Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation
Theorem
Let:
- $(1): \quad y + p y' + q y = 0$
be a constant coefficient homogeneous linear second order ODE.
Let $m_1$ and $m_2$ be the roots of the auxiliary equation $m^2 + p m + q = 0$.
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:
\(\ds m_1\) | \(=\) | \(\ds -\frac p 2 + i \sqrt {q - \frac {p^2} 4}\) | ||||||||||||
\(\ds m_2\) | \(=\) | \(\ds -\frac p 2 - i \sqrt {q - \frac {p^2} 4}\) |
As $p^2 < 4 q$ we have that:
- $\sqrt {q - \dfrac {p^2} 4} \ne 0$
and so:
- $m_1 \ne m_2$
Let:
\(\ds m_1\) | \(=\) | \(\ds a + i b\) | ||||||||||||
\(\ds m_2\) | \(=\) | \(\ds a - i b\) |
where $a = -\dfrac p 2$ and $b = \sqrt {q - \dfrac {p^2} 4}$.
\(\ds y_a\) | \(=\) | \(\ds e^{m_1 x}\) | ||||||||||||
\(\ds y_b\) | \(=\) | \(\ds e^{m_2 x}\) |
are both particular solutions to $(1)$.
We can manipulate $y_a$ and $y_b$ into the following forms:
\(\ds y_a\) | \(=\) | \(\ds e^{m_1 x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{\paren {a + i b} x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{a x} e^{i b x}\) | ||||||||||||
\(\text {(3)}: \quad\) | \(\ds \) | \(=\) | \(\ds e^{a x} \paren {\cos b x + i \sin b x}\) | Euler's Formula |
and:
\(\ds y_b\) | \(=\) | \(\ds e^{m_2 x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{\paren {a - i b} x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{a x} e^{-i b x}\) | ||||||||||||
\(\text {(4)}: \quad\) | \(\ds \) | \(=\) | \(\ds e^{a x} \paren {\cos b x - i \sin b x}\) | Euler's Formula: Corollary |
Hence:
\(\ds y_a + y_b\) | \(=\) | \(\ds e^{a x} \paren {\cos b x + i \sin b x} + e^{a x} \paren {\cos b x - i \sin b x}\) | adding $(3)$ and $(4)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 e^{a x} \cos b x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {y_a + y_b} 2\) | \(=\) | \(\ds e^{a x} \cos b x\) |
\(\ds y_b - y_a\) | \(=\) | \(\ds e^{a x} \paren {\cos b x - i \sin b x} - e^{a x} \paren {\cos b x + i \sin b x}\) | subtracting $(4)$ from $(3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 e^{a x} \sin b x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {y_b - y_a} 2\) | \(=\) | \(\ds e^{a x} \sin b x\) |
Let:
\(\ds y_1\) | \(=\) | \(\ds \frac {y_a + y_b} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{a x} \cos b x\) | ||||||||||||
\(\ds y_2\) | \(=\) | \(\ds \frac {y_b - y_a} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{a x} \sin b x\) |
We have that:
\(\ds \frac {y_1} {y_2}\) | \(=\) | \(\ds \frac {e^{a x} \cos b x} {e^{a x} \sin b x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cot b x\) |
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 $(1)$.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.17$: The Homogeneous Equation with Constant Coefficients