Lesson · Second-Order Linear Equations · The characteristic equation

The Characteristic Equation

Guessing the solution of a second-order equation does not scale, but one guess does: y = e^(rt). Its rate and the rate of its rate are multiples of itself, so the equation collapses to a quadratic in r after the exponential is divided out. The quadratic is called the characteristic equation, its discriminant names the kind of root, and the next three lessons turn each kind of root into a solution.

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y = e^(rt), y′ = r·e^(rt), y″ = r²·e^(rt) (a·r² + b·r + c)·e^(rt) = 0 e^(rt) is never 0, so a·r² + b·r + c = 0 disc = b² − 4ac picks the kind of root roots: −b ± √disc, over 2a

One substitution, one quadratic

Core ideas

What this lesson establishes

Three ideas: why the exponential is the guess, why the equation becomes a quadratic, and what the discriminant announces.

The exponential is the function whose rates are copies of itself

The rate of e^(rt) is r·e^(rt), a number times the same function, and the rate of that is r²·e^(rt). An equation that adds multiples of y, y′ and y″ therefore adds multiples of one function, and that is the only kind of guess for which the sum can collapse.

The exponential factors out, and a quadratic remains

Substituting gives a·r²·e^(rt) + b·r·e^(rt) + c·e^(rt) = 0, which is (a·r² + b·r + c)·e^(rt) = 0. Because e^(rt) is never zero, the bracket has to be, and that is the characteristic equation. It is an equation in the number r, with no t in it.

The discriminant says which of three things happens

b² − 4ac positive means two real roots, zero means one repeated root, negative means a complex pair. A positive discriminant that is not a perfect square gives roots with a surd in them, such as (1 ± √5)/2; the lab prints the surd and does not round it.

The material

From the equation to its quadratic

The substitution in full, the quadratic formula from Algebra, and the four presets that show each kind of root.

Separable Equations, Growth and Decay met y′ = k·y and its solution e^(kt), and the previous course leaned on that exponential at every step. Here the equation is a·y″ + b·y′ + c·y = 0, with a, b and c constants and a not zero, and the same function is the natural thing to try. The only unknown is the number r in the exponent.

y = e^(rt) y′ = r·e^(rt) y″ = r²·e^(rt) a·r²·e^(rt) + b·r·e^(rt) + c·e^(rt) = 0 (a·r² + b·r + c)·e^(rt) = 0

Theorem · The characteristic equation

The function e^(rt) solves a·y″ + b·y′ + c·y = 0 exactly when r solves a·r² + b·r + c = 0.

From the substitution above, e^(rt) solves the equation exactly when (a·r² + b·r + c)·e^(rt) is zero for every t.

The factor e^(rt) is positive for every t, so the product is zero exactly when the bracket is. The bracket has no t in it, so it is zero exactly when r is a root.

□

The rate of the exponential, (e^(rt))′ = r·e^(rt), is the claim Rates of Change and the Derivative demonstrated with rounded quotients in “The Exponential and Its Rate”; the proof above uses it and nothing else new. What is left is the algebra of a quadratic, which is “The Quadratic Formula” in Algebra's Quadratics and Complex Numbers.

r = (−b ± √(b² − 4ac)) / (2a) disc = b² − 4ac

The four presets

Each preset is an equation and its quadratic. The lab prints the discriminant, the kind of root and the roots, all exactly. For y″ + 3y′ + 2y = 0 the discriminant is 9 − 8 = 1, a perfect square, so the roots are rational. For y″ + 4y′ + 4y = 0 it is 16 − 16 = 0, one root twice. For y″ + 2y′ + 5y = 0 it is 4 − 20 = −16, so the roots are the complex pair −1 ± 2i.

Example · A discriminant that is positive and not a square

For y″ − y′ − y = 0 the discriminant is 1 + 4 = 5. Both roots are real, and they are (1 ± √5)/2. The lab prints that surd exactly.

Rounded, the roots are ≈ 1.61803 and ≈ −0.618034; the rounded numbers are for the eye only, and the exact surd is what the next lessons would carry into a solution.

Every equation of this form has a characteristic quadratic, and every root of it gives one exponential solution. What is still to decide is how to build the full set of solutions from the roots, and that depends on the kind of root: one lesson each, starting with two real roots.

Interactive

The characteristic equation

Roots, the general solution and the constants, exactly

The characteristic quadratic of a·y″ + b·y′ + c·y = 0

Choose a preset, or type a, b and c. The lab prints the discriminant, the kind of root and the roots exactly. Before each preset, compute b² − 4ac yourself and predict the kind.

Equation

a (coefficient of y″)

b (coefficient of y′)

c (coefficient of y)

Initial values y(0) y′(0) (may be empty)

View

Discriminant b² − 4ac—

Kind of roots—

Roots—

General solution—

Constants—

Solution—

Wronskian W(0)—

Three rational coefficients; the initial values are two numbers, y(0) then y′(0).

Method

From an equation to its roots

Five moves. The substitution is always the same, so the work is in the quadratic.

Read a, b and c with their signs

Take them from a·y″ + b·y′ + c·y = 0. For y″ − y′ − y = 0 they are 1, −1 and −1; a missing term has coefficient 0.

Write the characteristic equation

Replace y″ by r², y′ by r and y by 1, keeping the coefficients: a·r² + b·r + c = 0, with a zero on the right.

Compute the discriminant

b² − 4ac. Its sign, and whether it is a perfect square, name the kind of root before any root is found.

Apply the quadratic formula

r = (−b ± √disc)/(2a). Keep a surd as a surd, and a negative discriminant as an i.

Check one root

Put it back into a·r² + b·r + c. It should be 0 exactly.

Substituting, dividing and classifying

Score 0 / 0

Choose the answer you can justify.

y″ + 3y′ + 2y = 0 by substitution

Substitute y = e^(rt), divide out the exponential, and solve.

y = e^(rt), y′ = r·e^(rt), y″ = r²·e^(rt) (r² + 3r + 2)·e^(rt) = 0 e^(rt) is never 0, so r² + 3r + 2 = 0 disc = 9 − 8 = 1 r = (−3 ± 1)/2 r = −1 or r = −2

Each root gives a solution: e^(−t) and e^(−2t). Check r = −1 in the quadratic: 1 − 3 + 2 = 0; check r = −2: 4 − 6 + 2 = 0.

The discriminant 1 is a perfect square, and the lab confirms the kind as two real roots. Putting the two solutions together into one answer is the work of the next lesson.

Common mistakes

Writing the characteristic equation as a·r² + b·r + c = yThe right side of the characteristic equation is zero, not y. For y″ + 3y′ + 2y = 0 and r = −1 the quadratic r² + 3r + 2 is 0, but y = e^(−t) is not 0. Setting the quadratic equal to y would ask a number to equal a function of t, and no r does that for every t. The bracket is multiplied by e^(rt), and it is the product that equals the zero on the right.

Reading the signs of b and c wronglyFor y″ − 5y′ + 6y = 0 the coefficients are 1, −5 and 6, and the roots are 2 and 3. Reading b as +5 solves r² + 5r + 6 = 0 instead and gives −2 and −3. Put −2 back into the true quadratic and it gives (−2)² − 5·(−2) + 6 = 20, not 0. One root put back into the quadratic catches the slip.

Taking a positive discriminant to mean rational rootsThe discriminant of y″ − y′ − y = 0 is 5, which is positive, but the roots (1 ± √5)/2 are irrational. Only a perfect-square discriminant gives rational roots. The lab prints the kind as “two real roots (irrational)” for exactly this case.

Check your understanding

✓

Finish when you can turn a·y″ + b·y′ + c·y = 0 into its quadratic and name the kind of root before you solve it.

You should be able to substitute y = e^(rt), obtain a·r² + b·r + c = 0, compute the discriminant, classify the roots as two real, one repeated or a complex pair, and solve exactly, keeping a surd as a surd.

Note: The next lesson takes the first kind of root, two real roots that differ, and turns them into C₁·e^(r₁t) + C₂·e^(r₂t) with the constants fitted from the starting value and the starting rate.