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Question Number 148466 by Ar Brandon last updated on 28/Jul/21

$$\mathrm{xdx}+\mathrm{ydy}=\mathrm{xdy}-\mathrm{ydx}$$

Answered by bramlexs22 last updated on 28/Jul/21

$$(x+y)dx=(x-y)dy$$$$\frac{dy}{dx}=\frac{x+y}{x-y}$$$$lety=ux\Rightarrow \frac{dy}{dx}=u+x\frac{du}{dx}$$$$\Rightarrow x\frac{du}{dx}+u=\frac{x(1+u)}{x(1-u)}$$$$\Rightarrow x\frac{du}{dx}=\frac{1+u-u+u^2}{1-u}$$$$\Rightarrow x\frac{du}{dx}=\frac{1+u^2}{1-u}$$$$\Rightarrow \int \frac{(1-u)}{1+u^2}du=\int \frac{dx}{x}$$$$\Rightarrow \arctan u-\frac{1}{2}\int \frac{d(u^2+1)}{u^2+1}=\ln \mid x\mid +c$$$$\Rightarrow \arctan \left(\frac{y}{x}\right)=\ln \mid Cx+\sqrt{u^2+1}\mid$$$$\Rightarrow y=x\tan (\ln \mid Cx+\sqrt{\frac{y^2+x^2}{x^2}}\mid )$$$$\Rightarrow y=x\tan (\ln \mid \frac{{Cx}^2+\sqrt{y^2+x^2}}{x}\mid )$$

Commented by Ar Brandon last updated on 28/Jul/21

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{method}$$

Commented by Ar Brandon last updated on 28/Jul/21

$$\Rightarrow y=x\tan (\ln \mid \frac{{Cx}^2+\sqrt{y^2+x^2}}{x}\mid )$$

Answered by Ar Brandon last updated on 28/Jul/21

$$\mathrm{xdx}+\mathrm{ydy}=\mathrm{xdy}-\mathrm{ydx}$$$$\frac{\mathrm{d}({\mathrm{x}}^2+{\mathrm{y}}^2)}{2\mathrm{dx}}={\mathrm{x}}^2\cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$$$$\mathrm{d}({\mathrm{x}}^2+{\mathrm{y}}^2)={2\mathrm{x}}^2\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$$$$\frac{\mathrm{d}({\mathrm{x}}^2+{\mathrm{y}}^2)}{{\mathrm{x}}^2+{\mathrm{y}}^2}=\frac{{2\mathrm{x}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$$$$\frac{\mathrm{d}({\mathrm{x}}^2+{\mathrm{y}}^2)}{{\mathrm{x}}^2+{\mathrm{y}}^2}=\frac{2}{1+{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2}\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$$$$\ln ({\mathrm{x}}^2+{\mathrm{y}}^2)=2\arctan \left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\mathrm{C}$$

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