This question was previously posted on MSE at Change of variable for differential equations.

Given the following differential equation \begin{equation} -y(\zeta) \left(\frac{d^2 y(\zeta)^{-1}}{d \zeta^2}+2 j c \frac{d y(\zeta)^{-1}}{d \zeta}\right)+c^2 (1+y(\zeta)^{2})=\\ \frac{y''(\zeta)}{y(\zeta)}-\frac{2 y'(\zeta)^2}{y(\zeta)^2}+\frac{2 c j y'(\zeta)}{y(\zeta)}+c^2 y(\zeta)^2+c^2=\\ \frac{d^{2}}{d \zeta^{2}} \log{y(\zeta)}-\left(\frac{d}{d \zeta} \log{y(\zeta)}\right)^{2}+2jc \frac{d}{d \zeta} \log{y(\zeta)}+c^2 y(\zeta)^2+c^2 =f(\zeta), \end{equation} where $c$ is a constant, while $j=\sqrt{-1}$, is there a change of variables that allows it to be transformed into the following form? \begin{equation} -\frac{1}{2 a(\zeta)}\left(\frac{d^2 a(\zeta)}{d \zeta^2}-\frac{1}{2 a(\zeta)}\left(\frac{d a(\zeta)}{d \zeta}\right)^2 \right)+\frac{c}{a(\zeta)^2}=\\ -\frac{d^{2}}{d \zeta^{2}} \log{\sqrt{a(\zeta)}}-\left(\frac{d}{d \zeta} \log{\sqrt{a(\zeta)}}\right)^{2}+\frac{c_{1}}{a(\zeta )^2}=\\ \frac{d^{2}}{d \zeta^{2}} \log{\frac{1}{\sqrt{a(\zeta)}}}-\left(\frac{d}{d \zeta} \log{\frac{1}{\sqrt{a(\zeta)}}}\right)^{2}+\frac{c_{1}}{a(\zeta)^2}= f(\zeta). \end{equation} If this change of variables exists, how is it possible to find it?

A first attempt is to use a generic change of variables to identify the function $F$ such that $a(\zeta)=F(y(\zeta))$. Proceeding in this way it is possible to transform the second equation in the following way \begin{equation} -y''(\zeta)\frac{F'(y(\zeta))}{2 F(y(\zeta))}+y'(\zeta)^2 \left(\frac{F'(y(\zeta))^2}{4 F(y(\zeta))^2}-\frac{F''(y(\zeta))}{2 F(y(\zeta))}\right)+\frac{c}{F(y(\zeta))^2}=f(\zeta). \end{equation} The initial problem then reduces to identifying the function $F$ such that the first and the third equations are equal. However, as of this point, I have no idea how to proceed. Also, I am not sure that changing variables $a(\zeta)=F(y(\zeta))$ is enough to solve the problem. The math.stackexchange user Sal pointed out that the equation involving $F$ has no $y'$ term, while the first equation does. Since $F$ is presumed to be only a function of $y$ they can't be equal. For instance we could have $F=F(y,y')$, but also $F=F(\zeta,y)$, $F=F(\zeta,y,y')$. For instance for $F=F(\zeta,y)$, I have \begin{equation} -\frac{y''(\zeta) F^{(0,1)}(\zeta,y(\zeta))}{2 F(\zeta,y(\zeta))}+y'(\zeta)^2 \left(\frac{F^{(0,1)}(\zeta,y(\zeta))^2}{4 F(\zeta,y(\zeta))^2}-\frac{F^{(0,2)}(\zeta,y(\zeta))}{2 F(\zeta,y(\zeta))}\right)+y'(\zeta) \left(\frac{F^{(0,1)}(\zeta,y(\zeta)) F^{(1,0)}(\zeta,y(\zeta))}{2 F(\zeta,y(\zeta))^2}-\frac{F^{(1,1)}(\zeta,y(\zeta))}{F(\zeta,y(\zeta))}\right)+\frac{F^{(1,0)}(\zeta,y(\zeta))^2}{4 F(\zeta,y(\zeta))^2}-\frac{F^{(2,0)}(\zeta ,y(\zeta))}{2 F(\zeta,y(\zeta))}+\frac{c}{F(\zeta,y(\zeta))^2}=f(\zeta). \end{equation} Assuming that $F=F(\zeta,y)$ is the right change of variable, what is the next step?

I'm really interested in solving this problem, so if anything is unclear, please don't hesitate to let me know so that I can improve the post. Any suggestion is welcome.

If the general solution of this problem is difficult to determine, if it exists at all, it is probably possible to determine a particular solution. For example if $y=\frac{1}{\sqrt{a(\zeta)}}$ the first and second equations are satisfied if \begin{equation} 2jc \frac{d}{d \zeta} \log{y(\zeta)}+c^2 y(\zeta)^2+c^2 =c_{1} y(\zeta)^4 \end{equation} Is there a way to find an analytical solution of this equation?