Repeated Roots
Differential Equations · Section 3.4 · generated practice set
One root, two solutions — where the extra factor of t comes from and how it changes the initial-condition algebra.
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Try one now
Solve $$y'' + 8y' + 16y = 0, \qquad y(0) = -2,\ y'(0) = 5,$$ writing it as $y = (c_1 + c_2t)e^{rt}$.
Solution
- $r^2 + 8r + 16 = (r + 4)^2 = 0$: a repeated root $r = -4$.
- A repeated root gives the second solution $te^{rt}$, so $y = (c_1 + c_2t)e^{-4t}$.
- $y(0) = c_1 = -2$.
- $y' = c_2e^{-4t} - 4(c_1 + c_2t)e^{-4t}$, so $y'(0) = c_2 - 4c_1 = 5 \Rightarrow c_2 = -3$.
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Method
When $b^2 - 4ac = 0$ there is one root $r = -b/2a$, and the second solution picks up a factor of $t$: $$y = (c_1 + c_2t)e^{rt}.$$ Reduction of order is where that extra $t$ comes from.
Definitions and theorems
Worked example
Solve $y'' - 6y' + 9y = 0$, $y(0) = 1$, $y'(0) = 5$.
- $r^2 - 6r + 9 = (r-3)^2 = 0 \Rightarrow r = 3$ (double).
- $y = (c_1 + c_2t)e^{3t}$.
- $y(0) = c_1 = 1$.
- $y' = c_2e^{3t} + 3(c_1 + c_2t)e^{3t}$, so $y'(0) = c_2 + 3 = 5 \Rightarrow c_2 = 2$.
- $y = (1 + 2t)e^{3t}$.
Common mistakes
- Writing $c_1e^{rt} + c_2e^{rt}$. That is one solution, not two — you need the $t$.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.