I am utterly confused by the following question:
When you use separation of variables, you get
$U=X(x)T(t)$
$\frac{T'}{T}=\nu \frac{X''}{X}$
This solves nicely to an exponential solution for T: $T(t)=C_\lambda e^{\lambda t}$ and using the suggested series solution for X:
$X''-\frac{\lambda}{\nu}X =0$
$\Sigma_{n=0}^\infty (a_{n+2}(n+1)(n+2)-a_n)x^n = 0$
which gives the form for the coefficients
$a_{2n+1}=\frac{(\lambda/\nu)^n a_1}{(2n+1)!}$
$a_{2n}=\frac{(\lambda/\nu)^n a_0}{(2n)!}$
When $\lambda=0$, you get $T=const$ and $X=bx+c$
The full solution is therefore (where I assume $\lambda$ takes on discrete values not yet found
$U(x,t)=bx+c + \Sigma_{\lambda \neq 0} e^{\lambda t}\Sigma_{n=0}^\infty \frac{(\lambda/\nu)^n a_0 x^{2n}}{(2n)!}+\frac{(\lambda/\nu)^n a_1 x^{2n+1}}{(2n+1)!}$
So from this I am very confused. It is impossible, if this form is correct, to have a power of $x^2$ in the initial condition without also having all even powers of x.
Would appreciate a hint, or if someone can see one of my assumptions/steps is incorrect.
You've tried to do an eigenvalue expansion but you haven't found the possible eigenvalues! The method goes like this: Suppose that $$T'/T = \nu X''/X = \lambda$$ then what can we deduce about $\lambda$? You should observe that the series solution you've obtained is just $X(x) = a_{0}\cos(\omega x) + a_{1}\sin(\omega x)$, where $\omega^{2} = \lambda / \nu$. Well, one of our boundary conditions is $U(0,t) = 0$ so that surely forces $a_{0}=0$. Then we have to have $U(L,t) = 0$ too. But arbitrary choices of $\lambda$ won't make this happen! You get a constraint on the eigenvalues here, and the general solution will be a linear combination of all possible eigensolutions. This gives you enough freedom to express initial conditions like your given one (but cross that bridge when you come to it).