# Characteristic equation with repeated roots

### Characteristic equation with repeated roots

#### Lessons

In the previous section we came up with a method to solve linear homogeneous constant coefficient second order differential equations:

$Ay''+By'+Cy=0$

By using the characteristic equation:

$Ar^2+Br+C=0$

$r=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$

But what if $B^2=4AC$?

$r=\frac{-B\pm\sqrt{B^2-4AC}}{2A}=\frac{-B}{2A}$

Throughout the videos it will be shown that our solutions are:

$y_1(x)=e^{\frac{-B}{2A}x}$
$y_2(x)=xe^{\frac{-B}{2A}x}$

Or in full generality:
$y(x)=c_1 e^{r_1 x}+c_2 xe^{r_2 x}$

Where $r_1=r_2=-\frac{B}{2A}$
• Introduction
What is the solution to the Characteristic Equation with Repeated Roots?

• 1.
Determining the Characteristic Equation with Repeated Roots
Find the particular solution to the following differential equation:

$y''-6y'+9y=0$

With initial values $y(0)=3, y'(0)=2$

• 2.
Find the particular solution to the following differential equation:

$4y''-12y'+9y=0$

With initial values $y(2)=e^3, y'(2)=\frac{1}{2}e^3$