0
$\begingroup$

I have tried to find some known ODEs before posting on this forum, but I did not find anything about this kind of ODE:

$y'(x)^2 + a(x)*y(x)^2 = 1$

with $a\in C^∞(\mathbb{R},\mathbb{R})$ and $y\in C^∞(\mathbb{R},\mathbb{R})$

Can anyone give me some advices to find an analytical solution of this ODE and to find its existence and uniqueness?

Thanks in advance,

$\endgroup$
6
  • 1
    $\begingroup$ What do you mean by "analytical"? $\endgroup$ Commented Oct 3, 2014 at 13:16
  • $\begingroup$ Is there a general solution of this ODE? $\endgroup$
    – Linkis
    Commented Oct 3, 2014 at 13:49
  • $\begingroup$ The situation becomes more tangible if $a(x)$ itself obeys a differential equation (possibly in terms of $y$). Do you have any structural information on $a(x)$? $\endgroup$ Commented Oct 3, 2014 at 15:22
  • 1
    $\begingroup$ What is * in your ODE? Is this multiplication sign? $\endgroup$ Commented Oct 4, 2014 at 12:44
  • $\begingroup$ @Frits Veerman: we know a and it depends on y. However, if we want to express a with y, we will introduce a second unknow that we can not remove. I can write the expression of the function a, if you want. $\endgroup$
    – Linkis
    Commented Oct 6, 2014 at 8:07

2 Answers 2

4
$\begingroup$

The change $y=w(x\int\sqrt{a(x)}dx)$ reduces to $w'^2+w^2=1/a(x)$. This is entry 1.370 in Kamke. Kamke gives a further change of the variable that reduces this to the Abel equation of the form $v'=P_3(x,v),$ where $\deg_vP_3=3$. Abel's equation does not have any reasonable closed form solution (whatever you mean by the "closed form").

Reference: E. Kamke. Differentialgleichungen. Losungsmethoden und Losungen,I. Gewohnliche Differentialgleichungen. 6-th edition, Leipzig 1959.

$\endgroup$
1
$\begingroup$

I very much doubt that there is a closed-form general solution. Even in the simple case $a(x) = x$, Maple finds no closed-form solution and no symmetries.

$\endgroup$
1
  • $\begingroup$ Thanks for helping me! I would have checked for this simple case. $\endgroup$
    – Linkis
    Commented Oct 6, 2014 at 8:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .