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+biz=a+bi

We can plot these numbers on the complex plane:

complex plane

Euler’s Formula:
eiθ=cos(θ)+isin(θ)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=0Ay''+By'+Cy=0

By using the characteristic equation:

Ar2+Br+C=0Ar^2+Br+C=0

Using the quadratic formula:

r=B±B24AC2Ar=\frac{-B\pm\sqrt{B^2-4AC}}{2A}

Let’s suppose that B2B^2 < 4AC4AC. Hence we’re in the realm of complex roots.

r=B±B24AC2Ar=\frac{-B\pm\sqrt{B^2-4AC}}{2A} =B2A±B24AC2A=\frac{-B}{2A} \pm \frac{\sqrt{B^2-4AC}}{2A}

Or alternatively:
r=λ±μir=\lambda \pm \mu i
r1=λ+μir_1=\lambda + \mu i
r2=λμir_2=\lambda - \mu i

And the general solution will be:

y(x)=eλx((c1)cos(μx)+(c2)(sin(μx))y(x)=e^{\lambda x} ((c_1) \cos ( \mu x)+(c_2)( \sin (\mu x))
  • Introduction
    a)
    A very brief run-down on Complex Numbers and Euler’s Formula

    b)
    Using the Characteristic Equation with Complex Roots


  • 1.
    Determining Complex Solutions to the Characteristic Equations
    Find the particular solution to the following differential equation:

    y2y+2y=0y''-2y'+2y=0

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

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

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

    With initial values y(π2)=1y( \frac{\pi}{2} )=-1, y(π2)=1y' (\frac{\pi}{2})=1

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

    y2y+4y=0y''-2y'+4y=0

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