Theorem: Canonical Form of Underdamped Oscillatory System

Consider a physical system $S$ whose behaviour can be described with the second order ODE in the form:
:$(1): \quad \dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
for $a, b \in \R_{>0}$.
Let $b < a$, so as to make $S$ underdamped.
Then the value of $x$ can be expressed in the form:
:$x = \dfrac {x_0 a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
:$\theta = \map \arctan {\dfrac b \alpha}$
This can be referred to as the '''canonical form''' of the solution of $(1)$.

Proof:
From Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation, the general solution of $(1)$ is:
:$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
This is a homogeneous linear second order ODE with constant coefficients.
Let $m_1$ and $m_2$ be the roots of the auxiliary equation:
:$m^2 + 2 b + a^2 = 0$
From Solution to Quadratic Equation with Real Coefficients:
{{begin-eqn}}
{{eqn | l = m_1
      | r = -b + i \sqrt {a^2 - b^2}
      | c = 
}}
{{eqn | l = m_1, m_2
      | r = -b - i \sqrt {a^2 - b^2}
      | c = 
}}
{{end-eqn}}
So from Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation:
:$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
The following assumptions are made:
:We may label a particular point in time $t = 0$ at which the derivative of $x$ {{WRT|Differentiation}} $t$ is itself zero.
:We allow that at this arbitrary $t = 0$, the value of $x$ is assigned the value $x = x_0$.
This corresponds, for example, with a physical system in which the moving body is pulled from its equilibrium position and released from stationary at time zero.
Differentiating $(1)$ {{WRT|Differentiation}} $t$ gives:
:$\quad x' = -b e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + e^{-b t} \paren {-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}$
Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:
{{begin-eqn}}
{{eqn | l = x_0
      | r = e^0 \paren {C_1 \cos 0 + C_2 \sin 0}
      | c = 
}}
{{eqn | r = C_1
      | c = 
}}
{{end-eqn}}
Setting the initial condition $x' = 0$ when $t = 0$:
{{begin-eqn}}
{{eqn | l = 0
      | r = -b e^0 \paren {C_1 \cos 0 + C_2 \sin 0} + e^0 \paren {-\alpha C_1 \sin 0 + \alpha C_2 \cos 0}
      | c = 
}}
{{eqn | r = -b C_1 + \alpha C_2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = C_2
      | r = \frac {b C_1} \alpha
      | c = 
}}
{{eqn | r = \frac {b x_0} \alpha
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = x
      | r = e^{-b t} \paren {x_0 \cos \alpha t + \frac {b x_0} \alpha \sin \alpha t}
      | c = 
}}
{{eqn | n = 2
      | r = e^{-b t} \dfrac {x_0} \alpha \paren {\alpha \cos \alpha t + b \sin \alpha t}
      | c = 
}}
{{end-eqn}}
From Multiple of Sine plus Multiple of Cosine: Cosine Form, $(2)$ can be expressed as:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac {x_0} \alpha e^{-b t} \paren {\sqrt {\alpha^2 + b^2} \map \cos {\alpha t + \arctan \dfrac {-b} \alpha} }
      | c = 
}}
{{eqn | r = \dfrac {x_0 \sqrt {\sqrt{a^2 - b^2}^2 + b^2} } \alpha e^{-b t} \map \cos {\alpha t + \arctan \dfrac {-b} \alpha}
      | c = 
}}
{{eqn | r = \dfrac {x_0 \sqrt {a^2 - b^2 + b^2} } \alpha e^{-b t} \map \cos {\alpha t + \arctan \dfrac {-b} \alpha}
      | c = 
}}
{{eqn | r = \dfrac {x_0 a} \alpha e^{-b t} \cos {\alpha t + \arctan \dfrac {-b} \alpha}
      | c = 
}}
{{eqn | r = \dfrac {x_0 a} \alpha e^{-b t} \cos {\alpha t - \arctan \dfrac b \alpha}
      | c = Tangent Function is Odd
}}
{{eqn | r = \dfrac {x_0 a} \alpha e^{-b t} \cos {\alpha t - \theta}
      | c = where $\theta = \arctan \dfrac b \alpha$
}}
{{end-eqn}}
{{qed}}