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Is there any way to solve the following system of non-linear differential equations exactly?

$x'(t) = x\times(y - \frac{1}{3(t + C)})$

$y'(t) = -\frac{1}{3}x^2 - \frac{y}{t + C}$

Here $x$ and $y$ are functions of $t$, and $C$ is some constant.

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  • $\begingroup$ What is $z$ ? The fact that you use $x(t)$ on LHS and not on RHS is misleading. Must the parenthesis on the first RHS be read as evaluation of $x$ at some funky point, or is it just multiplication ? $\endgroup$ May 12, 2016 at 7:15
  • $\begingroup$ That $z$ was supposed t o be a $y$ and the parenthesis on the RHS was supposed to indicate multiplication. I can see why this was very unclear. Edited to fix. $\endgroup$ May 12, 2016 at 7:18

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By changing the origin of time, we may assume $C=0$. One special family of solutions is $x = 0, y = c/t$. But the general solution is

$$\eqalign{x(t) &= 12\,{\frac {a b^2{t}^{b-1}}{{a}^{2}{t}^{2\,b}+ 12\,{b}^{2}}} \cr y(t) &={\frac {144\,{b}^{5}- a^4 \left( b+2/3 \right) {t}^{4\,b}- 16\,a^2 b^2 {t}^{2\,b}-96\,{b}^{4}}{ \left( a^2 {t}^{2\,b}+ 12\,{b}^{2} \right) ^{2}t}} }$$

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    $\begingroup$ Perhaps a stupid question, but is there a source where the result is found, or is there a general method to apply here? $\endgroup$ May 12, 2016 at 9:08
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    $\begingroup$ I used Maple. Basically, I think it solves the first equation for $y$ and substitutes that into the second equation, obtaining a second-order nonlinear equation for $x$, then it uses symmetries to solve that. The equation for $x$ is a special case of the third Painlevé equation. $\endgroup$ May 12, 2016 at 17:03

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