Homogeneous

Consider the second order homogeneous ODE

$\displaystyle \frac{d^2y}{dx^2} + a\frac{dy}{dx} + by = 0$

We normally take an educated guess at a solution, e.g. $y=e^{mx}$ , where m is some constant. Thus

$\displaystyle \frac{dy}{dx} = me^{mx} \quad\quad \frac{d^2y}{dx^2} = m^2e^{mx}$

Subbing this back into our original homogeneous equation and then dividing by $e^{mx}$ gives us

$\displaystyle (m^2 + am + b)$	$\displaystyle = 0$ axuillary equation
$\displaystyle (m-m_1)(m-m_2)$	$\displaystyle = 0$ & are roots

There are three cases to consider, when $m_1$

are