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3.2: Complex Roots of the Characteristic Equation

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We have already addressed how to solve a second order linear homogeneous differential equation with constant coefficients where the roots of the characteristic equation are real and distinct. We will now explain how to handle these differential equations when the roots are complex. The example below demonstrates the method.

Example 3.2.1

Solve

y4y+13y=0.

Solution

As before we assume that y=ert is a solution. We have

y=rerty=r2ert.

Substituting back into the original differential equation gives

r2ert4rert+13ert=0

r24r+13=0 dividing by ert

This quadratic does not factor, so we use the quadratic formula and get the roots

r=2+3i and r=23i

We can conclude that the general solution to the differential equation is

y=a1e(2+3i)t+a2e(23i)t=a1e2te3it Using the rule of exponents =e2t(a1e3it+a2e3it) Factoring out the e2t.

Although this gives the general solution, it it not satisfactory since the solution involves complex exponents. To deal with this we use Euler's formula

eiq=cosq+isinq.

This gives

y=e2t[a1(cos(3t)+isin(3t))+a2(cos(3t)+isin(3t)).]

Since the cosx is an even function and sinx sin x is an odd function, we get

y=e2t[a1(cos(3t)+isin(3t))+a2(cos(3t)isin(3t))]

or

y=e2t[(a1+a2)cos(3t)+(a1a2)isin(3t)].

Finally let

c1=a1+a2andc2=i(a1a2)

and we get

y=e2t[c1cos(3t)+c2sin(3t)].

General Solution

In general if

ay+by+cy=0

is a second order linear differential equation with constant coefficients such that the characteristic equation has complex roots

r=l+miandr=lmi

Then the general solution to the differential equation is given by

y=elt[c1cos(mt)+c2sin(mt)]

Example 3.2.2: Graphical Solutions

Solve

y10y+29=0given y(0)=1,y(0)=3

Solution

The characteristic equation is

r210r+29=0

which has roots

r=5+2iandr=52i.

The general solution is

y=e5t[c1cos(2t)+c2sin(2t)].

We use the initial values to find the constants. Plug in y(0)=1

1=1[c1(1)+c2(0)]

so that c1=1. We have

y=5e5t[cos(2t)+c2sin(2t)]+e5t[2sin(2t)+2c2cos(2t)].

Plugging in y(0)=3

3=5[1+0]+1[0+2c2]=5+2c2.

Hence c2=1.

The final solution is

y=e5t[cos(2t)sin(2t)].

Let's investigate the graphs of solutions for the general solutions in Equation ???

  • For l=0, the graph is periodic.
  • For l>0, the amplitude increases exponentially
  • For l<0, the amplitude approaches 0 exponentially.

The three types are pictured below.

comple1.gifcomple2.gif

Contributors and Attributions


This page titled 3.2: Complex Roots of the Characteristic Equation is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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