Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root 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.


Then:

$y = e^{m_1 x}$ is a solution to $(1)$

if and only if

$m_1$ is a root of the auxiliary equation $m^2 + p m + q = 0$


Proof

Consider the equation:

$y = e^{m_1 x}$

Differentiating with respect to $x$:

\(\ds y\) \(=\) \(\ds e^{m_1 x}\)
\(\ds y'\) \(=\) \(\ds m_1 e^{m_1 x}\)
\(\ds y\) \(=\) \(\ds {m_1}^2 e^{m_1 x}\)


Sufficient Condition

Let $y = e^{m_1 x}$ be a solution to $(1)$.

Substituting for $y$ and its derivatives in $(1)$:

\(\ds y + p y' + q y\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds {m_1}^2 e^{m_1 x} + p m_1 e^{m_1 x} + q e^{m_1 x}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren { {m_1}^2 + p m_1 + q} e^{m_1 x}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds {m_1}^2 + p m_1 + q\) \(=\) \(\ds 0\) as $e^{m_1 x}$ is never zero

That is, by definition, $m_1$ is a root of $m^2 + p m + q = 0$.

$\Box$


Necessary Condition

Let $m_1$ be a root of $m^2 + p m + q = 0$.

Thus:

\(\ds {m_1}^2 + p m_1 + q\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren { {m_1}^2 + p m_1 + q} e^{m_1 x}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds {m_1}^2 e^{m_1 x} + p m_1 e^{m_1 x} + q e^{m_1 x}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds y + p y' + q y\) \(=\) \(\ds 0\) substituting for $y$ and its derivatives

Thus $y = e^{m_1 x}$ satisfies $(1)$.

$\blacksquare$


Sources

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