Solve the hyperbolic equation no homogeneous
$\frac{\partial^2u}{\partial x^2}-6\frac{\partial^2u}{\partial x\partial y}+8\frac{\partial^2u}{\partial y^2}=1$
My attempt:
As we have a hyperbolic equation by hypothesis then we have two family of characteristic.
I found that family of characteristic
$w_2(x,y)=y-4x$
$w_1(x,y)=y-2x$
Here i'm stuck. Can someone help me?
Hint: First, we solve the homogeneous problem. Attempt to "factor" the differential operator $\frac{\partial^2u}{\partial x^2}-6\frac{\partial^2u}{\partial x\partial y}+8\frac{\partial^2u}{\partial y^2}=0$ into: $(\frac{\partial}{\partial x}-4\frac{\partial}{\partial y})(\frac{\partial }{\partial x}-2\frac{\partial }{\partial y})u(x,y)=0$. As we know, the transport equation of canonical form $\frac{\partial u}{\partial x}-c\frac{\partial u}{\partial y} = 0 $ has general solution $u(x,y) = \phi(y+cx)$, for a smooth enough function $\phi$. (This can be re-derived using method of characteristics.) So, this amounts to solving two linear PDEs: $\frac{\partial u}{\partial x}-4\frac{\partial u}{\partial y}=0$ and $\frac{\partial u}{\partial x}-2\frac{\partial u}{\partial y}=0$. These have solutions $u_1(x,y) = \phi(y+4x)$ and $u_2(x,y) = \psi(y+2x)$. Since the equation is linear, we can use the principle of superposition to say that the sum is the solution, i.e. $u(x,y) = u_1(x,y) +u_2(x,y) = \phi(y+4x)+\psi(y+2x)$. This solves the homogenous problem. To solve the inhomogeneous problem, we have at least a couple options. We can use what is known as "Duhamel's principle". However, since $f(x,y)=1$, it is straightforward to "guess" what the inhomogeneous solution will be, it should be $\frac{x^2}{2}$. Thus, the solution to the inhomogeneous problem should be: $u(x,y) =\phi(y+4x)+\psi(y+2x)+\frac{x^2}{2}$, for $\phi, \psi$ smooth enough.