How can I solve $\tiny\begin{cases} \dfrac{dx}{dt} = t^2 x^2 y\\ \dfrac{dy}{dt} = -t^2 y^2 x \end{cases}$? I have:
$$\begin{cases}
\dfrac{dx}{dt} = t^2 x^2 y\\
 \dfrac{dy}{dt} = -t^2 y^2 x
\end{cases}$$
How do I go about solving this system of equations?

I tried separation of variables however because of the extra $y$ term in the $\frac{dx}{dt}$ and $x$ term in $\frac{dy}{dt}$ I am unsure how to start.
 A: As suggested by Jack, you could consider eliminating $t$ from both equations leaving you with the following:
$$ t^2 = \frac{1}{x^2y}\frac{dx}{dt} = -\frac{1}{y^2x}\frac{dy}{dt}$$
$$ \frac{dy}{dx} = -\frac{y}{x}$$ which is now variable separable. 
A: $$\frac{dy}{dt} \frac{dt}{dx} = \frac{-t^2y^2x}{t^2x^2y}$$
$$ \frac{dy}{dx} = -\frac{y}{x}$$
A: Note that:
$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
Hence, you obtain the following differential equation:
$$\frac{dy}{dx}=-\frac{y}{x}$$
Now, separate the variables:
$$\int \frac{1}{y}~dy=-\int \frac{1}{x}~dx$$
From here on, it should be relatively straightforward.
A: From the two equations,
$$\frac{dx}x+\frac{dy}y=0,$$ so that $xy=3C$.
Then
$$\frac{dx}x=3Ct^2$$ and $$x=C'e^{Ct^3},y=C''e^{-Ct^3}$$ with $C'C''=3C$.
A: First observe that
$$
\frac{d}{dt}(xy)=x\frac{dy}{dt}+y\frac{dx}{dt}=0. 
$$
Thus $x(t)y(t)=c_1$, for some constant $c_1\in\mathbb R$. The system now becomes
$$\begin{cases}
\dfrac{dx}{dt} =  c_1t^2x\\
 \dfrac{dy}{dt} = -c_1t^2y.
\end{cases}$$
General solution of the system is hence
$$
x(t)=c_2\mathrm{e}^{c_1t^3/3},\quad y(t)=c_3\mathrm{e}^{-c_1t^3/3},
$$
where $c_2c_3=c_1$.
