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

Theorem
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:

As $p^2 < 4 q$ we have that:
 * $\sqrt {q - \dfrac {p^2} 4} \ne 0$

and so:
 * $m_1 \ne m_2$

Let:

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:

are both particular solutions to $(1)$.

We can manipulate $y_a$ and $y_b$ into the following forms:

and:

Hence:

Let:

We have that:

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 $(1)$.