ODE/PDE- eighenvalues question

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Homework Statement


X''+\lambda*X=0 0<x<2pi

X(0)=X(2pi) and X'(0)=X'(2pi)


Homework Equations





The Attempt at a Solution


My only problem is the case when \lambda=0
in such case the general solution is X(x)= C1+C2*x

Now, after applying the BCs, this is what I get:

C1=C2*2pi
and

C2=C2

now, what should C2 be ? zero or any value ?
if C2 is zero, we have the trivial solution only. However, if C2 is any number, we have a nontrivial solution.

how can I solve this question?


Thanks.
 
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After applying the first BC, I get C1=C1+C2*2*pi, not C1=C2*2*pi.
 
Roni1985 said:
C1=C2*2pi

This should be

C_1 = C_1 + 2\pi C_2.

How does the solution with \lambda\neq 0 behave in the limit as \lambda \rightarrow 0? Is this consistent with your solution for \lambda = 0?
 
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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