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

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

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 = 0 (1) with f(D) = a0Dn + … + an-1D + an, (2) a differential operator. 3

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

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

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

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

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

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

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

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

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

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

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

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

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.