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

xdx+ydy=xdy−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) ∣)

Commented byAr Brandon last updated on 28/Jul/21

Thanks for your method

Commented byAr Brandon last updated on 28/Jul/21

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

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