How do I approach this question?

I thought solving by change of variables means that we need to find new variables that are functions of x and t?
Am I meant to use the form u=e^bt * g(x) to find what b and g(x) is by partially differentiate and plug into the equation? In that cue, I got up to b^2 g(x) + 2b g'(x) + g''(x) = 0, which I don't know how to proceed further.
Any help is appreciated.

Solve the second order ODE, should be prerequisite.

choose an ansatz and solve the auxiliary equation.

i solved AE and made u = (e^-x + xe^-x) (e^-t + xe^-t) and I almost solved it, could you please give me a bit more hint?(I omitted constants etc)

How did you arrive to that? What are the initial conditions?? (since you are missing arbitrary constants, but you say you omitted them? -- I just mean it's simpler to see if you're correct if you keep them in to show us, given that there are no initial conds.)

Note that the AE is m^2 + 2\beta m + \beta^2 = 0 \implies m = - \beta \implies g(x) =

i think it's safe to say that my initial working was totally wrong and no there wasn't any initial conditions.
I just thought that x and t play the same role in this question so whatever we have for x, we need to have for t right?
I don't really understand how you arrived to that Auxiliary equation if I am honest.
is the final answer e^beta t times by your g(x)?

$x,t$ are both independent variables here, but they are not the same and 'duplicating' one expression in terms of the other variable seems unjustified (if that's what you did)

Have you not covered the theory behind solving 2nd order ODE's?

I refer you to this post where I cover the main basics and ideas: https://www.thestudentroom.co.uk/showthread.php?t=5147458#post75629378

In your case, your 2nd order ODE is $\dfrac{d^2 g}{dx^2} + 2 \beta \dfrac{dg}{dx} + \beta^2 = 0$ and you are solving for $g(x)$.

yeah I covered 2nd ordinary differential equations before but i am not entire sure how it can be used in pde though

$u_{xx} = \dfrac{\partial^2}{\partial x^2}(e^{\beta t} g(x)) = e^{\beta t}\dfrac{\partial^2}{\partial x^2} g(x)$

But since $g(x)$ is a function of $x$ only, you can just rewrite it as $e^{\beta t}\dfrac{d^2}{dx^2} g(x)$.

So you end up with no 'partial' derivatives overall.

I think i finally got it now, thank you very much!
On a side note, because I am somewhat curious, if we neglect the form the question gave us, $u=e^{-\beta x}e^{-\beta t}(A+Bt)$ would work too?

$u(x,t) = e^{\beta x} e^{-\beta t}(A+Bt)$ would indeed work as well, there is nice symmetry about.