Given some differential equation/s, is it always possible to construct a Lagrangian, which reproduces the equation/s after applying the Euler-Lagrange equation?
As a specific example, consider a complex function $E(t)$, where $t$ is real. Then the equation: $$\dfrac{\partial^{2}E}{\partial t^{2}}+AE=BE^{2}E^{*}\ ,$$ where $A$ and $B$ are constants, has a simple Lagrangian if $A$ and $B$ are real, namely: $$L=\dfrac{\partial E}{\partial t}\dfrac{\partial E^{*}}{\partial t}-AEE^{*}+\dfrac{B}{2}E^{2}(E^{*})^{2}\ ,$$ where we consider $E$ and $E^{*}$ as separate generalized coordinates. But, if $A$ and $B$ are complex, then I can't seem to find a Lagrangian?
If I understand the specifics of your question, you want $$\dfrac{\partial^{2}E}{\partial t^{2}}+AE-BE^{2}E^{*}=0\\ \dfrac{\partial^{2}E^*}{\partial t^{2}}+A^*E^*-B^*E^{*2}E=0 $$ to be the extremal conditions of the same action, even though they are the complex conjugates of each other.
That is, performing the trivial and irrelevant integration by parts with respect to t, you essentially want both equations to be $$ \frac{\delta L }{\delta E} =0 , \qquad \frac{\delta L }{\delta E^*} =0. $$ But, in order for these two to be mutual complex conjugates, you need L to be real, so
$$L=\dfrac{\partial E}{\partial t}\dfrac{\partial E^{*}}{\partial t}-AEE^{*}+\dfrac{B}{2}E^{2}(E^{*})^{2}\ ,$$ with real coefficients A,B.
The pair of equations you posit then can only be the E-L equations of a lagrangian if the coeffs A,B are real.