Solution of Constant Coefficient Homogeneous LSOODE

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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 $m^2 + p m + q = 0$.


Then $(1)$ has the general solution:

$y = \begin{cases} C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\ & \\ \paren {C_1 + C_2 x} e^{m_1 x} & : p^2 = 4 q \\ & \\ e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q \end{cases}$

where:

$a + i b = m_1$
$a - i b = m_2$


Proof

Real Roots of Auxiliary Equation

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 $(1)$.

$\Box$


Equal Real Roots of Auxiliary Equation

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 one (repeated) root, that is:

$m_1 = m_2 = -\dfrac p 2$


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

$y_1 = e^{m_1 x}$

is a particular solution to $(1)$.


From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:

$\map {y_2} x = \map v x \, \map {y_1} x$

where:

$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$

is also a particular solution of $(1)$.


We have that:

\(\ds \int P \rd x\) \(=\) \(\ds \int p \rd x\)
\(\ds \) \(=\) \(\ds p x\)
\(\ds \leadsto \ \ \) \(\ds e^{-\int P \rd x}\) \(=\) \(\ds e^{-p x}\)
\(\ds \) \(=\) \(\ds e^{2 m_1 x}\)


Hence:

\(\ds v\) \(=\) \(\ds \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x\) Definition of $v$
\(\ds \) \(=\) \(\ds \int \dfrac 1 {e^{2 m_1 x} } e^{2 m_1 x} \rd x\) as $y_1 = e^{m_1 x}$
\(\ds \) \(=\) \(\ds \int \rd x\)
\(\ds \) \(=\) \(\ds x\)


and so:

\(\ds y_2\) \(=\) \(\ds v y_1\) Definition of $y_2$
\(\ds \) \(=\) \(\ds x e^{m_1 x}\)


From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:

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

$\Box$


Complex Roots of Auxiliary Equation

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 $b = \sqrt {q - \dfrac {p^2} 4}$.


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

\(\ds y_a\) \(=\) \(\ds e^{m_1 x}\)
\(\ds y_b\) \(=\) \(\ds e^{m_2 x}\)

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 $(3)$
\(\ds \) \(=\) \(\ds 2 e^{a x} \sin b x\)
This article, or a section of it, needs explaining.
In particular: isn't this $-2 i e^{a x} \sin b x$? Where did the i go?
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\(\ds \leadsto \ \ \) \(\ds \frac {y_b - y_a} 2\) \(=\) \(\ds e^{a x} \sin b x\)


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

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