Case Two: Repeated Roots
Chapter One Hundred Twenty
Syllabus topic Module 2, "2.3 Linear Differential Equations with Constant Coefficients"
Pages 267 to 268 of 303
In one line
A repeated root gives an exponential and x times the same exponential, and the x is not a fudge.
The problem
Suppose the auxiliary equation has a repeated root m. Then the trial solution e to the mx is found once, and the two terms of the complementary function would be A e to the mx plus B e to the mx, which is (A plus B) e to the mx: one constant, not two.
A second-order equation needs two. So one solution is missing.
The answer
y = (A + B x) e^(m x)
The second solution is x times the first.
Where the x comes from
Not from guessing. Here are two derivations, and the first is the one to give in an examination.
By a limit. Take two distinct roots m and m plus h, so the complementary function is A e to the mx plus B e to the (m plus h)x. Choose the constants as minus B over h and plus B over h, which is legitimate because they are arbitrary.
Then the solution is B(e to the (m+h)x minus e to the mx) over h, which as h tends to zero is B times the derivative of e to the mx with respect to m, which is B x e to the mx.
So x e to the mx is what the second solution becomes as the two roots merge, and that is why the x appears.
By substitution. For a repeated root the operator is (D minus m) squared. Try y = u e to the mx. By the shift relation of the D-laws chapter, (D minus m) squared acting on u e to the mx is e to the mx times D squared acting on u.
So the equation becomes D squared u = 0, which says u double prime = 0, so u is a linear function of x: u = A plus Bx.
That derivation is shorter and it also generalises: for a root repeated three times the equation becomes D cubed u = 0, so u is a quadratic.
Worked
Solve y'' minus 4y' plus 4y = 0.
The auxiliary equation is m squared minus 4m plus 4 = 0, that is (m minus 2) squared = 0, so m = 2 twice.
d2y/dx2 - 4 dy/dx + 4y = 0
y = C1 e^(2x) + C2 x e^(2x)
Worked, a negative repeated root
Solve y'' plus 6y' plus 9y = 0.
The auxiliary equation is (m plus 3) squared = 0, so m = minus 3 twice.
d2y/dx2 + 6 dy/dx + 9y = 0
y = C1 e^(-3x) + C2 x e^(-3x)
The x grows and the exponential decays, and the exponential wins: the solution rises to a peak and then dies away. That is critical damping, the fastest a system can settle without overshooting, and it is what a well-adjusted door closer does.
Case Two: Repeated Roots
Worked, a root repeated three times
Solve y''' minus 3y'' plus 3y' minus y = 0.
The auxiliary equation is m cubed minus 3m squared plus 3m minus 1 = 0, which is (m minus 1) cubed = 0, so m = 1 three times.
By the substitution derivation, u must satisfy D cubed u = 0, so u is a quadratic in x.
d3y/dx3 - 3 d2y/dx2 + 3 dy/dx - y = 0
y = C1 e^x + C2 x e^x + C3 x^2 e^x
Three constants for a third-order equation, and the powers of x run 0, 1, 2.
The general pattern
A root repeated k times contributes k terms: the exponential multiplied by 1, by x, by x squared, and so on up to x to the power k minus 1.
| Roots | Complementary function |
|---|---|
| 2, 2 | (A + Bx) e^(2x) |
| 2, 2, 2 | (A + Bx + Cx^2) e^(2x) |
| 1, 2, 2 | A e^x + (B + Cx) e^(2x) |
| 0, 0 | A + Bx |
The last row is the case where the repeated root is zero, so the exponential is 1 and the answer is a straight line. That is the solution of y'' = 0, which it obviously should be.
d2y/dx2 = 0
y = C1 + C2 x
Check yourself
d2y/dx2 - 2 dy/dx + y = 0
y = C1 e^x + C2 x e^x
d2y/dx2 + 4 dy/dx + 4y = 0
y = C1 e^(-2x) + C2 x e^(-2x)
d3y/dx3 - 4 d2y/dx2 + 4 dy/dx = 0
y = C1 + C2 e^(2x) + C3 x e^(2x)
The third has roots 0, 2 and 2, so a constant from the zero root and two terms from the repeated one.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.