I am wondering how I finish off the solution to the following 2nd Order differential equation:
$$3u_{tt} + 2u_{xt} - 8u_{xx} = \cos(x + t)$$
I know that I have to factorize it and get the following:
$$(3d_{t} - 4d_{x})(d_{t} + 2d_{x})u = cos(x + t)$$
Then I solve for the homogeneous solution:
$$(3d_{t} - 4d_{x})(d_{t} + 2d_{x})u = 0$$
By letting: $v = 3u_{t} - 4u_{x}$
I get the general solution as:
$$u(x, t) = f(x - 2t) + g(3x - 4t)$$
for the homogeneous part.
How do I find a solution that includes the inhomogeneous part of the equation?
Could somebody provide a step-by-step method as I am struggling with this.
Thanks!