How to solve the following PDE for A and B?

Question

I have a PDE of the form $$ \frac{dA}{dt}- A(t) \frac{dB}{dt} x^2 - 2 \gamma x^2 A(t)B(t) - \sigma^2 A(t)B(t) + 2 \sigma^2 x^2 A(t) B(t)^2 = 0. $$ where $\sigma,\gamma$ are constants (I believe we can just set them to $1$ to simplify) and $A,B$ are functions of $t$. What methods can we use to solve for the forms of $A(t)$ and $B(t)$?

Answer 1

If the equation $$ \frac{dA}{dt}- A(t) \frac{dB}{dt} x^2 - 2 \gamma x^2 A(t)B(t) - \sigma^2 A(t)B(t) + 2 \sigma^2 x^2 A(t) B(t)^2 $$ holds for all $x$, then it holds for $x=0$ and therefore from one side $$ \frac{dA}{dt}= \sigma^2 A(t) B(t) $$ which gives $$ \frac{dA}{A}= \sigma^2 B(t)dt \Rightarrow\ln A=\sigma^2 \int B(t)dt\\\Rightarrow A(t)=\exp\left({\sigma^2 \int B(t)dt}\right) $$

From the other side $$\frac{dB}{dt} = 2 \gamma B(t) + 2 \sigma^2 B(t)^2 $$ so $$B(t)=-\frac{\gamma e^{C \gamma +2 \gamma t}}{\sigma ^2 e^{C \gamma +2 \gamma t}\pm 1}$$ where $C$ is the integration constant to be determined using initial condition for $B$.

Finally, integration gives $$A(t)=\exp\left({\sigma^2 \int B(t)dt}\right)= \widetilde C e^{ -\frac{1}{2}\ln \left(1\pm\sigma ^2 e^{C \gamma +2 \gamma t}\right) }= \frac{\widetilde C}{\sqrt{1\pm\sigma ^2 e^{c_1 \gamma +2 \gamma t}}} $$ where $\widetilde C $ is an integration constant, to be determined from initial condition for $A$.