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
$$\left({x}+{y}\right){dx}=\left({x}−{y}\right){dy} \\ $$$$\:\frac{{dy}}{{dx}}=\:\frac{{x}+{y}}{{x}−{y}} \\ $$$$\:{let}\:{y}={ux}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{u}\:+\:{x}\:\frac{{du}}{{dx}} \\ $$$$\Rightarrow{x}\:\frac{{du}}{{dx}}+{u}=\:\frac{{x}\left(\mathrm{1}+{u}\right)}{{x}\left(\mathrm{1}−{u}\right)} \\ $$$$\Rightarrow{x}\:\frac{{du}}{{dx}}=\:\frac{\mathrm{1}+{u}−{u}+{u}^{\mathrm{2}} }{\mathrm{1}−{u}} \\ $$$$\Rightarrow{x}\:\frac{{du}}{{dx}}=\:\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\mathrm{1}−{u}} \\ $$$$\Rightarrow\int\:\frac{\left(\mathrm{1}−{u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:=\:\int\:\frac{{dx}}{{x}} \\ $$$$\Rightarrow\mathrm{arctan}\:{u}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}\:=\:\mathrm{ln}\:\mid{x}\mid\:+\:{c} \\ $$$$\Rightarrow\mathrm{arctan}\:\left(\frac{{y}}{{x}}\right)=\mathrm{ln}\:\mid{Cx}+\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}\mid \\ $$$$\Rightarrow{y}={x}\:\mathrm{tan}\:\left(\mathrm{ln}\mid{Cx}+\sqrt{\frac{{y}^{\mathrm{2}} +{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\:\mid\right) \\ $$$$\Rightarrow{y}={x}\:\mathrm{tan}\left(\mathrm{ln}\mid\:\frac{{Cx}^{\mathrm{2}} +\sqrt{{y}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{x}}\:\mid\right) \\ $$
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
$$\Rightarrow{y}={x}\:\mathrm{tan}\left(\mathrm{ln}\:\mid\frac{{Cx}^{\mathrm{2}} +\sqrt{{y}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{x}}\:\mid\right) \\ $$
Answered by Ar Brandon last updated on 28/Jul/21
$$\mathrm{xdx}+\mathrm{ydy}=\mathrm{xdy}−\mathrm{ydx} \\ $$$$\frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)}{\mathrm{2dx}}=\mathrm{x}^{\mathrm{2}} \centerdot\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \\ $$$$\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)=\mathrm{2x}^{\mathrm{2}} \mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \\ $$$$\frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{2x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \\ $$$$\frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{1}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{2}} }\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \\ $$$$\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)=\mathrm{2arctan}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\mathrm{C} \\ $$