Characteristic Equation
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 7.1: “Second-Order Linear Equations” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/7-1-second-order-linear-equations |
| Textbook used in class | Stewart, Calculus, Section 17.1: “Second-Order Linear Equations” (Examples 1-4) |
Quick Reference
For $ay'' + by' + cy = 0$ with constant $a, b, c$, try $y = e^{rx}$. Substituting gives the characteristic equation: $$ar^2 + br + c = 0.$$
| Discriminant $b^2 - 4ac$ | Roots | General solution |
|---|---|---|
| $> 0$ | $r_1 \neq r_2$ (real) | $y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ |
| $< 0$ | $r = \alpha \pm \beta i$ (complex) | $y = e^{\alpha x}(c_1\cos\beta x + c_2\sin\beta x)$ |
| $= 0$ | $r_1 = r_2 = r$ (repeated) | $y = (c_1 + c_2 x)\,e^{rx}$ |
Motivation
Exponential functions $e^{rx}$ are the natural trial solutions for constant-coefficient equations because they reproduce themselves under differentiation: $(e^{rx})'' = r^2 e^{rx}$, $(e^{rx})' = r e^{rx}$. Substituting into $ay''+by'+cy=0$ pulls out a factor of $e^{rx}$ and leaves the polynomial $ar^2+br+c=0$. Solving that polynomial completely determines all solutions of the DE.
Three Cases
Case 1 -- Two distinct real roots: $r_1$ and $r_2$ are real and unequal. Both $e^{r_1 x}$ and $e^{r_2 x}$ are solutions. Their Wronskian is $(r_2-r_1)e^{(r_1+r_2)x} \neq 0$, so they are linearly independent.
Case 2 -- Complex conjugate roots: $r = \alpha \pm \beta i$ with $\beta > 0$. The complex solutions $e^{(\alpha+\beta i)x}$ and $e^{(\alpha-\beta i)x}$ are valid, but Euler’s formula converts them to the real pair $e^{\alpha x}\cos\beta x$ and $e^{\alpha x}\sin\beta x$, which are real-valued and linearly independent.
Case 3 -- Repeated root: $r_1 = r_2 = r = -b/(2a)$. The equation gives only one exponential solution $e^{rx}$. The second solution is found by “reduction of order” to be $xe^{rx}$. (These are linearly independent because their Wronskian is $e^{2rx} \neq 0$.)
Worked Example
Solve $y'' - 5y' + 6y = 0$.
Characteristic equation: $r^2 - 5r + 6 = 0 \Rightarrow (r-2)(r-3) = 0 \Rightarrow r = 2, 3$.
Two distinct real roots. General solution: $$y = c_1 e^{2x} + c_2 e^{3x}.$$
Solve $y'' + 4y' + 13y = 0$.
Characteristic equation: $r^2 + 4r + 13 = 0$.
Discriminant: $16 - 52 = -36 < 0$. Quadratic formula: $r = \dfrac{-4 \pm 6i}{2} = -2 \pm 3i$.
So $\alpha = -2$, $\beta = 3$. General solution: $$y = e^{-2x}(c_1\cos 3x + c_2\sin 3x).$$
Solve $y'' - 6y' + 9y = 0$.
Characteristic equation: $r^2 - 6r + 9 = (r-3)^2 = 0 \Rightarrow r = 3$ (repeated).
General solution: $$y = (c_1 + c_2 x)\,e^{3x}.$$
for complex roots $\alpha \pm \beta i$, the solution is $c_1 e^{\alpha x}\cos\beta x + c_2 e^{\alpha x}\sin\beta x$, so the $e^{\alpha x}$ is included only if $\alpha \neq 0$. The $e^{\alpha x}$ factor is always present, even when $\alpha = 0$. When $\alpha = 0$, $e^{\alpha x} = e^0 = 1$, so the solution reduces to $c_1\cos\beta x + c_2\sin\beta x$. There is no special case -- the formula $e^{\alpha x}(c_1\cos\beta x + c_2\sin\beta x)$ covers both situations. Forgetting the $e^{\alpha x}$ when $\alpha \neq 0$ is a common sign that the student copied the $\alpha = 0$ special case instead of the general formula.
Common Misconceptions
the characteristic equation comes from differentiating the original DE with respect to $x$.
This is the action-view-of-function error. The characteristic equation $ar^2 + br + c = 0$ is obtained by substituting the trial solution $y = e^{rx}$ into $ay'' + by' + cy = 0$ and factoring out $e^{rx}$, which is never zero. The equation $ar^2 + br + c = 0$ is purely algebraic: it is a condition on the exponent $r$ that makes $e^{rx}$ a solution of the DE. No differentiation of the equation itself is involved; the derivation is entirely about finding which values of $r$ make the substitution work.
Leveled Practice
Problem 1. Solve $y'' + y' - 2y = 0$.
Show answer
$r^2 + r - 2 = (r+2)(r-1) = 0$, so $r = -2, 1$.
$y = c_1 e^{-2x} + c_2 e^{x}$.
Problem 2. Solve $y'' + 6y' + 9y = 0$.
Show answer
$r^2 + 6r + 9 = (r+3)^2 = 0$, so $r = -3$ (repeated).
$y = (c_1 + c_2 x)e^{-3x}$.
Problem 3. Solve $y'' + 9y = 0$ with $y(0) = 2$, $y'(0) = -3$.
Show answer
$r^2 + 9 = 0$, $r = \pm 3i$ ($\alpha = 0$, $\beta = 3$).
$y = c_1\cos 3x + c_2\sin 3x$.
$y(0) = c_1 = 2$. $y' = -3c_1\sin 3x + 3c_2\cos 3x$. $y'(0) = 3c_2 = -3 \Rightarrow c_2 = -1$.
$y = 2\cos 3x - \sin 3x$.