Heat in a Rod Fundamental Solution

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Homework Statement



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The Attempt at a Solution


So I know that I must have boundary conditions u(0,t) = 0 and ux(L,t) = 0. My textbook recommends reducing the given boundary conditions to homogeneous ones by subtracting the steady state solution. But, I thought these were already homogenous boundary conditions (are they?). Is my steady state condition v''(x) = 0, but, then by the boundary conditions I know that this must be a trivial v. Am I thinking about this incorrectly?
 
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So I thought I might have to write something of the form:
Assume the solution can be written u(x,t) = X(x)T(t). Thus, by the heat equation u_t = a^2 u_{xx}, we wind up with two linear differential equations. Namely, X&#039;&#039; + qX = 0 and T&#039; + a^2 q T = 0. Now I have to find which values of q make q an eigenvalue of the eigenfunction. We test three cases: q = 0; q > 0; q < 0.

q = 0:
We must have that X = C_1 x + C_2, but by the boundary conditions, this forces both of the arbitrary constants to be zero.
q < 0:
X = C_1 sinh(\sqrt{-q}x) + C_2 cosh(\sqrt{-q}x), which implies that C_2 = 0, and, when one takes the derivative, we also have C_1 = 0 by the assumption that q was nonzero.

q > 0:
X = C_1 cos(\sqrt{q}x) + C_2 sin(\sqrt{q}x), which implies that C_1 = 0, and, again, when one takes the derivative, we force the other constant to zero. This means that I am only getting trivial solutions here. What other approaches can I try? Or have I done something wrong?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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