Answer to Question #259555 in Differential Equations for josh

Question

xydx(2x^2+3y^2-20)dy=0

Answer 1

$xydx+(2x^2+3y^2-20)dy=0$

integrating factor:

$\mu (y)=y^3$

$xy^4dx+(2x^2+3y^2-20)y^3dy=0$

$f(x,y)=\int y^4xdx=y^4x^2/2+g(y)$

$(y^4x^2/2+g(y))_y=2y^3x^2+dg(y)/dy$

$2y^3x^2+dg(y)/dy=(2x^2+3y^2-20)y^3$

$dg(y)/dy=(3y^2-20)y^3$

$g(y)=\int(3y^2-20)y^3dy=y^6/2-5y^4$

$f(x,y)=y^6/2-5y^4+y^4x^2/2$

$f(x,y)=c_1$

$y^6/2-5y^4+y^4x^2/2=c_1$