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I would like to solve the following system of two coupled nonlinear ODEs

$$\begin{aligned} \dot x &= \frac yx\\ \dot y &= \left( \frac{1-x}{x} \right) y \end{aligned}$$

What I did was the following. From the 1st ODE, assuming $x \neq 0$, I got $y = x \dot x$ and, hence,

$$y = \frac 12 \frac{\mathrm d}{\mathrm d t} \left(x^2 \right)$$

and, plugging into the 2nd ODE,

$$\frac{\mathrm d^2}{\mathrm d t^2} \left(x^2 \right) = \left( \frac{1-x}{x} \right) \frac{\mathrm d}{\mathrm d t} \left(x^2 \right)$$

which eventually yields the following nonlinear 2nd order ODE

$$x \ddot x + \dot x \left( \dot x + x - 1 \right) = 0$$

which Wolfram Alpha can solve. However, this looks messy. I am looking for cleaner solutions.

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2 Answers 2

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From the quotient of both equations you get $$ \frac{dy}{dx}=1-x $$ so that $y=C+x-\frac12x^2$ and then $$ \dot x=\frac{C}x+1-\frac12x $$ which can be solved as separable DE.

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  • $\begingroup$ Much cleaner, indeed. Thank you. $\endgroup$ Oct 13, 2019 at 17:58
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you can simplify more straight forwardly $$ \dot{y} = (1-x)\frac{y}{x} = (1-x)\dot{x} $$ so we have $$ \frac{d}{dt}\left(y - x + \frac{x^2}{2}\right) = 0 $$ or $$ y - x + \frac{x^2}{2} = A $$ then we have $$ y = A + x - \frac{x^2}{2} $$ you can solve from there - not pretty, but clearer.

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  • $\begingroup$ Very nice. Thank you. The other answer was posted a few seconds earlier, so I accepted that one. $\endgroup$ Oct 13, 2019 at 18:02

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