Section 3.3 Second Order Linear Homogeneous DEs with Constant
Coefficients; Case complex Roots Revised
Key Terms/ Ideas:
• Complex Roots;
• Complex exponential function
• Euler’s Formula
•General Solution for characteristic equation with complex roots
Here we will consider a second order linear homogeneous equation with constant
coefficients that has the form
ay" + by' + cy = 0
where a, b, and c are given constants and the roots of the characteristic equation
ar2 + br + c = 0 are complex.
Thus the roots are
where b2 – 4ac < 0. Define the roots as follows:
r1 = λ + i and r2 = λ - i, with μ ≠ 0.
This implies we should consider a pair of solutions
where we see that the exponential function has a complex exponent.
This pair of solutions forms a Fundamental Set of Solutions to ay" + by' + cy = 0 since the
roots of the characteristic polynomial are different. (See previous proof for the real case
with unequal roots.) Then the general solution is
y(t )  C1y1(t )  C2 y2 (t )  C1e(  i)t  C2e( i)t  C1et ei t  C2et e i t
BUT… these are complex functions; we need to get real functions.
Here we treat the complex function ei t and equivalently ei t . From calculus we know that
Only odd
powers.
Only even
powers.
So let x = it (that is i times t)and we have
Now recall that even powers of i are -1 and odd powers of i are –i. So next we split this
series into even powered terms of t and odd powered terms of t to get
cos(t)
So we have eit = cos(t) + i sin(t)
and
eit
= cos(t) + i sin(t)
i sin(t)
This called Euler’s formula.
So we can express the two complex solutions in the form
y1(t )  et eit  et  cos(t   i sin  t )
Explain this part.
y2 (t )  et eit  et cos(t   i sin  t )
Since we are dealing with a homogeneous linear DE we can use the principle of
superposition: so adding or subtracting solutions gives another solution.
We can “neglect” the coefficients 2 and 2i by the principle of superposition and
we get the pair of distinct solutions
So we have that the real valued general solution of DE ay" + by' + cy = 0, when the
roots of the characteristic equation are complex values r1 = λ + i and r2 = λ - i with
 ≠ 0, is
Recall
Example: Solve y'' - 2y' + 6y = 0
Characteristic equation is r2 – 2r + 6 = 0
From the quadratic formula the roots are
So λ = 1 and
Integral curves
Thus the solutions are
Next suppose we have initial conditions y(0) = 3, y'(0) = 1; find c1 and c2.
y(0) = 3 
3 = c1 e0 cos(0) + c2 e0 sin(0) 
y'(0) = 1; we first need to compute y'.
0
Now set y' = 1 and t = 0;
3 = c1
Make sure you use the product rule
for derivatives.
0
Thus
Finally we have the solution of the IVP is
Describe the long term behavior of the solution?
B&D 3.3 #8
3 exp(t) cos(sqrt(5) t)+(-2/sqrt(5)) exp(t) sin(sqrt(5) t)
3 exp(t) cos(sqrt(5) t)+(-2/sqrt(5)) exp(t) sin(sqrt(5) t)
800
40
20
600
0
400
-20
200
-40
-60
0
-80
-200
0
0.5
1
1.5
2
2.5
3
3.5
t
3 exp(t) cos(sqrt(5) t)+(-2/sqrt(5)) exp(t) sin(sqrt(5) t)
4
0
1
2
3
t
4
5
6
2000
Behavior: Sinusoid of
increasing amplitude as t→∞.
1000
0
-1000
-2000
-3000
0
1
2
3
4
t
5
6
7
8
Recall
Example: Solve
9y'' +25y = 0
Characteristic equation is 9r2 +25 = 0
From the quadratic formula the roots are
So λ = 0 and
Thus the solutions are
For initial conditions y(0) = A, yꞌ(0) = B, as long as c1 and c2 are not both zero, describe
the long term behavior of the solution of the IVP.
2 cos(5/3 t)-3 sin(5/3 t)
4
For example with c1 = 2, c2 =-3
3
2
1
0
Behavior: Sinusoid of
fixed amplitude.
-1
-2
-3
-4
0
1
2
3
4
5
6
7
8
9
10
Example: Suppose we had a second order linear DE with constant coefficients where the
roots of the characteristic polynomial are r = -0.85 ± 2i.
(a) Write the expression for the general solution.
(b) If we had initial conditions so that c1 and c2 are not both zero, describe the long term
behavior of the particular solution of the IVP.
The general solution:
Long term behavior:
Oscillating with decreasing amplitude
1
For example with c1 = -2, c2 = -1
0.5
0
-0.5
-1
-1.5
-2
-2.5
0
1
2
3
4
5
6
7
8
9
10
Example: Solve the IVP 5y'' + 2y' + 7y = 0, y(0) = 2, y'(0) = 1
The characteristic equation is 5r2 + 2r + 7 = 0 , with roots
The solution is
Using the given initial conditions, we get
0
y(0) = 2  2 = c1
y'(0) = 1 ; compute y' first;
Set t = 0 and y’ = 1.
0

So we have the solution of the IVP:
Describe the solution as t gets large.
B&D Sec. 3.3 #24
For the IVP 5y'' + 2y' + 7y = 0, y(0) = 2, y'(0) = 1 explain how you would determine the
smallest time T so that |y (t)| ≤ 0.1 for all t > T.
Outline a plan to accomplish this!
Graph the solution.
2*exp(-t/5)*cos(sqrt(34)/5*t)+(7/sqrt(34))*exp(-t/5)*sin(sqrt(34)/5*t)
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
0
2
4
6
8
10
12
14
16
18
20
Draw horizontal lines.
2*exp(-t/5)*cos(sqrt(34)/5*t)+(7/sqrt(34))*exp(-t/5)*sin(sqrt(34)/5*t)
2.5
2
1.5
1
y = 0.1
0.5
0
-0.5
y = -0.1
-1
-1.5
0
2
4
6
8
10
12
14
16
18
20
0.15
0.1
0.05
0
-0.05
-0.1
-0.15
-0.2
14.5
15
15.5
T is approximately 14.5.
16
16.5
17