# Solution of Constant Coefficient Homogeneous LSOODE

## Theorem

Let:

$(1): \quad y'' + p y' + q y = 0$

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:

$m_1 \ne m_2$
 $\ds y_1$ $=$ $\ds e^{m_1 x}$ $\ds y_2$ $=$ $\ds e^{m_2 x}$

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.

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

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:

$m_1 = m_2 = -\dfrac p 2$
$y_1 = e^{m_1 x}$

is a particular solution to $(1)$.

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

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 $y_2$ $\ds$ $=$ $\ds x e^{m_1 x}$
$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$

$\Box$

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

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

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.

$y_1 = \dfrac {y_a + y_b} 2$
$y_2 = \dfrac {y_b - y_b} 2$

are both particular solutions to $(1)$.

$y = C_1 e^{a x} \cos b x + C_2 e^{a x} \sin b x$

or:

$y = e^{a x} \paren {C_1 \cos b x + C_2 \sin b x}$

is the general solution to $(1)$.

$\blacksquare$