Solving a differential equation related to quasihomogeneous polynomials I have these equations: $$\frac{dx}{dt}=ax^3+by^2 \quad \text{and} \quad \frac{dy}{dt}=x^2y$$
I need to somehow find an expression for $y(x)$, and I have been given a hint that I can use the fact that these correspond to "quasihomogeneous" polynomials. I understand that this means that the degree of the polynomials $P(x,y):=ax^3+by^2$ and $Q(x,y):=x^2y$ have some special relaltion.
Once checked this relation,  to obtain an expression is straightforward, but again I am not sure how to do this. Sorry for the vagueness of this explanation but I am confused and I can not find any references or examples.
If you have some other method to deal with the system I will be glad to learn them as well.
Thank you!
 A: Note that
$$\frac{dy}{dx}=\frac{x^2 y}{ax^3+by^2}. \tag{1}$$
@DogeChan's comment tells you how I would typically solve such a differential equation (as an exact equation). However, I present below the approach which I think your professor wanted you to use based on the context you provided. Let
$$f(x,y)=\frac{x^2 y}{ax^3+by^2}.$$
We call the ODE $y'(x)=f(x,y)$ quasihomogeneous if for all $\lambda$ and for some $\alpha,\beta\neq 0$ we have
$$f(\lambda^{\alpha}x,\lambda^{\beta} y)=\lambda^{\beta-\alpha}f(x,y).$$
Once you find such $\alpha$ and $\beta$, make the change of variables $y=x^{\beta/\alpha}z$ to get a separable ODE.

Alternatively, you can derive the change of variables by computing the degree of the rational function
$$f(x,y)=\frac{Q(x,y)}{P(x,y)}=\frac{x^2 y}{ax^3+by^2}.$$
in $x$ and $y$. The degree of $f$ in $x$ and $y$ (denote by $\deg_x f$ and $\deg_y f$ respectively) is by convention given by the maximum of the degrees of $Q$ and $P$ in $x$ and $y$ respectively, where $Q$ and $P$ have no common factors. The change of variable is then given by
$$y=x^{\deg_y f/\deg_x f} z.$$
So in your case $\deg_x f=\max\{\deg_x Q,\deg_x P\}=\max\{2,3\}=3$ and $\deg_y f=\max\{\deg_y Q,\deg_y P\}=\max\{1,2\}=2$, therefore the appropriate change of variable is $y=x^{3/2}z$.
For more information on this method, see e.g. this answer and this answer.
A: The hint of the term 'quasihomogeneous' suggests that one should at least investigate the substitution
$u=x^{3/2}$ so that the first equation becomes $\dfrac{dx}{dt}=au^2+by^2$.
Completing the change of variables gives
$$ \frac{du}{dt}=\frac{3a}{2}u^{7/3}+\frac{3b}{2}u^{1/3}y^2\text{ and }\frac{dy}{dt}=u^{4/3}y $$
Therefore
$$ \frac{du}{dy}=\frac{3a}{2}\cdot\frac{u}{y}+\frac{3b}{2}\cdot\frac{y}{u} $$
which is homogeneous.
