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Question Number 96051 by i jagooll last updated on 29/May/20

(x−y) dx + (x^2 +y^2 ) dy = 0

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

Commented by bobhans last updated on 29/May/20

(dy/dx) = ((y−x)/(x^2 +y^2 )) . set y = vx (dy/dx) = v + x (dv/dx) ⇒ v+x (dv/dx) = ((x(v−1))/(x^2 (1+v^2 ))) x (dv/dx) = ((v−1)/(x+xv^2 ))−v = ((v−1−vx−xv^3 )/(x+xv^2 )) tobe continu

$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{y}−\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:.\:\mathrm{set}\:\mathrm{y}\:=\:\mathrm{vx}\: \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{v}\:+\:\mathrm{x}\:\frac{\mathrm{dv}}{\mathrm{dx}}\:\Rightarrow\:\mathrm{v}+\mathrm{x}\:\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}\left(\mathrm{v}−\mathrm{1}\right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{v}^{\mathrm{2}} \right)} \\ $$$$\mathrm{x}\:\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:\frac{\mathrm{v}−\mathrm{1}}{\mathrm{x}+\mathrm{xv}^{\mathrm{2}} }−\mathrm{v}\:=\:\frac{\mathrm{v}−\mathrm{1}−\mathrm{vx}−\mathrm{xv}^{\mathrm{3}} }{\mathrm{x}+\mathrm{xv}^{\mathrm{2}} } \\ $$$$\mathrm{tobe}\:\mathrm{continu} \\ $$

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