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Note that: The solution must be non zero
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Answer 1

Explanation:

Answer 2

`y'' + \omega^2 y = 0`

has characteristic equation

`r^2 + \omega^2 = 0`

with roots at
`r = \pm√(-\omega^2) = \pm|\omega|i`, hence the characteristic solution is

`y = C_1 e^(i|\omega|x) + C_2 e^(-i|\omega|x)`

or equivalently,

`y = C_1 \cos(|\omega|x) + C_2 \sin(|\omega|x)`

With the given boundary conditions, we require

`y(0) = 0 \implies C_1 = 0`

and

`y'(\pi) = 0 \implies -|\omega| C_1 \sin(|\omega|\pi) + |\omega| C_2 \cos(|\omega|\pi) = 0`

With
`C_1=0`, the second condition reduces to

`|\omega| C_2 \cos(|\omega|\pi) = 0`

Assuming
`C_2\\eq0` (because we don't want the trivial solution
`y=0`), it follows that

`\cos(|\omega|\pi) = 0 \implies |\omega|\pi = \pm\frac\pi2 + 2n\pi \implies |\omega| = 2n\pm\frac12`

where
`n` is an integer. In order to ensure
`|\omega|\ge0`, we must have
`n\ge1` if
`|\omega|=2n-\frac12`, or
`n\ge0` if
`|\omega|=2n+\frac12`.