Is there any way of solving the following second-order ODE $$y''+\frac{(y'+2ax)^2+4b^2}{2y}+\frac{10}{3}a=0,$$ where $a$ and $b$ are some constant? If we know that one solution exists, how would it help to possibly find another solution?

Any help will be appreciated.

  • $\begingroup$ OK - I apologize and remove my comment. $\endgroup$ – Dominic van der Zypen Jul 12 at 10:33
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    $\begingroup$ There is no reason to think there is a closed-form solution. $\endgroup$ – Gerald Edgar Jul 12 at 12:46

Here are two rather special solutions:

$$ \eqalign{y \left( x \right) &=-\frac{a{x}^{2}}{3}-{\frac {3{b}^{2}}{4a}}\cr y(x) &= -\frac{3a x^2}{2} - \frac{6 b^2}{a}\cr} $$

I don't think there is any way to use these to get more closed-form solutions.

Of course there are series solutions.

  • $\begingroup$ I really appreciate it. Could you please explain how do you obtain the solutions? $\endgroup$ – Masoud Jul 12 at 18:20
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    $\begingroup$ Guess there's a solution that's a quadratic and see what the coefficients have to be. $\endgroup$ – Robert Israel Jul 12 at 22:55

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