On how to solve the ODE $v'-\frac{1}{v}=x$...

Question

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.

Answer 1

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 $y$.

Answer 2

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 :

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