Laplace Transform of Squared Function

Question

I have the following differential equation that I am trying to solve: $$g(u,v,t) = a + bg(u,v,t) - c\frac{dg(u,v,t)}{dt} - k\times g(u,v,t)\frac{\partial(g(u,v,t))}{\partial t},$$ where $a,b,c,k$ are constants, and the variables $u$ and $v$ are independent of $t$.

I replaced $g(u,v,t)\frac{\partial(g(u,v,t))}{\partial t}$ with $\frac{1}{2}\frac{d(g(u,v,t))^2}{dt}$ in the last term on the RHS.

My initial idea was to compute the Laplace transform on both sides: $$G(u,v,s) = \frac{a}{s} + bG(u,v,s) - csG(u,v,s) - k/2\times s\mathcal{L}(g^2(u,v,t)), $$ where $s$ is the Laplace transform variable and $G(s)$ is the Laplace transform of $g(u,v,t)$. But now I cannot solve for G because of the $\mathcal{L}(g^2)$ term.
I am stuck now and will appreciate any ideas to proceed further.

Answer 1

Laplace transform generally doesn't help with nonlinear equations.

Since $u$ and $v$ don't appear anywhere except in $g(u,v,t)$, you might as well get rid of them and just write $g(t)$.

This is a separable differential equation.

Answer 2

This equation can be rewritten in the form $$ \frac{dg}{dt}=\frac{(b-1)g+a}{kg+c}. $$ This is a first order differential equation that can be solved by quadratures. Can you go on from here?