x(x-3)dy+ydx=0

Question

diferantial equation

Answer 1

Separation of variables.

x(x-3)dy+ydx=0,

x(x-3)dy=-ydx,

dy/y=-dx/(x(x-3)).

If -1/(x(x-3))=A/x+B/(x-3), then:

-1=A(x-3)+Bx=Ax-3A+Bx, so:

-3A=-1, A=⅓.

And A+B=0, so B=-⅓.

dy/y=⅓(dx/x-dx/(x-3)).

Integrate both sides:

ln|y|=⅓(ln|x|-ln(x-3))+C=ln|∛(x/(x-3))+C.

Constant C can be incorporated into the log if C=ln(a).

So y=a ∛(x/(x-3)), or y³=bx/(x-3) where b=a³, a constant.