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Differential Equations · Second-Order Equations

Characteristic Roots — Practice Problems and Worked Solutions

Turn constant-coefficient ODEs into algebra.

The formulas you need here

Characteristic equation

a y^{''} + b y^{'} + cy = 0 \Rightarrow a r^{2} + b r + c = 0

What it says: Guess y = e^{rx} and the differential equation becomes an ordinary quadratic.
When to reach for it: Second-order linear equations with constant coefficients.
Watch out: Three cases: distinct real roots give two exponentials, a repeated root needs an extra factor of x, and complex roots a ± bi give e^{ax}(C₁cos bx + C₂sin bx).

y^{''} - 3 y^{'} + 2 y = 0 \Rightarrow r = 1, 2 \Rightarrow y = C_{1} e^{x} + C_{2} e^{2 x}

Two distinct real roots, so two independent exponentials.

17 worked problems

Every step is generated and checked by a computer algebra system.

1.
Find the distinct characteristic roots.
$1 y^{''} + (- 3) y^{'} + (2) y = 0$
Answer: ${1, 2}$
Step-by-step solution
1. $1 r^{2} + (- 3) r + (2) = 0$
2. $r \in {1, 2}$
2.
Find the distinct characteristic roots.
$1 y^{''} + (1) y^{'} + (- 2) y = 0$
Answer: ${- 2, 1}$
Step-by-step solution
1. $1 r^{2} + (1) r + (- 2) = 0$
2. $r \in {- 2, 1}$
3.
Find the distinct characteristic roots.
$1 y^{''} + (- 5) y^{'} + (6) y = 0$
Answer: ${2, 3}$
Step-by-step solution
1. $1 r^{2} + (- 5) r + (6) = 0$
2. $r \in {2, 3}$
4.
Find the distinct characteristic roots.
$2 y^{''} + (- 2) y^{'} + (- 4) y = 0$
Answer: ${- 1, 2}$
Step-by-step solution
1. $2 r^{2} + (- 2) r + (- 4) = 0$
2. $r \in {- 1, 2}$
5.
Find the distinct characteristic roots.
$1 y^{''} + (4) y^{'} + (4) y = 0$
Answer: ${- 2}$
Step-by-step solution
1. $1 r^{2} + (4) r + (4) = 0$
2. $r \in {- 2}$
6.
Find the distinct characteristic roots.
$1 y^{''} + (- 2) y^{'} + (5) y = 0$
Answer: ${1 - 2 i, 1 + 2 i}$
Step-by-step solution
1. $1 r^{2} + (- 2) r + (5) = 0$
2. $r \in {1 - 2 i, 1 + 2 i}$
7.
Find the distinct characteristic roots.
$1 y^{''} + (- 4) y^{'} + (3) y = 0$
Answer: ${1, 3}$
Step-by-step solution
1. $1 r^{2} + (- 4) r + (3) = 0$
2. $r \in {1, 3}$
8.
Find the distinct characteristic roots.
$1 y^{''} + (2) y^{'} + (- 3) y = 0$
Answer: ${- 3, 1}$
Step-by-step solution
1. $1 r^{2} + (2) r + (- 3) = 0$
2. $r \in {- 3, 1}$
9.
Find the distinct characteristic roots.
$1 y^{''} + (- 7) y^{'} + (12) y = 0$
Answer: ${3, 4}$
Step-by-step solution
1. $1 r^{2} + (- 7) r + (12) = 0$
2. $r \in {3, 4}$
10.
Find the distinct characteristic roots.
$1 y^{''} + (5) y^{'} + (6) y = 0$
Answer: ${- 3, - 2}$
Step-by-step solution
1. $1 r^{2} + (5) r + (6) = 0$
2. $r \in {- 3, - 2}$
11.
Find the distinct characteristic roots.
$1 y^{''} + (- 6) y^{'} + (8) y = 0$
Answer: ${2, 4}$
Step-by-step solution
1. $1 r^{2} + (- 6) r + (8) = 0$
2. $r \in {2, 4}$
12.
Find the distinct characteristic roots.
$1 y^{''} + (0) y^{'} + (- 4) y = 0$
Answer: ${- 2, 2}$
Step-by-step solution
1. $1 r^{2} + (0) r + (- 4) = 0$
2. $r \in {- 2, 2}$
13.
Find the distinct characteristic roots.
$1 y^{''} + (- 1) y^{'} + (- 6) y = 0$
Answer: ${- 2, 3}$
Step-by-step solution
1. $1 r^{2} + (- 1) r + (- 6) = 0$
2. $r \in {- 2, 3}$
14.
Find the distinct characteristic roots.
$2 y^{''} + (- 6) y^{'} + (4) y = 0$
Answer: ${1, 2}$
Step-by-step solution
1. $2 r^{2} + (- 6) r + (4) = 0$
2. $r \in {1, 2}$
15.
Find the distinct characteristic roots.
$1 y^{''} + (6) y^{'} + (9) y = 0$
Answer: ${- 3}$
Step-by-step solution
1. $1 r^{2} + (6) r + (9) = 0$
2. $r \in {- 3}$
16.
Find the distinct characteristic roots.
$1 y^{''} + (- 2) y^{'} + (2) y = 0$
Answer: ${1 - i, 1 + i}$
Step-by-step solution
1. $1 r^{2} + (- 2) r + (2) = 0$
2. $r \in {1 - i, 1 + i}$
17.
Find the distinct characteristic roots.
$1 y^{''} + (3) y^{'} + (- 10) y = 0$
Answer: ${- 5, 2}$
Step-by-step solution
1. $1 r^{2} + (3) r + (- 10) = 0$
2. $r \in {- 5, 2}$

The formulas you need here

Characteristic equation

17 worked problems

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