let $y(x)$ be a continuous function $0\le y(x) \le1$ over the support $x_\min\le x \le x_\max$, for $x_\min < x_\max$ strictly positive values.

let $y'(x)$ be the first order derivative of $y(x)$.

let $y(x)^k$ be the $k$-th power of the function $y(x)$.

Solve the Cauchy problem:

$$a(x)y'(x) = b(x)y(x)^{-3} - cy(x)^{-2} - y(x)^{-1}$$

$$y(x_0) = y_0$$

where $a(x)=a_1 + a_2x$ with $a_1>0, a_2<0$ and $b(x)=b_1 + b_2x$ with $b_1<0, b_2>0$

and $x_0$, $y_0$, $c$ are positive constant real values.

I am interested in a solution such that $y'(x)<0$. I proved existence and uniqueness, but I am looking for the analytic expression of $y(x)$.

  • $\begingroup$ I guess this won't help, but what do you mean by "particular"? $\endgroup$ – Maesumi Jul 11 '13 at 13:46
  • $\begingroup$ the particular solution, so out of the general solution, the only trajectory that satisfies the boundary condition $\endgroup$ – user85884 Jul 11 '13 at 14:02
  • $\begingroup$ I can prove the existence and uniqueness of a solution y(x) such that y'(x)>0. But I want to find the analytical expression for the solution (if any) $\endgroup$ – user85884 Jul 11 '13 at 14:04
  • $\begingroup$ Nonlinear equation with two generic functions involved... I would not expect any explicit solutions, unless $a$ and $b$ are known functions, and pretty special at that. $\endgroup$ – 40 votes Jul 11 '13 at 23:41





This belongs to an Abel equation of the second kind.

Let $u=x-\dfrac{y^2}{b_2}-\dfrac{cy}{b_2}+\dfrac{b_1}{b_2}$ ,

Then $x=u+\dfrac{y^2}{b_2}+\dfrac{cy}{b_2}-\dfrac{b_1}{b_2}$


$\therefore u\left(\dfrac{du}{dy}+\dfrac{2y}{b_2}+\dfrac{c}{b_2}\right)=\dfrac{a_2y^3}{b_2}\left(u+\dfrac{y^2}{b_2}+\dfrac{cy}{b_2}-\dfrac{b_1}{b_2}\right)+\dfrac{a_1y^3}{b_2}$




My academic interest are PDEs and ODEs. Personally, I'd like to say that there would be explicit solutions for such nonlinear ODE in the first place.

  • $\begingroup$ This does not answer the question. It is more appropriate as a comment. $\endgroup$ – Daryl Jul 13 '13 at 10:21
  • $\begingroup$ thanks, this does not answer the question tough and I am still looking for $\endgroup$ – user85884 Jul 13 '13 at 11:40

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