Linear Second Order ODE/y'' + 4 y = 0

Theorem

The second order ODE:

$(1): \quad y' ' + 4 y = 0$

has the general solution:

$y = C_1 \cos 2 x + C_2 \sin 2 x$

Proof 1

It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.

Its auxiliary equation is:

$(2): \quad: m^2 + 4 = 0$

From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:

$m_1 = 2 i$

$m_2 = -2 i$

These are complex and unequal.

So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:

$y = C_1 \cos 2 x + C_2 \sin 2 x$

$\blacksquare$

Proof 2

This is an instance of:

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

which yields:

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

where $k = 2$.

Hence the result.

$\blacksquare$