# Characteristic equation with complex roots

### Characteristic equation with complex roots

#### Lessons

Complex Numbers and Euler’s Formula:

A complex number is a number of the form:

$z=a+bi$

We can plot these numbers on the complex plane:

Euler’s Formula:
$e^{i \theta}=\cos (\theta) + i\sin(\theta)$

Characteristic Equation with Complex Roots:

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:

$Ar^2+Br+C=0$

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

Let’s suppose that $B^2$ < $4AC$. Hence we’re in the realm of complex roots.

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

Or alternatively:
$r=\lambda \pm \mu i$
$r_1=\lambda + \mu i$
$r_2=\lambda - \mu i$

And the general solution will be:

$y(x)=e^{\lambda x} ((c_1) \cos ( \mu x)+(c_2)( \sin (\mu x))$
• 1.
a)
A very brief run-down on Complex Numbers and Euler’s Formula

b)
Using the Characteristic Equation with Complex Roots

• 2.
Determining Complex Solutions to the Characteristic Equations
Find the particular solution to the following differential equation:

$y''-2y'+2y=0$

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

• 3.
Find the particular solution to the following differential equation:

$y''+4y'+8y=0$

With initial values $y( \frac{\pi}{2} )=-1$, $y' (\frac{\pi}{2})=1$

• 4.
Find the particular solution to the following differential equation:

$y''-2y'+4y=0$

With initial values $y( 0)=4$, $y' (0)=5$