Solving this 2nd Order non-homogeneous PDE

Question

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!

Answer 1

Hints:

1. 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: $$ \left(\,3\frac{\partial^2}{\partial x^2} - 10\frac{\partial^2}{\partial x\,\partial t} - 3\frac{\partial^2}{\partial t^2}\,\right)u = \sin(x+t) $$
2. 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.
3. 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)
4. Find a particular solution of the full equation. Hint: When you differentiate $\sin$ twice, you get $\sin$ again, up to a sign change...
5. Complete the solution by adding the solution you found in 3. to the solution you found in 4.