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I've been having trouble finding the general solution of $v$ for $v'-\frac{1}{v}=x$. I've attempted various substitutions in attempts of obtaining separation of variables or recognizable form to apply the method of the integrating factor. A couple of substitutions I've attempted: $$\alpha=\frac{1}{v}$$ $$\beta=\frac{1}{\alpha^2}$$ I tried others but threw out the scratch paper (yeah...would've helped now to see the other substitutions that didn't work so I don't reattempt them...). Does anyone have an idea of how to get this DE into a form we can play with? Please don't supply the final to the DE. I'm simply looking for direction down the proper path.

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    $\begingroup$ Are you sure that this is the ODE to solve ? If it is, it looks like a monster (at least to me). $\endgroup$ – Claude Leibovici Feb 27 '16 at 8:11
  • $\begingroup$ What you probably want is: Abel differential equation of the second kind, with g(x) = 1, and f(x) = x. eqworld.ipmnet.ru/en/solutions/ode/ode0125.pdf $\endgroup$ – Paichu Feb 27 '16 at 8:42
  • $\begingroup$ @Claude Leibovici: Anyway, this still be a very lucky case of Abel equation of the second kind. $\endgroup$ – doraemonpaul Mar 5 '16 at 15:20
  • $\begingroup$ Sorry for such a late response. That was indeed the ODE and the Airy equation was indeed the method to solve it... -_- $\endgroup$ – user146925 Mar 20 '16 at 1:21
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Making the change of independent function $v=f+\frac{x^2}{2}$, we get $$f'_x=\frac{1}{f+\frac{x^2}{2}}$$ Next interpret this as an equation for $x$ as a function of $f$: $$x'_f=\frac{x^2}{2}+f.$$ This is a Riccati equation solvable in terms of Airy functions: indeed, setting $x(f)=-2y'/y$, we get Airy equation $2y''+f y=0$ for $y$.

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  • $\begingroup$ Thank you!!! This helped out tremendously. I apologize for taking so long to thank you and mark this as the correct answer. $\endgroup$ – user146925 Mar 20 '16 at 1:22
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A change of function leads to an Riccati ODE. The usual method of solving leads to an ODE of Bessel kind. Finally the general solution is expressed on parametric form :

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While I was typing my answer, "Start wearing purple" posted his answer which is based on the same method.

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