Question: Given,

$U\_t-U\_{xx}-2U\_x=0$

Using method of separation of variables, find ALL possible solution.

My answer :

$U(x,t)=X(x)T(t)$

$T'(t)/T(t)$=$[X''(x)+2X'(x)]/X(x)$=$\lambda$

From here I'm stuck. Could someone show me some working steps for me to proceed further.

Answer: 
The simplest example is probably the polynomial $X^8+1$, and now I’ll explain why.

It’s a fact that every finite abelian extension of $\mathbb Q$ is contained in a cyclotomic extension $\mathbb Q(\zeta\_m)$, $\zeta\_m$ being a primitive $m$-th root of unity. The Galois group of $\mathbb Q(\zeta\_m)$ over $\mathbb Q$ is the group of units of the ring $\mathbb Z/m\mathbb Z$. With this information, you can get any finite abelian group as the quotient of a $(\mathbb Z/m\mathbb Z)^\times$, and so an extension of $\mathbb Q$ with that Galois group.

Now, $\mathbb Z/16\mathbb Z$ has for its units the odd numbers modulo $16$, as a multiplicative group, and you easily check that this has the shape $\mathbb Z/2\mathbb Z\times\mathbb Z/4\mathbb Z$, generators being $-1$ and $5$. The polynomial for the primitive sixteenth roots of unity is $(X^{16}-1)/(X^8-1)=X^8+1$, and there you have it.
Actually, the simplest example is $\Bbb Q(\zeta\_{15})$, since cyclotomic extensions are well-known.

A quick calculation shows $\operatorname{Gal}(\Bbb Q(\zeta\_{15})/\Bbb Q) = (\Bbb Z/15\Bbb Z)^\times = (\Bbb Z/3\Bbb Z)^\times \times (\Bbb Z/5\Bbb Z)^\times = C\_2 \times C\_4$.