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I wonder if the ODE

$y''+e^{y}=a$

can be solved explicitly. For $a=0$, it is well-known that there is a two-parameter family of explicit solutions

$y=\ln(2)-2\ln(\cosh(cx+d))+2\ln(c)$, $c,d \in R$. Are there explicit solutions for $a>0$?

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  • $\begingroup$ What is $u$? Is $y=u$? $\endgroup$ – Alexandre Eremenko Mar 12 '16 at 5:14
  • $\begingroup$ Yes, I revised the question. $\endgroup$ – User4966 Mar 12 '16 at 8:15
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Multiply by $y'$, we get $((y')^2/2+e^y-ay)'=0$, so $(y')^2/2+e^y-ay=c$, $y'=f(y)$, where $f(y)=\pm \sqrt{2c+2ay-2e^y}$, so $dx/dy=1/f(y)$, $x$ is antiderivative of $1/f(y)$. I doubt that this antiderivative is expressed in elementary functions for general $a,c$.

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