Case c.

Case c.

Where

$m_1 = m_2$

. Thus

$\displaystyle y_1 = e^{m_1x}$

So we need another solution. We will prove this in a later section.

$\displaystyle y_2(x)$	$\displaystyle = xe^{m_1x}$
$\displaystyle y(x)$	$\displaystyle = Ay_1(x) + By_2(x)$ general solution
$\displaystyle y(x)$	$\displaystyle = (A+Bx)e^{m_1x}$

Example:

$\displaystyle \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y$	$\displaystyle = 0$
$\displaystyle m^2 + 2m + 1$	$\displaystyle = (m+1)^2 = 0$ auxillary equation
$\displaystyle m$	$\displaystyle = -1$
$\displaystyle y_1(x)$	$\displaystyle = e^{-x}$
$\displaystyle y_2(x)$	$\displaystyle = xe^{-x}$
$\displaystyle y(x)$	$\displaystyle = (A+Bx)e^{-x}$
general solution

We can check that $y_2$

$y_2$

is a solution.

$\displaystyle \frac{dy_2}{dx}$	$\displaystyle = e^{-x}-xe^{-x} = (1-x)e^{-x}$
$\displaystyle \frac{d^2y_2}{dx^2}$	$\displaystyle = -e^{-x}-e^{-x}+xe^{-x} = (x-2)e^{-x}$
$\displaystyle \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y$	$\displaystyle = 0$
original ODE
$\displaystyle [(x-2) + 2(1-x) + x]e^{-x}$	$\displaystyle = 0$
0	$\displaystyle = 0$