xdx-ydy-xdy-ydx-

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 (dy/dx)= ((x+y)/(x−y)) let y=ux ⇒(dy/dx) = u + x (du/dx) ⇒x (du/dx)+u= ((x(1+u))/(x(1−u))) ⇒x (du/dx)= ((1+u−u+u^2 )/(1−u)) ⇒x (du/dx)= ((1+u^2 )/(1−u)) ⇒∫ (((1−u))/(1+u^2 ))du = ∫ (dx/x) ⇒arctan u−(1/2)∫((d(u^2 +1))/(u^2 +1)) = ln ∣x∣ + c ⇒arctan ((y/x))=ln ∣Cx+(√(u^2 +1))∣ ⇒y=x tan (ln∣Cx+(√((y^2 +x^2 )/x^2 )) ∣) ⇒y=x tan(ln∣ ((Cx^2 +(√(y^2 +x^2 )))/x) ∣)

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

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

⇒y=x tan(ln ∣((Cx^2 +(√(y^2 +x^2 )))/x) ∣)

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

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

xdx+ydy=xdy−ydx ((d(x^2 +y^2 ))/(2dx))=x^2 ∙(d/dx)((y/x)) d(x^2 +y^2 )=2x^2 d((y/x)) ((d(x^2 +y^2 ))/(x^2 +y^2 ))=((2x^2 )/(x^2 +y^2 ))d((y/x)) ((d(x^2 +y^2 ))/(x^2 +y^2 ))=(2/(1+((y/x))^2 ))d((y/x)) ln(x^2 +y^2 )=2arctan((y/x))+C

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

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