+\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }} \\ $$$$\frac{{y}}{{x}}={v}\rightarrow\:\:\frac{{dy}}{{dx}}={x}.\frac{{dv}}{{dx}}+{v} \\ $$$${v}+{x}\frac{{dv}}{{dx}}=\frac{\mathrm{1}−\mathrm{2}{v}−{v}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{v}+{v}^{\mathrm{2}} } \\ $$$${x}\frac{{dv}}{{dx}}=\frac{\mathrm{1}−\mathrm{2}{v}−{v}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{v}+{v}^{\mathrm{2}} }−{v} \\ $$$${x}\frac{{dv}}{{dx}}=\frac{\mathrm{1}−\mathrm{2}{v}−{v}^{\mathrm{2}} −{v}+\mathrm{2}{v}^{\mathrm{2}} −{v}^{\mathrm{3}} }{\mathrm{1}−\mathrm{2}{v}+{v}^{\mathrm{2}} } \\ $$$$\frac{\left(\mathrm{1}−\mathrm{2}{v}+{v}^{\mathrm{2}} \right){dv}}{\mathrm{1}−\mathrm{3}{v}+{v}^{\mathrm{2}} −{v}^{\mathrm{3}} }=\frac{{dx}}{{x}} \\ $$$${wait} \\ $$$$ \\ $$$$ \\ $$

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