1. Chapter 5: Linear Equations with Constant Coefficients
MATH 374 Lecture 15
Chapter 5: Linear Equations with Constant Coefficients

2. Note
In this chapter of our class notes, all differential operators and differential equations will have constant coefficients! 2

3. 5.1: The Auxiliary Equation: Distinct Roots
Any nth order linear homogeneous differential equation with constant coefficients can be written in the form:
f(D)y = (1)
with
f(D) = a0Dn + … + an-1D + an, (2)
a differential operator. 3

4. The Auxiliary Equation
f(D)y = 0 (1)
The Auxiliary Equation
Definition: For equation (1), we call the equation f(m) = 0 the auxiliary equation of (1).
(Boyce and DiPrima call f(m) = 0 the characteristic equation of (1).) 4

5. f(D)y = 0 (1)
General Solution to (1)
Theorem 5.1: For the nth order linear homogeneous differential equation (1),
if the roots m1, m2, … , mn of the auxiliary equation f(m) = 0 are all real and distinct,
then the n functions are linearly independent solutions of (1) and the general solution to (1) is:
where c1, c2, … , cn are arbitrary constants. 5

6. f(D)y = 0 (1)
Proof of Theorem 5.1
From Corollary 1 of Theorem 4.7, if mi is a root of f(m) = 0, then f(D)emi= 0.
Since the Wronskian of is non-zero (check),
these functions are linearly independent.
It follows from Theorem 4.4 that the general solution to (1) is of the form (3).  6

7. Example 1: Solve y’’ + 2y’ = 0.
Solution: Rewrite the differential equation as (D2 + 2D)y = 0.
Find the roots of the auxiliary equation:
m2 + 2m = 0 ) m(m+2) = 0
) m = 0 or m = -2.
Hence the general solution is y = c1e0·x + c2e-2x = c1 + c2e-2x. 7

8. Example 2: Solve (D3+3D2-4D-12)y = 0
Solution: The auxiliary equation is: m3+3m2-4m-12 = 0.
Factor with synthetic division.
(Possible rational roots are §1, §2, §3, §4, §6, §12 – see Boyce and DiPrima p. 230.)
Coefficients of powers of m
in decreasing order, including
zero coefficients. 1 3 -4
-12 8

9. Example 2: Solve (D3+3D2-4D-12)y = 0
Solution: The auxiliary equation is: m3+3m2-4m-12 = 0.
Factor with synthetic division.
(Possible rational roots are §1, §2, §3, §4, §6, §12 – see Boyce and DiPrima p. 230.)
Possible
Root 2 1 3 -4
-12 9

10. Example 2: Solve (D3+3D2-4D-12)y = 0
Solution: The auxiliary equation is: m3+3m2-4m-12 = 0.
Factor with synthetic division.
(Possible rational roots are §1, §2, §3, §4, §6, §12 – see Boyce and DiPrima p. 230.) 2 1 3 -4
-12 10

11. Example 2: Solve (D3+3D2-4D-12)y = 0
Solution: The auxiliary equation is: m3+3m2-4m-12 = 0.
Factor with synthetic division.
(Possible rational roots are §1, §2, §3, §4, §6, §12 – see Boyce and DiPrima p. 230.) 2 1 3 -4
-12 11

12. Example 2: Solve (D3+3D2-4D-12)y = 0
Solution: The auxiliary equation is: m3+3m2-4m-12 = 0.
Factor with synthetic division.
(Possible rational roots are §1, §2, §3, §4, §6, §12 – see Boyce and DiPrima p. 230.) 2 1 3 -4
-12 5 12

13. Example 2: Solve (D3+3D2-4D-12)y = 0
Solution: The auxiliary equation is: m3+3m2-4m-12 = 0.
Factor with synthetic division.
(Possible rational roots are §1, §2, §3, §4, §6, §12 – see Boyce and DiPrima p. 230.) 2 1 3 -4
-12 10 5 13

14. Example 2: Solve (D3+3D2-4D-12)y = 0
Solution: The auxiliary equation is: m3+3m2-4m-12 = 0.
Factor with synthetic division.
(Possible rational roots are §1, §2, §3, §4, §6, §12 – see Boyce and DiPrima p. 230.) 2 1 3 -4
-12 10 5 6 14

15. Example 2: Solve (D3+3D2-4D-12)y = 0
Solution: The auxiliary equation is: m3+3m2-4m-12 = 0.
Factor with synthetic division.
(Possible rational roots are §1, §2, §3, §4, §6, §12 – see Boyce and DiPrima p. 230.) 2 1 3 -4
-12 10 12 5 6 15

16. Example 2: Solve (D3+3D2-4D-12)y = 0
Solution: The auxiliary equation is: m3+3m2-4m-12 = 0.
Factor with synthetic division.
(Possible rational roots are §1, §2, §3, §4, §6, §12 – see Boyce and DiPrima p. 230.)
Therefore
m3+3m-4m-12 = (m-2)(m2+5m+6)
= (m-2)(m+2)(m+3)
) (m-2)(m+2)(m+3) = 0
) m = 2, -2, -3 2 1 3 -4
-12 10 12 5 6 16
It follows that the general solution is y = c1e2x + c2e-2x + c3e-3x.