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Given $3u_{tt} + 10u_{xt} + 3u_{xx} = \sin(x+t)$ find the general solution.

I have yet to solve any inhomogeneous second order PDE (or even first order ones at that). For homogeneous PDE of same order, I managed to solve them by factoring the operators and so forth. Being new to PDEs (self studying via Strauss PDE book) I lack the intuition to find a clever way of solving these, however from my experience with ODEs I reckon there is a way to solve these by first solving the associated homogeneous first by factoring operators and so forth and stuff.. but not finding much progress on incorporating the $\sin(x+t)$ term.

Any help & direction to solving this would be greatly appreciated.

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Did you try to change the coordinates? For example, write $u(t,x)=v(\phi(t,x))$ where $\phi$ is a linear bijective transformation such that $3u_{tt}+10u_{xt}+3u_{xx}$ is simpler. – Davide Giraudo Sep 30 '11 at 17:27
I tried but failed to find a suitable candidate for the change as the $sin(x+t)$ and the various partial terms were throwing me off (It was easily doable in first order but second order adds more to take into factor) – Room Sep 30 '11 at 17:54
Put $u(t,x)=v(at+bx,x)$. The partial derivatives are simpler. – Davide Giraudo Sep 30 '11 at 18:20

I'm sure you have some change of variables formulas in your PDE course, meant to simplify the given equation. I don't remember the formulas, but there is a way to avoid them. The equation can be written in the following way:

$$ A u = \sin(t+x) $$ where $A$ is the operator

$$ A= 3\frac{\partial^2}{\partial t^2}+10\frac{\partial^2}{\partial t \partial x} +3 \frac{\partial^2}{\partial x^2}.$$

In this case, the operator can be factored (just like a binomial expression) in the following way:

$$ A =\left( \frac{\partial }{\partial t}+3\frac{\partial}{\partial x}\right)\left(3\frac{\partial}{\partial t}+\frac{\partial }{\partial x} \right).$$

Now, change the variables such that each of the two factors will be a partial derivation.

$$ \frac{\partial}{\partial y}=\left( \frac{\partial }{\partial t}+3\frac{\partial}{\partial x}\right), \frac{\partial}{\partial z}=\left(3\frac{\partial}{\partial t}+\frac{\partial }{\partial x} \right)$$

For this, change the variables to $\displaystyle \begin{cases} t=y+3z \\ x=3y+z \end{cases}$ and define $w(y,z)=u(t,x)$. Then we have

$$\frac{\partial^2w }{\partial y \partial z}=Au= \sin(x+t)=\sin(4y+4z)$$

Integrate two times, with respect to $y,z$ and you will find $w$. From there it is easy to get to $u$.

This might not be the most efficient method (for an exam, for example), but it might show from where the change of variables formulas come from. I did this in my exam, once, because I didn't remember the formulas. (I got zero points, because my teacher didn't like the method and there were some small mistakes). It is best to learn all the cases for change of variable, because that gains you time in an exam, but is also important to know from where those change of variable formulas come from.

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