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

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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:

$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$


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 real roots:

\(\ds m_1\) \(=\) \(\ds -\frac p 2 + \sqrt {\frac {p^2} 4 - q}\)
\(\ds m_2\) \(=\) \(\ds -\frac p 2 - \sqrt {\frac {p^2} 4 - q}\)

As $p^2 > 4 q$ we have that:

$\sqrt {\dfrac {p^2} 4 - q} \ne 0$

and so:

$m_1 \ne m_2$


From Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation:

\(\ds y_1\) \(=\) \(\ds e^{m_1 x}\)
\(\ds y_2\) \(=\) \(\ds e^{m_2 x}\)

are both particular solutions to $(1)$.


We also have that:

\(\ds \frac {y_1} {y_2}\) \(=\) \(\ds \frac {e^{m_1 x} } {e^{m_2 x} }\)
\(\ds \) \(=\) \(\ds e^{\paren {m_1 - m_2} x}\)
\(\ds \) \(\ne\) \(\ds 0\) as $m_1 \ne m_2$

Thus $y_1$ and $y_2$ are linearly independent.


It follows from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution that:

$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$

is the general solution to $(1)$.

$\blacksquare$


Sources

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