Theorem: Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation

{{:Solution of Constant Coefficient Homogeneous LSOODE}}
Let $p^2 < 4 q$.
Then $(1)$ has the general solution:
:$y = e^{a x} \paren {C_1 \cos b x + C_2 \sin b x}$
where:
:$m_1 = a + i b$
:$m_2 = a - i b$

Proof:
Consider the auxiliary equation of $(1)$:
:$(2): \quad m^2 + p m + q$
Let $p^2 < 4 q$.
From Solution to Quadratic Equation with Real Coefficients, $(2)$ has two complex roots:
{{begin-eqn}}
{{eqn | l = m_1
      | r = -\frac p 2 + i \sqrt {q - \frac {p^2} 4}
}}
{{eqn | l = m_2
      | r = -\frac p 2 - i \sqrt {q - \frac {p^2} 4}
}}
{{end-eqn}}
As $p^2 < 4 q$ we have that:
:$\sqrt {q - \dfrac {p^2} 4} \ne 0$
and so:
:$m_1 \ne m_2$
Let:
{{begin-eqn}}
{{eqn | l = m_1
      | r = a + i b
}}
{{eqn | l = m_2
      | r = a - i b
}}
{{end-eqn}}
where $a = -\dfrac p 2$ and $b = \sqrt {q - \dfrac {p^2} 4}$.
From Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation:
{{begin-eqn}}
{{eqn | l = y_a
      | r = e^{m_1 x}
}}
{{eqn | l = y_b
      | r = e^{m_2 x}
}}
{{end-eqn}}
are both particular solutions to $(1)$.
We can manipulate $y_a$ and $y_b$ into the following forms:
{{begin-eqn}}
{{eqn | l = y_a
      | r = e^{m_1 x}
      | c = 
}}
{{eqn | r = e^{\paren {a + i b} x}
      | c = 
}}
{{eqn | r = e^{a x} e^{i b x}
      | c = 
}}
{{eqn | n = 3
      | r = e^{a x} \paren {\cos b x + i \sin b x}
      | c = Euler's Formula
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = y_b
      | r = e^{m_2 x}
      | c = 
}}
{{eqn | r = e^{\paren {a - i b} x}
      | c = 
}}
{{eqn | r = e^{a x} e^{-i b x}
      | c = 
}}
{{eqn | n = 4
      | r = e^{a x} \paren {\cos b x - i \sin b x}
      | c = Euler's Formula: Corollary
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = y_a + y_b
      | r = e^{a x} \paren {\cos b x + i \sin b x} + e^{a x} \paren {\cos b x - i \sin b x}
      | c = adding $(3)$ and $(4)$
}}
{{eqn | r = 2 e^{a x} \cos b x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {y_a + y_b} 2
      | r = e^{a x} \cos b x
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = y_b - y_a
      | r = e^{a x} \paren {\cos b x - i \sin b x} - e^{a x} \paren {\cos b x + i \sin b x}
      | c = subtracting $(4)$ from $(3)$
}}
{{eqn | r = 2 e^{a x} \sin b x
      | c = {{explain| isn't this $-2 i e^{a x} \sin b x$? Where did the i go?}}
}}
{{eqn | ll= \leadsto
      | l = \frac {y_b - y_a} 2
      | r = e^{a x} \sin b x
      | c = 
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = y_1
      | r = \frac {y_a + y_b} 2
}}
{{eqn | r = e^{a x} \cos b x
      | c = 
}}
{{eqn | l = y_2
      | r = \frac {y_b - y_a} 2
}}
{{eqn | r = e^{a x} \sin b x
      | c = 
}}
{{end-eqn}}
We have that:
{{begin-eqn}}
{{eqn | l = \frac {y_1} {y_2}
      | r = \frac {e^{a x} \cos b x} {e^{a x} \sin b x}
}}
{{eqn | r = \cot b x
}}
{{end-eqn}}
As $\cot b x$ is not zero for all $x$, $y_1$ and $y_2$ are linearly independent.
From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE:
:$y_1 = \dfrac {y_a + y_b} 2$
:$y_2 = \dfrac {y_b - y_b} 2$
are both particular solutions to $(1)$.
It follows from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution that:
:$y = C_1 e^{a x} \cos b x + C_2 e^{a x} \sin b x$
or:
:$y = e^{a x} \paren {C_1 \cos b x + C_2 \sin b x}$
is the general solution to $(1)$.