$y'' + \lambda\cdot y = 0$ with initial conditions $y'(0) = 0$ and $y(\pi) = 0$

Question

Here's the question:

$$y'' + \lambda \cdot y = 0 $$

Initial Conditions:

$$ y'(0) = 0$$ $$ y(\pi) = 0$$

Here's what I did:

---Write the characteristic equation--- $$m^2 + \lambda = 0$$

--- Solve for m --- $$m = \pm \lambda^{1/2} i$$

At which point it is my understanding when one has imaginary values, we get the following:

$$y = c_1\cos((\lambda)^{1/2}x) + c_2\sin((\lambda)^{1/2}x)$$

--- Differentiating---

$$y' = -c_1(\lambda)^{1/2}\sin((\lambda)^{1/2}x) + c_2(\lambda)^{1/2}\cos((\lambda)^{1/2}x)$$

--- Using initial condition $y'(0) = 0$ ---

$$ 0 = y'(0) = -c_1(\lambda)^{1/2}\sin(0) + c_2(\lambda)^{1/2}\cos(0)$$ $$ 0 = c_2(\lambda)^{1/2}$$ $$ 0 = c_2$$

---Using other initial condition $ y(\pi) = 0$ --- $$ 0 = y(\pi) = c_1\cos((\lambda)^{1/2}\pi) + 0\cdot \sin((\lambda)^{1/2}\pi)$$ $$ 0 = c_1\cos((\lambda)^{1/2}\pi)$$

We don't want $c_1$ to be equal to zero, therefore we need $\cos((\lambda)^{1/2}\pi)$ to equal zero.

$\cos(\theta) = 0$ when $\theta = n\pi$ for $n = 0, 1, 2, ...$

Therefore, set $(\lambda)^{1/2}\pi$ equal to $n\pi$. Solving for $\lambda$ we get $$\lambda = n^2$$

However, my answers say

$$\lambda = (1/2 + n)^2$$ for $ n = 0,1,2,...$

Where did I go wrong?

Answer 1

$\cos(\theta) = 0$ when $\theta = n\pi$ for $n = 0, 1, 2, \dots$

Oh really? ;-)