Obtaining the general solution of a parabolic PDE

Question

I am trying to obtain the canonical form of this PDE:
$$(1+\sin(x))u_{xx} + 2\cos(x)u_{xy} + (1- \sin(x))u_{yy} – u_y – \cos^2(x) = 0 $$
Since the discriminant is equal to zero, the euqation is a parabolic equation.
We have to find two functions $\zeta(x,y)$ and $\eta(x,y)$. Since the equation is parabolic and the equation of the characteristics is:
$$\frac{dy}{dx}= \frac{\cos(x)}{1+\sin(x)}$$
I found one of them as:
$$\eta(x,y)= y-\ln(1+\sin(x))$$
and I chose the other one to be :
$$\zeta(x,y) = x$$
It is easy to check that:
$$J= \zeta_x\eta_y – \zeta_y\eta_x = 1$$Then I took $v(\zeta(x,y),\eta(x,y)) = u(x,y)$
After doing calculations I found the below equation:
$$(1+\sin(x))v_{\zeta\zeta} – \cos^2 (x) =0$$
How can I find a general solution for the original PDE?

Answer 1

Since $ζ=x$: $$v_{ζζ}=\frac{\cos^2ζ}{1+\sinζ}=1-\sinζ$$ $$v_ζ=ζ+\cos ζ+F(η)$$ $$v=\frac{ζ^2}{2}+\sinζ+ζF(η)+G(η)$$ $$u=\frac{x^2}{2}+\sin x+xF(y−\ln(1+\sin x))+G(y−\ln(1+\sin x))$$