L}}{\delta u} \equiv v_t + x^2 v_{xx} -x^2(1 + 2 u)v_x + 2 (x + 2 x u - 1)v = 0.$$ Letting $v = \varphi(u)$ one obtains: $$\begin{aligned}
 \frac{\delta {\cal L}}{\delta u}\bigg|_{v = \varphi(u)}
 & = \varphi'(u) [u_t + x^2 u_{xx} - x^2 (1 + 2 u) u_x]\notag\\[1.5ex]
 & +
\varphi''(u)  x^2 u_x^2 + 2 (x + 2 x u - 1) \varphi(u).\notag
 \end{aligned}$$ Writing the quasi self-adjointness condition (\[qsa.2\]) in the form $$\frac{\delta {\cal L}}{\delta u}\bigg|_{v = \varphi(u)}
 = \lambda [- u_t + x^2 u_{xx} + (x^2 + 4 x + 2 x^2 u) u_x
 + 4 x (u + u^2)]$$ and comparing the coefficients for $u_t$ in both sides one obtains $\lambda = - \varphi'(u),$ so that the quasi self-adjointness condition takes the form $$\begin{aligned}
 & \varphi'(u) [u_t + x^2 u_{xx} - x^2 (1 + 2 u) u_x] +
\varphi''(u)  x^2 u_x^2 + 2 (x + 2 x u - 1) \varphi(u)\notag\\[1.5ex]
 & = \varphi'(u) [u_t - x^2 u_{xx} - (x^2 + 4 x + 2 x^2 u) u_x
 - 4 x (u + u^2)].\notag
 \end{aligned}$$ Comparing the coefficients for $u_{xx}$ in both sides we obtain $\varphi'(u) = 0.$ Then the above equation becomes $(x + 2 