Characteristic Roots and Free Response

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The idea

Constant coefficients make exponential trial solutions natural: every derivative of $e^{rt}$ is a constant multiple of it. Substitution in $ay''+by'+cy=0$ gives the characteristic polynomial $ar^2+br+c=0$ , with $a\ne0$ .

Visual guide

VISUAL GUIDERepeated roots need a second independent shape

e⁻ᵗ
te⁻ᵗ
(1 + t)e⁻ᵗ

For the repeated root −1, e⁻ᵗ and te⁻ᵗ are independent solutions. Their sum (1 + t)e⁻ᵗ has initial displacement 1 and initial velocity 0. The t factor is essential to match both conditions.

Method and assumptions

Distinct real roots give $c_1e^{r_1t}+c_2e^{r_2t}$ . A repeated root gives $(c_1+c_2t)e^{rt}$ . Roots $\alpha\pm i\beta$ give $e^{\alpha t}(c_1\cos\beta t+c_2\sin\beta t)$ . Apply initial conditions after constructing an independent basis.

Worked example: repeated decay

$y''+2y'+y=0$ has $(r+1)^2=0$ . Thus $y=(c_1+c_2t)e^{-t}$ . Conditions $y(0)=1,y'(0)=0$ give $c_1=c_2=1$ . The extra factor $t$ provides the missing independent solution.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

A characteristic equation has a repeated root r=−2. Which pair gives the homogeneous solution basis?

e⁻²ᵗ and e²ᵗ

e⁻²ᵗ and t e⁻²ᵗ

e⁻²ᵗ and e⁻²ᵗ

Hint 1 · Find a starting point

Repeating the same function does not add an independent solution.

Hint 2 · Take the next step

A repeated root introduces a factor of t.

Show the reasoning

Answer: e⁻²ᵗ and t e⁻²ᵗ

The solution is (C₁+C₂t)e⁻²ᵗ; the two displayed terms are independent.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: growing oscillation

$y''-2y'+5y=0$ has roots $1\pm2i$ . With $y(0)=0,y'(0)=2$ , the solution is $e^t\sin2t$ . Its oscillation does not make it stable: the exponential envelope grows.

Interpreting the result

The sign of the real parts controls exponential growth or decay. Purely imaginary roots produce bounded oscillation in this simple second-order homogeneous case, but do not attract neighboring solutions.

Practice

Solve the characteristic polynomial $r^2-3r+2=0$ .
Write the family for $y''+9y=0$ .
What factor is needed for a repeated root $r=2$ ?

Show worked solutions

Roots $1,2$ give $c_1e^t+c_2e^{2t}$ .
$c_1\cos3t+c_2\sin3t$ .
The second solution is $te^{2t}$ , independent of $e^{2t}$ .

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

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