University:
Georgia Southern University
Course:
MATH 3230 | Ordinary Differential Equations
Academic year:

2024
Views:

231

Pages:

8
Author:

Jalen W.

Repeated Real Roots Suppose our characteristic equation had repeated real roots, such as with our final example from Section 4.3, y 00 − 6y 0 + 9y = 0 → r 2 − 6r + 9 = 0 → (r − 3)2 = 0. I This has only one real root, r = 3. I Then certainly y1 (x) = e 3x is a solution to the ODE. I There should be a second solution y2 (x) so that S = {y1 (x), y2 (x)} is linearly independent. I CLAIM: y2 (x) = xe 3x is a solution. I y20 (x) = e 3x + 3xe 3x , y200 (x) = 6e 3x + 9xe 3x Repeated Real Roots Let’s try plugging in y2 (x) = xe 3x and its derivatives. y 00 − 6y 0 + 9y = [6e 3x + 9xe 3x ] − 6[e 3x + 3xe 3x ] + 9[xe 3x ] = 0, as desired. So y2 (x) = xe 3x is a solution to the ODE as well. Then, the general solution to the ODE is just y (x) = C1 e 3x + C2 xe 3x Fact 1: Repeated Real Roots Suppose r = r1 is a real root of multiplicity k (i.e. it is a root that is repeated k times) of the characteristic equation to some homogeneous linear ODE with constant coefficients. Then, {y1 (x) = e r1 x , y2 (x) = xe r1 x , . . . , yk (x) = x k−1 e r1 x } is a linearly independent set of k solutions to the ODE. Special Note about this Chapter Since we are only dealing with second order homogeneous linear ODEs with constant coefficients... I Char. Eq.: ar 2 + br + c = 0 I Only has two roots TOTAL, since second degree polynomial. I Thus, either you have two distinct (not necessarily REAL) roots, or you have one root of multiplicity 2. Example Find a general solution to the ODE. y 00 + 4y 0 + 4y = 0 Example Find a general solution to the ODE. 16y 00 + 8y 0 + y = 0 Example Find a general solution to the ODE. y 00 − y = 0 Example Find a general solution to the ODE. y 00 + y = 0 Hint: You can’t do this one with current knowledge.

Characteristic Equations with Repeated Real Roots

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