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I am trying to solve the following differential equation:
\[ \frac{1}{\sqrt{1+y}} \frac{dx}{dy} - \frac{2\sqrt{1+y}}{x} = 2(x+5). \]

After performing the substitution:
\[ p = \sqrt{1+y}, \quad y = p^2 - 1, \quad dy = 2p , dp, \]
the equation transforms into:
\[ \frac{dx}{dp} = 4p^2\left(\frac{p}{x} + x + 5\right). \]

I attempted the following approaches, but they didn’t work:

  1. Separation of variables: The equation mixes $x$ and $p$ in a way that prevents easy separation.
  2. Substitutions: I tried substituting $u = x^2$ to simplify the nonlinear terms, but the resulting equation was still complex: \[ \frac{du}{dp} = \frac{4p^3}{u} + 4p^2 + 20p^2\sqrt{u}. \] However, I couldn't find a way to proceed further.

I suspect that the solution might involve more advanced techniques, possibly using special functions or another substitution that I haven’t considered yet.

Could someone guide me on how to approach this equation or suggest resources for solving nonlinear differential equations of this type?

Any help or hints would be greatly appreciated!

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  • $\begingroup$ After trying for a minute or so, Mathematica cannot do anything with this differential equation. So, most likely, it cannot be solved in reasonable form. $\endgroup$
    – Iosif Pinelis
    Commented Nov 19 at 19:13
  • $\begingroup$ I am sure you can find some special function solution, if it has been studied before. Perhaps looking at something in modular forms and functions, sometimes these sort of ODEs pop up there like Weirestrass P-function. $\endgroup$
    – Alan
    Commented Nov 19 at 20:00
  • $\begingroup$ Where did you encounter this equation? $\endgroup$
    – LSpice
    Commented Nov 20 at 0:20

1 Answer 1

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Well, here's something I thought right now, so it might be wrong (or obviously wrong).

You have $$dx/dp=4p^2(p/x+x+5)=4p^2h(p,x)$$ Let's assume we want to have such a function $h(p,x):=h(p/x)$. Then change variables: $p/x=z$, differentiate: $dz=(xdp-pdx)/x^2$.

Now how to proceed? Not sure... Just an idea; flimsy I know.

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  • $\begingroup$ Ah yes, there's this method of $N(x,y)dx+M(x,y)dy=dF$ which is learnt in undergraduate, but I can't remember how this method conditues. $\endgroup$
    – Alan
    Commented Nov 19 at 20:10
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    $\begingroup$ Since this is not an answer, it might have been better to post it as a comment. $\endgroup$
    – LSpice
    Commented Nov 20 at 0:21

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