1
One root gives only one exponential
With a zero discriminant the formula returns the same root twice. C₁·e^(rt) + C₂·e^(rt) is (C₁ + C₂)·e^(rt), which has one constant: it fits a starting value and cannot also fit a starting rate.
Lesson · Second-Order Linear Equations · The characteristic equation
Repeated Roots
When the discriminant is zero the quadratic has one root, and e^(rt) is only one solution, which would leave the equation with a family of one constant. The second solution is t·e^(rt), and the lab verifies it exactly, so the general solution is (C₁ + C₂·t)·e^(rt) and the two constants can again be fitted from a start and a rate.
disc = 0: one root r = −b/(2a)
e^(rt) and t·e^(rt) are both solutions
y = (C₁ + C₂·t)·e^(rt)
y(0) = C₁
y′(0) = r·C₁ + C₂
A repeated root needs a factor of t
Core ideas
What this lesson establishes
Three ideas: why the usual form falls short, which second function takes its place, and how the constants are read.
2
The second solution is t times the first
The function t·e^(rt) has the rate (1 + r·t)·e^(rt) and a second rate (2r + r²·t)·e^(rt). Substituted into the equation, the result is a multiple of a·r² + b·r + c plus a multiple of 2a·r + b, and both are zero when r is the repeated root.
3
The constants read off simply at t = 0
Because t = 0 in (C₁ + C₂·t)·e^(rt) leaves C₁, the starting value is y(0) = C₁. The starting rate adds the two sources, y′(0) = r·C₁ + C₂, since (C₁ + C₂·t)·e^(rt) has the rate (r·C₁ + C₂ + r·C₂·t)·e^(rt).
The material
Where the second solution comes from
Why a single exponential is not enough, the substitution that shows t·e^(rt) works, and a fit with both constants.
When the discriminant b² − 4ac is 0, the quadratic formula gives r = −b/(2a) once. For y″ + 4y′ + 4y = 0 the quadratic is r² + 4r + 4 = (r + 2)², and r = −2 is its only root. That gives e^(−2t), and its multiples, a family with one constant.
A family with one constant cannot fit two starting values. The equation y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = 1 needs C = 1 from the first and, since (C·e^(−2t))′ = −2·C·e^(−2t), −2·C = 1 from the second. The two demands disagree, and a second solution must be found.
Theorem · The second solution for a repeated root
If r is the only root of a·r² + b·r + c = 0, then t·e^(rt) is a solution of a·y″ + b·y′ + c·y = 0, and the general solution is (C₁ + C₂·t)·e^(rt).
The product rule gives (t·e^(rt))′ = (1 + r·t)·e^(rt) and (t·e^(rt))″ = (2r + r²·t)·e^(rt).
Substituting, a·y″ + b·y′ + c·y is [(a·r² + b·r + c)·t + (2a·r + b)]·e^(rt). The first bracket is zero because r is a root. The second is zero because a repeated root is r = −b/(2a). So the residual is zero.
That t·e^(rt) is the second solution is verified here. That these two are all the solutions is stated, and “Superposition and the Wronskian” shows why they can fit any starting values.
□
A concrete check
For y″ + 4y′ + 4y = 0 and y = t·e^(−2t) the rates are y′ = (1 − 2t)·e^(−2t) and y″ = (4t − 4)·e^(−2t). Then y″ + 4y′ + 4y = [(4t − 4) + 4(1 − 2t) + 4t]·e^(−2t) = 0, because the t terms are 4 − 8 + 4 = 0 and the constants are −4 + 4 = 0.
y = t·e^(−2t)
y′ = (1 − 2t)·e^(−2t)
y″ = (4t − 4)·e^(−2t)
(4t − 4) + 4·(1 − 2t) + 4t = 0
Fitting both constants
With y(0) = 1 and y′(0) = 1, the first condition gives C₁ = 1. The second is −2·C₁ + C₂ = 1, so C₂ = 3, and the solution is (1 + 3t)·e^(−2t). The factor 1 + 3t grows, but e^(−2t) shrinks faster, so the solution rises at first and then decays, a shape that one exponential cannot make.
Example · A root of zero, repeated
For y″ = 0 the quadratic is r² = 0 with r = 0 twice, and the general solution is (C₁ + C₂·t)·e^(0·t) = C₁ + C₂·t, a straight line. That is right: a function with a zero second rate has a constant first rate, so it is a line.
With y(0) = 1 and y′(0) = 2 the line is 1 + 2t, and the lab prints the constants C₁ = 1, C₂ = 2.
The repeated root is a boundary case. A tiny change in b makes the discriminant positive or negative and the form of the solution changes, though the solution itself changes only a little. It is also the case the next course uses, as the dividing line between motion that overshoots and motion that does not.
Interactive
The characteristic equation
Roots, the general solution and the constants, exactly
Method
Solving with a repeated root
Five moves. The one that is new is the second solution.
1
Confirm the discriminant is zero
b² − 4ac = 0. Then the single root is r = −b/(2a).
2
Write both solutions
e^(rt) and t·e^(rt). The second is the first with a factor of t.
3
Write the general solution
y = (C₁ + C₂·t)·e^(rt). The factor C₁ + C₂·t is a line in t.
4
Fit the constants
C₁ = y(0) and C₂ = y′(0) − r·C₁. Substitute the numbers; keep the fractions.
5
Substitute back
Differentiate twice by the product rule and check the residual. A forgotten factor of t fails here.
Repeated roots and the second solution
Score 0 / 0Choose the answer you can justify.
y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = 1
One root, the second solution checked, and both constants fitted.
(r + 2)² = 0, r = −2 twice
y = t·e^(−2t)
y′ = (1 − 2t)·e^(−2t)
y″ = (4t − 4)·e^(−2t)
(4t − 4) + 4(1 − 2t) + 4t = 0
y(0) = C₁ = 1
y′(0) = −2·C₁ + C₂ = 1, C₂ = 3
y = (1 + 3t)·e^(−2t)
The lab prints the same solution expanded, as 3·t·e^(−2t) + e^(−2t). The fifth line is the exact check: the t terms cancel and so do the constants, and the lab's status line reports a residual of 0.
Note that C₂ is 3 and not 1. The starting rate is 1, but −2·C₁ already contributes −2, so the second term has to supply 3.
Common mistakes
1
Using C·e^(rt) alone, with one initial valueWith one root, y = C·e^(rt) looks like the answer, and y(0) = 1 fixes C = 1. For y″ + 4y′ + 4y = 0 that is e^(−2t), whose rate at 0 is −2. A start with y′(0) = 1 is not met, and no choice of C does it. The second solution t·e^(−2t) supplies the missing freedom, giving (1 + 3t)·e^(−2t).
2
Reading C₂ as the starting rateSince y′(0) = r·C₁ + C₂, the constant C₂ is the starting rate minus r·C₁. For y(0) = 1, y′(0) = 1 and r = −2, that is C₂ = 1 − (−2)(1) = 3. Taking C₂ = 1 gives (1 + t)·e^(−2t), whose rate at 0 is −1, not 1.
3
Writing the second solution as another copy of e^(rt)Both roots equal −2, so it is tempting to write C₁·e^(−2t) + C₂·e^(−2t). That is (C₁ + C₂)·e^(−2t), one function with one constant, and the lab's constants for y(0) = 1, y′(0) = 1 are C₁ = 1, C₂ = 3 on e^(−2t) and t·e^(−2t). The residual of the copy is 0, which is why it looks right; what fails is the fit, since e^(−2t) has y′(0) = −2 and not 1.
Check your understanding
✓
Finish when you can solve an equation whose characteristic quadratic has a repeated root, and fit both constants.
You should be able to recognise a zero discriminant, write (C₁ + C₂·t)·e^(rt), check t·e^(rt) by substitution, and solve C₁ = y(0) and r·C₁ + C₂ = y′(0) exactly.
Note: Real roots, distinct or repeated, give solutions that do not oscillate. When the discriminant is negative the roots are a complex pair, and the exponential and a sine and cosine appear together: that is “Complex Roots and Oscillation”.