Vtyjkyuk

Question: Find the solution to the differential equation $(2x^2+xy-2y^2)dx+(3x^2+2xy)dy=0$

The question prompts me to let $G=(2x^2+xy-2y^2)dx+(3x^2+2xy)dy$ and prove that $e^fracyxfracGx$ is an exact differential. But what is an exact differential anyways? Is it to multiply $e^fracyx$ into $fracGx$ and differentiate $e^fracyx(3x^2+2xy)$ by $x$ and $e^fracyx(2x^2+xy-2y^2)$ by $y$ and prove that both functions are the same?

I'm stuck on this question and would appreciate some help. Thanks!

The ODE is homogeneous ODE of order one. This is because the coefficients of $dx$ and $dy$ are both homogeneous two variables functions of the same order. I suggest you write the ODE as $$y'=frac2t^2-t-23+2t=f(t), ~~~(xneq 0, t=y/x)$$ and then solve the well-known ODE: $$fracdtf(t)-t=fracdxx$$ by seperation method!

Hint.

Note that

$$
fracpartialpartial yleft(frace^fracyx left(2 x^2+x y-2 y^2right)xright) = frace^fracyx left(3 x^2-3 x y-2 y^2right)x^2\
fracpartialpartial xleft(frace^fracyx left(3 x^2+2 x yright)xright) = frace^fracyx left(3 x^2-3 x y-2 y^2right)x^2
$$

Let
$$P=2x^2+xy-2y^2,quad Q=3x^2+2xy.$$
Then integrating factor is
$$mu=frac2xP+yQ=frac12x^2 y+x^3$$
Solution of differential equation
$$(2x^2+xy-2y^2)dx+(3x^2+2xy)dy=0$$
is
$$ln(x^2+2xy)+fracyx=C$$