Sturm-Liouville, find normalized eigenfunction

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?

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 $$\dfrac{c_n^2}{2} = 1$$

It's there to normalize their inner product: $$=\int_0^1 u_n u_m =\delta_{nm}$$ 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 $$v_n \cdot v_m = \delta_{nm}$$