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Question Number 102382 by Ar Brandon last updated on 08/Jul/20

$${\mathrm{x}}^2\cdot \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2$$

Answered by PRITHWISH SEN 2 last updated on 08/Jul/20

$$\mathrm{put}\mathrm{y}=\mathrm{vx}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}$$$$\mathrm{and}\mathrm{the}\mathrm{equation}\mathrm{changes}\mathrm{to}$$$$\frac{\mathrm{dv}}{1+{\mathrm{v}}^2}=\frac{\mathrm{dx}}{\mathrm{x}}$$$${\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\mathrm{lnCx}\mathrm{C}=\mathrm{const}.$$$$\mathbf{y}=\mathbf{x}\tan \{\mathbf{ln}\mid \mathbf{Cx}\mid \}$$

Commented by Ar Brandon last updated on 08/Jul/20

Oh thanks, I didn't notice that.

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