I have the following system of ordinary differential equations: \begin{align*} x'&=\frac{1}{4}x^2+\frac{3}{4}y^2-2x\\ y'&=\frac{1}{12}x^2+\frac{1}{4}y^2-\frac{2}{3}y, \end{align*} with boundary condition $x(0)=y(0)=1$. Note that $\frac{1}{2}x'-\frac{3}{2}y'=y-x$.

After some googling I found that this is an Autonomous Algebraic Ordinary Differential Equation (AODE), and there exists a large body of literature for finding exact solution for these type of ODEs. I was wondering if someone could point me in the right direction for solving this specific AODEs.

  • $\begingroup$ Indeed, thank you for noticing I adapted it. $\endgroup$ – Darkwizie Oct 5 '18 at 12:15
  • $\begingroup$ I hope it does! $\endgroup$ – Darkwizie Oct 5 '18 at 12:17
  • $\begingroup$ I am mainly interested in which techniques COULD be applied to find a solution, I can try to apply these methods myself but when I google myself I am overwhelmed by overly complicated methods and have no idea which of them could work well. $\endgroup$ – Darkwizie Oct 5 '18 at 12:24

You found that $$ x=3y+Ce^{-t}. $$ Inserting into the second equation results in $$ y'=y^2+\frac12Ce^{-t}y-y+\frac1{12}C^2e^{-2t} $$ This is now a Riccati equation where you can use the standard substitution $y=-\frac{u'}{u}$ to obtain a second order linear ODE.

  • $\begingroup$ There were still some minor issues with the question, I have now rectified them, I am taking a close look at your answer. $\endgroup$ – Darkwizie Oct 5 '18 at 14:39
  • $\begingroup$ Ok, now there are no immediate simplifications. Already this case with simplifications did not lead far, the more general case should not be easier. Except that one can compute the power series expansion for algebraic ODE to any degree I see no directly algebraic solution path. $\endgroup$ – LutzL Oct 5 '18 at 14:45
  • $\begingroup$ Do you have a (basic) reference for the power series expansion approach? $\endgroup$ – Darkwizie Oct 5 '18 at 14:59

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