I have a problem with solution this hyperbolic equation:
$$u_{xx}-u_{xy}-6u_{yy}=y\cos(x).$$
I found two characteristics: $$C_1=y-2x$$ $$C_2=y+3x.$$
I know that $u(x,y)=F(y-2x)+G(y+3x)-\sin(x)-y\cos(x)$ but I don't understand the part with $-\sin(x)-y\cos(x)$. Do you know how to calculate a particular solution ?
As
$$ u_{xx}-u_{xy}-6u_{yy}=y\cos(x)\equiv (\partial_x+2\partial_y)(\partial_x-3\partial_y)u = y\cos x $$
this PDE can be solved making
$$ \cases{ (\partial_x-3\partial_y)u = v\\ (\partial_x+2\partial_y)v = y\cos x } $$
A good method to solve this system of PDEs is the method of characteristics applied first to $(\partial_x+2\partial_y)v = y\cos x$ and then to $(\partial_x-3\partial_y)u = v$.
Regarding your question as how to determine a particular solution, one way is to apply a guessing method. Having in mind that the PDEs are linear with constant coefficients, the task is relatively easy. For instance proposing $u_p = (a_1 x+ b_1 y + c_1)\sin x + (a_2 x+ b_2 y + c_2)\cos x$ as a possible generic particular solution, after substitution into $(\partial_x+2\partial_y)v - y\cos x = 0$ we quickly verify that for $a_1=c_1=a_2=b_2=0, b_1 = 1, c_2 = 2$, $u_p = 2\cos x+y\sin x$ can be used as a particular solution. The same procedure can be used with $(\partial_x-3\partial_y)u = v$