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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.

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    Have you tried cancelling out variable $t$? – 2017-01-07
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    @Jack how can i justify that? and how would i go about doing it – 2017-01-07
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    @RobertThompson The chain rule, perhaps? $\frac{dy}{dx} = \frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$. – 2017-01-07
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    @arthur good idea, thanks – 2017-01-07

5 Answers 5

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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 $c_2c_3=c_1$.

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    $x=e^{t^3}, y=2e^{-t^3}$ is allowed by your general solution, though it is not a solution of the original ODE. – 2017-01-07
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    I added a sentence: *where $c_2c_3=c_1$* – 2017-01-07
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    Great, you are coming closer to a clone of my answer. – 2017-01-07
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    Not really. In my answer I DO NOT divide by quantities which may vanish. – 2017-01-07
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    Instead, you multiply by quantities which may vanish. Not more rigorous. – 2017-01-07
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    Which quantity may vanish? – 2017-01-07
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    The same as mine. – 2017-01-07
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    There is nothing wrong with multiplication by zero. Division by zero is the problem – 2017-01-07
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    Stupid me, your are right. I forgot that $1=2\iff0\cdot1=0\cdot2$. – 2017-01-07
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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.

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$$\frac{dy}{dt} \frac{dt}{dx} = \frac{-t^2y^2x}{t^2x^2y}$$

$$ \frac{dy}{dx} = -\frac{y}{x}$$

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    Don't forget the minus sign. – 2017-01-07
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    @Rumplestillskin thanks :D – 2017-01-07
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    @ramanujan thank you, i can't believe i didn't think of this – 2017-01-07
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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.

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    thank you, i realise i was being silly in not noticing this – 2017-01-07
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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 $C'C''=3C$.

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    Why $\frac{dx}{x}=-\frac{dy}{y}$? – 2017-01-07
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    @Ramanujan: because $t^2xy\,dt=t^2xy\,dt$, maybe. – 2017-01-07
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    Are you sure you are allowed to divide by $x$ and by $y$? – 2017-01-07
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    @YiorgosS.Smyrlis: I am sure I am not when $x=0$ or $y=0$. – 2017-01-07