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=f(\zeta) \end{equation} Is there a change of variables that allows the differential equation to be transformed into the following form? In the above equation c is a constant, while $j=\sqrt{-1}$. \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}=f(\zeta) \end{equation} If this variable change 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. As pointed out by Sal the change of variables for the independent variable $\zeta$ cannot help us to solve the problem.
As an example we consider the change of variables $a(\zeta)=y(\zeta)^{-2}$. This change of variables is not correct in fact the second equation becomes: \begin{equation} \frac{y''(\zeta)}{y(\zeta)}-\frac{2 y'(\zeta)^2}{y(\zeta)^2}+c y(\zeta)^4=f(\zeta) \end{equation} which is different from the first equation although two terms are common to both equations.