I have the following system of differential equations:

$$ \begin{align*} \dot x &= -2y\\ \dot y x-\dot x y &= 1+y^2 \end{align*} $$

with the usual initial conditions $x(0)=x_0$ and $y(0)=y_0$. I don't know how to solve that, but when I plug it into Mathematica, it manages to give me an answer in terms of elementary functions. One of the solutions is:

$$ \begin{align*} x(t) &= \frac{t^2 y_0^4-2 t^2 y_0^2+t^2+2 t x_0 \left(y_0^2-1\right) y_0+x_0^2 y_0^2-x_0^2}{x_0 \left(y_0^2-1\right)} \\ y(t) &= \sqrt{\frac{t^2 y_0^4-2 t^2 y_0^2+t^2+2 t x_0 \left(y_0^2-1\right) y_0+x_0^2 y_0^2}{x_0^2}} \end{align*} $$

I don't believe Mathematica has access to black voodoo magic, so what's the trick here?


Plug first equation value of $\dot x $ into second and separate the variables

$$ \int \frac{dy}{1-y^2} = \frac{dx}{x} $$

which can be solved in terms of elementary functions.


The above treats $ \dot y = \dfrac {dy}{dx} $. But treating as $t$ time independent variable $ \dot y=y′(x) \dot x, $ it becomes :

$$ \int \frac{2\, y\,dy}{y^2-1} = \int \frac{dx}{x} $$

as suggested by mickep below.

  • $\begingroup$ I think you forgot that it is $\dot y=\frac{dy}{dt}$ in the left-hand side of the second equation, and not $y'(x)$. Using the chain rule, one can of course write $\dot y=y'(x)\dot x$. Then, the differential equation becomes $$\frac{2y}{y^2-1}\,dy=\frac{1}{x}\,dx.$$ $\endgroup$ – mickep Oct 20 '15 at 16:02
  • $\begingroup$ Yes thanks, that was a hint which basically can tackle $y$, shall convert to answer. $\endgroup$ – Narasimham Oct 21 '15 at 1:18

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