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?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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