The Characteristic Equation

Guess an exponential and the calculus becomes algebra.

12 problems · about 24 minutes · A-REI.B.4, F-IF.B.6

What this lesson teaches

The student solves constant-coefficient equations using the characteristic equation.

Guessing y = e^(rt) turns a constant-coefficient equation into a polynomial in r.
Distinct real roots give two exponentials; a repeated root needs an extra factor of t.
Complex roots give oscillation, with the real part setting growth or decay.

Straightforward practice. Get the method working first.

5 problems

y″ + 5y′ + 4y = 0. The roots are −1 and what other value?
Answer -4
Why −4.
What do complex characteristic roots produce?
Answer Oscillation, with the real part setting growth or decay.
Why Oscillating solutions.
y″ + 3y′ + 2y = 0. The roots are −1 and what other value?
Answer -2
Why They sum to −3.
y″ − 4y = 0. The roots are 2 and what other value?
Answer -2
Why r² = 4.
y″ + y = 0. Are the roots real? 1 yes, 0 no.
Answer 0
Why r² = −1.

The same ideas with more to keep track of.

3 problems

A repeated root r. What extra factor multiplies the second solution? Enter 1 for t, 2 for t squared.
Answer 1
Why A single factor of t.
y″ + 6y′ + 9y = 0. What is the repeated root?
Answer -3
Why A perfect square.
Complex roots with real part 0. Does the solution grow, decay or oscillate steadily? Enter 1 grow, 2 decay, 3 steady oscillation.
Answer 3
Why No exponential envelope.

Mixed problems. Work out what kind of question it is before you start.

4 problems

Match each root situation to the shape of the solution.
Answer Two distinct negative real roots → A sum of two decaying exponentials; A repeated root → An exponential and t times that exponential; Complex roots → Oscillation, possibly inside an exponential envelope
Why Complex roots are the only ones that oscillate.
y″ − 9y = 0. The roots are −3 and what other value?
Answer 3
Why r² = 9.
The Roots: y″ + 7y′ + 12y = 0. The roots are −3 and what other value?
Answer -4
Why −4.
The Long Run: Both characteristic roots are negative. What value does the solution approach as t grows large?
Answer 0
Why 0.