Solving this 2nd Order non-homogeneous PDE I am trying to solve the following equation:
$$3u_{xx} - 10u_{xt} - 3u_{tt} = \sin(x + t)$$
I know that the left hand side is a quadratic equation which I have to factorise. Then I let one of the factors equal to $v$ and solve the first order non-homogeneous PDE. But I don't know how to do this. 
I know the solution to this equation is:
$u(x, t) = f(3x - t) + g(x - 3t) + \frac{1}{16}\sin(x + t)$
Thanks for your help!
 A: Hints:


*

*Write the partial derivatives as operators, as follows:
$$
u_{xx} = \frac{\partial^2}{\partial x^2} u
$$
$$
u_{xt} = \frac{\partial^2}{\partial x\,\partial t} u
$$
$$
u_{tt} = \frac{\partial^2}{\partial t^2} u
$$
so the PDE looks like this:
$$
\left(\,3\frac{\partial^2}{\partial x^2} - 10\frac{\partial^2}{\partial x\,\partial t} - 3\frac{\partial^2}{\partial t^2}\,\right)u = \sin(x+t)
$$

*Factorise the differential operator being applied to u. In other words, you want to write the operator on the left as
$$
\left(a\frac{\partial}{\partial x} + b\frac{\partial}{\partial t}\right)
\left(c\frac{\partial}{\partial x} + d\frac{\partial}{\partial t}\right)
$$
for some well-chosen $a,b,c,d$, which are constants in this problem.

*Solve the factorised PDE, ignoring the so-called non-homogeneous part, i.e., ignoring the $\sin(x+t)$. This is because the general solution to a linear PDE is the sum of the general solution of the homogeneous equation and a particular solution of the full equation. (Read the previous sentence a few times to fully grasp what it's saying)

*Find a particular solution of the full equation. Hint: When you differentiate $\sin$ twice, you get $\sin$ again, up to a sign change...

*Complete the solution by adding the solution you found in 3. to the solution you found in 4.

