I want to solve this system \begin{align*}\tag{*} x'(s)=x^2(s)+y(s), y'(s)=x(s)y(s) \end{align*} with initial conditions $$x(0)=t, y(0)=t,$$ where $t\not=0.$

With the help of Maple, the solution is

$$ x \left( s \right) ={\frac {2\,st+2\,t}{-{s}^{2}t-2\,st+2}}, ~y \left( s \right) =2\,{\frac {t}{-{s}^{2}t-2\,st+2}}. $$

But I want to know the details of the whole process of solution.

I have tried to convert this system into a single ODE $$x''(s)=3x(s)x'(s)-x^3(s). \tag{**}$$ But this ODE is semilinear, which is beyond my capability to solve. Then I tried to deduced the order by transformation $$p=x'(s),~~ p\frac{dp}{dx}=x''(s),$$ which simplified $(**)$ to $$\frac{dp}{dx}=3x-\frac{x^3}{p},\tag{***}$$ which is also difficult for me to solve.

Can anyone help me to solve $(*)$ , $(**)$ or $(***)$ ?


1 Answer 1


Divide the two equations. You get $${dx\over dy}={x\over y}+{1\over x}.$$ Multiply by $x$ and set $x^2=u$. You get $${1\over 2}{du\over dy}={u\over y}+1,$$ a linear equation.

  • $\begingroup$ Of course, there is always the possibility that you are dividing by zero :( $\endgroup$
    – Igor Rivin
    Mar 19, 2014 at 19:00
  • 2
    $\begingroup$ The separate discussion of this special case is, as we usually say, left as an exercise to the reader. $\endgroup$ Mar 19, 2014 at 19:04
  • $\begingroup$ @Michael Renardy: Thanks a lot. Your answer is very helpful. $\endgroup$
    – azhi
    Mar 20, 2014 at 1:12
  • $\begingroup$ Sorry, after multiplying by $x$ won't you get ${x{dx\over dy}}={{x^2}\over y}+1$? How does the LHS convert to what you have in your second equation after setting $x^2 = u$? $\endgroup$ May 19, 2015 at 21:45
  • 1
    $\begingroup$ @MattPhillips by differentiating $x^2=u$ by $y$ you get $2x\frac{dx}{dy}=\frac{du}{dy}$ this is part of standard u-substitution $\endgroup$
    – Rick
    Jul 20, 2015 at 19:59

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