I came across this question in the book "One Thousand Exercises in Probability" by Grimmett and Stirzaker: it is a question number 5 in section 13.8.
My natural approach would be to apply Ito's Lemma to $F(X_t, t)=X_t e^{\beta t}$, yielding: $\frac{\partial F}{\partial t}=\beta X_t e^{\beta t}$, $\frac{\partial F}{\partial X_t}=e^{\beta t}$, $\frac{\partial^2 F}{\partial X_t^2}=0$, so that (we can assume that $X_0=0$):
$$ F(X_t, t)=X_te^{\beta t}=\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+ \frac{\partial F}{\partial X_t}*(-\beta X_t)+0 \right)dt+\int_{h=0}^{h=t}\frac{\partial F}{\partial X_t}dW_h=\\= \int_{h=0}^{h=t}\left(\beta X_h e^{\beta h}-\beta X_he^{\beta h} \right)dt+\int_{h=0}^{h=t}e^{\beta h}dW_h=\\=0+\int_{h=0}^{h=t}e^{\beta h}dW_h $$
So that the final solution is:
$$X_t= \int_{h=0}^{h=t}e^{\beta (h-t)}dW_h $$
The book gives a different solution, as follows: it first states that "We may assume that Gaussian white noise $G_t=\frac{dW_t}{dt}$ exists in sufficiently many senses to appear as an integrand". It then goes on to say:
"The question permits us to use $e^{\beta t}$ as an integrating factor, to give, formally:
$$e^{\beta t} X_t=\int_0^te^{\beta t} \frac{dW_s}{ds}ds=e^{\beta t}W_t-\beta \int_0^te^{\beta s}W_sds$$
on integrating by parts. This is the required result".
I suppose the two solutions should be equivalent, and so one could show that one equals the other.
Question 1: are the solutions equivalent?
Question 2: what would be the advantage of representing the solution as in the book, as opposed to the first solution presented above?
Observe that thanks to the integration by parts formula $$\int_0^t e^{\beta h}dW_h+\int_0^t W_hde^{\beta h}=e^{\beta t}W_t$$ but the second integral in LHS is $\beta\int_0^tW_h e^{\beta h} dh$ so in effect both representations agree.