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

$$(\mathrm{x}-\mathrm{y})\mathrm{dx}+({\mathrm{x}}^2+{\mathrm{y}}^2)\mathrm{dy}=0$$

Commented by bobhans last updated on 29/May/20

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

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