For real coefficients $a,b$ solve the PDE $$v^3v_x+2(ax+by)v_y=bv, \qquad v(0,y)=f(y).$$
I set a characteristic space $(y_0,s)$
$x(s=0)=0$
$y(s=0)=y_0$
$v(s=0)=f(y_0)$
And write out the characteristic equations:
$dx/ds=v^3\implies x=v^3s$;
$dy/ds=2ax+2by\implies y=y_0e^{2bs}-ax/b$;
$dv/ds=bv\implies v=f(y_0)e^{bs}$;
Now i can find $s$ and $y_0$: $s=x/v^3$, $y_0=(y+ax/b)e^{-2bx/v^3}$.
And i get that $v=f[(y+ax/b)e^{-2bx/v^3}]e^{bx/v^3}$, which is not solvable, because $v$ is a function of an arbitrary function which contains $v$ itself. How do i find the general solution then? Or is this expresseion still considered to be the general solution , even though $v$ is a function of itself?
$$v^3v_x+2(ax+by)v_y=bv$$ Your equations written on an equivalent form : $$\frac{dx}{v^3}=\frac{dy}{2(ax+by)}=\frac{dv}{bv}=ds$$ A first characteristic equation comes from solving $\frac{dx}{v^3}=\frac{dv}{bv}$ : $$v^3-3bx=c_1$$ A second characteristic equation comes from solving $\frac{dx}{v^3}=\frac{dy}{2(ax+by)}$
With $v^3=c_1+3bx\quad;\quad \frac{dx}{c_1+3bx}=\frac{dy}{2(ax+by)}\quad;\quad \frac{dy}{dx}=\frac{2(ax+by)}{c_1+3bx}$
This is a first order linear ODE. Solving it leads to : $y=\frac{a}{b^2}(c_1+2bx)+c_2(c_1+3bx)^{2/3}$
$\frac{y-\frac{a}{b^2}(c_1+2bx)}{(c_1+3bx)^{2/3}}=c_2\quad$ and with $c_1=v^3-3bx\quad$ we get $\quad \frac{y-\frac{a}{b^2}(v^3-bx)}{(v^3)^{2/3}}=c_2$ $$\frac{1}{v^2}\left(y-\frac{a}{b^2}(v^3-bx) \right)=c_2$$
The general solution of the PDE on the form of implicit equation $c_2=\Phi(c_1)$ is : $$\boxed{\frac{1}{v^2}\left(y-\frac{a}{b^2}(v^3-bx) \right)=\Phi\left(v^3-3bx\right)}$$ $\Phi$ is an arbitrary function, to be determined according to the boundary condition: $v(0,y)=f(y)$
$\frac{1}{f(y)^2}\left(y-\frac{a}{b^2}(f(y)^3) \right)=\Phi\left(f(y)^3\right)$
Let $f(y)^3=X\quad;\quad f(y)=X^{1/3}\quad;\quad y=f^{-1}\left(X^{1/3}\right)$
$f^{-1}$ means the inverse function. $$\Phi(X)=\frac{1}{X^{2/3}}\left(f^{-1}\left(X^{1/3}\right)-\frac{a}{b^2}X \right)$$ Now the function $\Phi(x)$ is determined. We put it into the above general solution where $X=(v^3-3bx)$ : $$\frac{1}{v^2}\left(y-\frac{a}{b^2}(v^3-bx) \right)=\frac{1}{(v^3-3bx)^{2/3}}\left(f^{-1}\left((v^3-3bx)^{1/3}\right)-\frac{a}{b^2}(v^3-3bx)\right)$$ This is the solution on the form of implicit equation. Honestly I am not sure that there is no mistake in the calculus because the result appears too complicated for an academic exercise.