Linear Second Order ODE/y'' + k^2 y = 0

Theorem
The second order ODE:
 * $(1): \quad y'' + k^2 y = 0$

has the solution:
 * $y = A \sin \left({k x + B}\right)$

or can be expressed as:
 * $y = C_1 \sin k x + C_2 \cos k x$

Proof
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:


 * $p \dfrac {\mathrm d p} {\mathrm d y} = -k^2 y = 0$

where $p = \dfrac {\mathrm d y} {\mathrm d x}$.

From:
 * First Order ODE: $y \, \mathrm d y = k x \, \mathrm d x$

this has the solution:


 * $p^2 = -k^2 y^2 + C$

or:
 * $p^2 + k^2 y^2 = C$

As the is the sum of squares, $C$ has to be positive for this to have any solutions.

Thus, let $C = \alpha^2$.

Then:

From Multiple of Sine plus Multiple of Cosine: Sine Form, this can be expressed as:


 * $y = C_1 \sin k x + C_2 \cos k x$