1. Auxiliary Equation with Repeated Roots
MATH 374 Lecture 16
Auxiliary Equation with Repeated Roots
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2. 5.2: The Auxiliary Equation: Repeated Roots
Given the nth order linear homogeneous differential equation with constant coefficients,
f(D)y = 0, (1)
suppose (D-b) appears as a factor of f(D), k times with 1 < k ≤ n.
Then the auxiliary equation f(m) = 0 has a root b with multiplicity k, hence k roots are equal to b. 2

3. f(D)y = 0, (1)
Repeated Roots Case
Trying to apply the same method as in Section 5.1, we will run into trouble, as we need n linearly independent solutions of (1) to find the general solution of (1), and
y = c1ebx + c2ebx + … + ckebx = cebx
yields only “one” solution. 3

4. f(D)y = 0, (1)
Repeated Roots Case
Now, since (D-b)k is a factor of f(D), it follows that
f(D) = g(D)(D-b)k (2)
where g(D) yields the other n-k factors of f(D).
Unfortunately, g(D) will give at most n-k solutions to (1), which means we will have at most n-k+1 < n (check!) linearly independent solutions of (1).
Here is a way to get n linearly independent solutions of (1)!! 4

5. Repeated Roots - General Solution
f(D)y = 0, (1)
Repeated Roots - General Solution
Theorem 5.2: For the linear homogeneous differential equation of order n, (1), let m1, m2, … , mL be real distinct roots of the auxiliary equation: f(m) = 0 with multiplicities k1, k2, … , kL, respectively.
Then the n functions:
are linearly independent solutions of (1) and the general solution
to (1) is: 5

6. f(D)y = 0, (1)
Proof of Theorem 5.2:
f(D) = g(D)(D-b)k (2)
If mi is a root of f(m) = 0, of multiplicity ki, then (2) implies f(D) = gi(D)(D-mi)ki.
It follows from Corollary 2 of Theorem 4.7 that for j = 0, 1, 2, … , ki – 1,
f(D)(xj emix) = g(D)(D-mi)ki (xj emix) = 0.
Since for each i = 1, 2, … , L, the functions
emix , xemix , … , xki – 1 emix
are linearly independent (HW problem # 6, p. 115),
it follows (check) that we have n linearly independent solutions to (1).  6

7. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 7

8. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 8

9. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 9

10. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 10

11. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 11

12. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 12

13. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 13

14. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 14

15. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 15

16. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 16

17. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 17

18. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 18

19. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 19

20. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 20

21. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 21

22. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 22

23. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 23

24. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 24

25. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 25

26. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 26

27. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 27

28. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 28

29. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1 29

30. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1 30

31. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1 31

32. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1 32

33. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1 33

34. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1 34

35. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1 35

36. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1 36

37. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1
m5 - 5m4 + 7m3 + m2 - 8m + 4 37

38. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1
m5 - 5m4 + 7m3 + m2 - 8m + 4
= (m-1)(m4 – 4m3 + 3m2 + 4m - 4) 38

39. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1
m5 - 5m4 + 7m3 + m2 - 8m + 4
= (m-1)(m4 – 4m3 + 3m2 + 4m - 4)
= (m-1)2 (m3 – 3m2 + 4) 39

40. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1
m5 - 5m4 + 7m3 + m2 - 8m + 4
= (m-1)(m4 – 4m3 + 3m2 + 4m - 4)
= (m-1)2 (m3 – 3m2 + 4)
= (m-1)2 (m+1) (m2 – 4m + 4) 40

41. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1
m5 - 5m4 + 7m3 + m2 - 8m + 4
= (m-1)(m4 – 4m3 + 3m2 + 4m - 4)
= (m-1)2 (m3 – 3m2 + 4)
= (m-1)2 (m+1) (m2 – 4m + 4) 41

42. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Solution: The auxiliary equation is:
m5 – 5m4 + 7m3 + m2 – 8m + 4 = 0.
Factor LHS with synthetic division. (Possible rational roots are §1, §2, §4) 1 -5 7 -8 4 -4 3 -3 -1
m5 - 5m4 + 7m3 + m2 - 8m + 4
= (m-1)(m4 – 4m3 + 3m2 + 4m - 4)
= (m-1)2 (m3 – 3m2 + 4)
= (m-1)2 (m+1) (m2 – 4m + 4) 42
= (m-1)2 (m+1) (m-2)2

43. Example 1: Solve y(5) – 5y(4) + 7y’’’ + y’’ – 8y’ + 4y = 0
Therefore, (m-1)2 (m+1) (m-2)2 = 0
) m = 1, 1, -1, 2, 2.
Thus, the general solution to this differential equation is
y = c1 ex + c2 x ex + c3 e-x + c4 e2 x + c5 x e2x. 43