Linear Second Order ODE/y'' + 2 b y' + a^2 y = 0/b less than a

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

has the general solution:
 * $y = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$

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

Its auxiliary equation is:
 * $(2): \quad: m^2 + 2 b m + a^2 = 0$

From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
 * $m_1 = - b + \sqrt {b^2 - a^2}$
 * $m_2 = - b - \sqrt {b^2 - a^2}$

As $b^2 < a^2$, this converts into:
 * $m_1 = - b + i \sqrt {a^2 - b^2}$
 * $m_2 = - b - i \sqrt {a^2 - b^2}$

These are complex and unequal.

So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
 * $y = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$