We have \( g(0) = g_0 \). The trial solution must fulfill this condition to be a proper solution of (16).
A possible way to ensure that \( g_t(0, P) = g_0 \), is to let \( F(N(x,P)) = x\cdot N(x,P) \) and \( A(x) = g_0 \). This gives the following trial solution: $$ \begin{equation} g_t(x, P) = g_0 + x \cdot N(x, P). \tag{17} \end{equation} $$