Auxiliary Equation with Repeated Roots

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Auxiliary Equation with Repeated Roots MATH 374 Lecture 16 Auxiliary Equation with Repeated Roots TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAAA

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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