Solution to the differential equation $(2x^2+xy-2y^2)dx+(3x^2+2xy)dy=0$

Question

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 $e^\frac{y}{x}\frac{G}{x}$ is an exact differential. But what is an exact differential anyways? Is it to multiply $e^\frac{y}{x}$ into $\frac{G}{x}$ and differentiate $e^\frac{y}{x}(3x^2+2xy)$ by $x$ and $e^\frac{y}{x}(2x^2+xy-2y^2)$ by $y$ and prove that both functions are the same?

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

Answer 1

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 $$y'=\frac{2t^2-t-2}{3+2t}=f(t), ~~~(x\neq 0, t=y/x)$$ and then solve the well-known ODE: $$\frac{dt}{f(t)-t}=\frac{dx}{x}$$ by seperation method!

Answer 2

Hint.

Note that

$$ \frac{\partial}{\partial y}\left(\frac{e^{\frac{y}{x}} \left(2 x^2+x y-2 y^2\right)}{x}\right) = \frac{e^{\frac{y}{x}} \left(3 x^2-3 x y-2 y^2\right)}{x^2}\\ \frac{\partial}{\partial x}\left(\frac{e^{\frac{y}{x}} \left(3 x^2+2 x y\right)}{x}\right) = \frac{e^{\frac{y}{x}} \left(3 x^2-3 x y-2 y^2\right)}{x^2} $$

Answer 3

Let $$P=2x^2+xy-2y^2,\quad Q=3x^2+2xy.$$ Then integrating factor is $$\mu=\frac{2}{xP+yQ}=\frac{1}{2x^2 y+{{x}^{3}}}$$ Solution of differential equation $$(2x^2+xy-2y^2)dx+(3x^2+2xy)dy=0$$ is $$\ln(x^2+2xy)+\frac{y}{x}=C$$