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I want to find the eigenvalues and normalized eigenfunctions of the problem $$-y'' = \lambda y, y'(0) = y(1) = 0. $$

By solving $r^2 + \lambda = 0$ I found the general solution $y(x) = c_1\cos(\sqrt{\lambda} x) + c_2 \sin(\sqrt{\lambda} x)$, and using that $y'(0) = 0$, I found that we must have $c_2 = 0$. Using $y(1) = 0$, I obtained the eigenvalues $\lambda_n = \frac{(2n+1)^2 \pi}{4}$.

Since $c_2=0$, we have $y(t) = c_1 \cos(\sqrt{\lambda} x)$ and that made me believe that the normalized eigenfunctions were $u_n(x) = \cos \left(\frac{(2n+1)\pi}{2}x\right),$ but apparently the correct answer is $u_n(x) = \sqrt{2} \cos \left(\frac{(2n+1)\pi}{2}x \right)$. Where did the $\sqrt{2}$ come from?

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Normalized eigenfunctions require

$$ \int_0^1 (u_n(x))^2 dx = 1 $$

To find the constant, note that the general solution of the eigenvalue problem is

$$ u_n(x) = c_n \cos\left(\frac{2n+1}{2}\pi x\right) $$

The normalization condition gives

$$ \int_0^1 c_n^2 \cos^2\left(\frac{2n+1}{2}\pi x\right)\ dx = 1 $$

You can solve the above integral to show that this condition is equivalent to $\dfrac{c_n^2}{2} = 1$

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It's there to normalize their inner product: $$<u_n,u_m>=\int_0^1 u_n u_m =\delta_{nm}$$ It's the same when you deal with "nice enough" matrices: you want the eigenvectors to form an ortonormal basis, i.e to have that $$v_n \cdot v_m = \delta_{nm}$$

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