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I've got a ordinary differential equation basically on this form: $$ \frac{(f'(x))^2 x^2}{\sinh^2\left(2\,q\,f(x)\right)}+K^2\sinh^2\left(2\,q\,f(x)\right)=\frac{a}{q^2} $$ where $q>0$, $f(x)>0 $, and everything here is real. I would like a solution valid for all $x$, not a particular $x$. Does someone have a clue as to how I can proceed with this? Thanks for any help!

Best regards, Jakob

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2 Answers 2

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Denote $u(x)=\sinh(2\,q\,f(x))\;$. Then $$ (u'(x))^2=4q^2(f'(x))^2\cosh^2(2\,q\,f(x))=4q^2(f'(x))^2(1-u^2(x)) $$ and the equation can be rewritten as $$ \frac{x^2u'^2}{4 q^2 u^2 (1-u^2)}+K^2 u^2=\frac{a}{q^2}. $$ The solution of which in the implicit form is $$ \int\frac{du}{u \sqrt{(1-u^2)(a-K^2 q^2 u^2)}}=\pm2\log|x|+C. $$

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  • $\begingroup$ after that, just for the joke, please, post solution to this equation which gives Mathematica 8 $\endgroup$
    – Norbert
    Jun 26, 2012 at 20:07
  • $\begingroup$ @Norbert It can be done only in implicit form. I edited the answer. $\endgroup$
    – Andrew
    Jun 26, 2012 at 20:11
  • $\begingroup$ @Norbert well, Mathematica do calculate the integral, but cannot get $u$ explicitly. $\endgroup$
    – Andrew
    Jun 26, 2012 at 20:17
  • $\begingroup$ I know that. Semi-explicit formula was horribly huge. That's why I asked you to post it :) $\endgroup$
    – Norbert
    Jun 26, 2012 at 20:23
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If you want explicit solutions, Maple 16 gives

f(x) = -1/8*(-2*ln(-a^(1/2)+2*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))*(q^2*K^2*a^(3/2)*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))+tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))*a^(5/2)-(q^2*K^2*a^4*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2+tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2*a^5-a^4*q^2*K^2-a^5)^(1/2))/a/(q^2*K^2*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2+a))+ln(a))/q,

f(x) = 1/8*(2*ln(-a^(1/2)+2*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))*(q^2*K^2*a^(3/2)*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))+tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))*a^(5/2)-(q^2*K^2*a^4*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2+tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2*a^5-a^4*q^2*K^2-a^5)^(1/2))/a/(q^2*K^2*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2+a))-ln(a))/q,

f(x) = -1/8*(-2*ln(-a^(1/2)+2*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))*(q^2*K^2*a^(3/2)*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))+tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))*a^(5/2)+(q^2*K^2*a^4*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2+tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2*a^5-a^4*q^2*K^2-a^5)^(1/2))/a/(q^2*K^2*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2+a))+ln(a))/q,

f(x) = 1/8*(2*ln(-a^(1/2)+2*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))*(q^2*K^2*a^(3/2)*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))+tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))*a^(5/2)+(q^2*K^2*a^4*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2+tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2*a^5-a^4*q^2*K^2-a^5)^(1/2))/a/(q^2*K^2*tanh(-2*ln(x)a^(1/2)+2_C1*a^(1/2))^2+a))-ln(a))/q

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