Linear Homogeneous 2nd-Order ODEs
Differential Equations · Section 3.1 · generated practice set
Characteristic roots, real and distinct, with initial conditions — and reading long-run behavior off the largest root.
Practice this set → Fresh numbers on every attempt. No account needed.
Try one now
Solve the initial value problem $$y'' - 4y' - 5y = 0,\qquad y(0) = -4,\quad y'(0) = -6,$$ and write it as $y = c_1e^{r_1t} + c_2e^{r_2t}$ with $r_1 < r_2$.
Solution
- Characteristic equation: $r^2 - 4r - 5 = 0$.
- Roots: $r = -1$ and $r = 5$ — real and distinct, so $y = c_1e^{-1t} + c_2e^{5t}$.
- $y(0) = c_1 + c_2 = -4$.
- $y'(0) = -1c_1 + 5c_2 = -6$.
- Solving the 2×2 system: $c_1 = -\frac{7}{3}$, $c_2 = -\frac{5}{3}$.
- At least one exponent is positive, so $|y| \to \infty$ as $t \to \infty$ (unless that coefficient happens to be 0).
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Method
For $ay'' + by' + cy = 0$ with constant coefficients, substituting $y = e^{rt}$ turns calculus into algebra.
$$ar^2 + br + c = 0$$
When the roots $r_1 \neq r_2$ are real, the general solution is $$y = c_1e^{r_1t} + c_2e^{r_2t}.$$
Initial value problems
$y(0) = \alpha$ and $y'(0) = \beta$ give the system $$c_1 + c_2 = \alpha, \qquad r_1c_1 + r_2c_2 = \beta,$$ which solves to $$c_1 = \frac{\beta - r_2\alpha}{r_1 - r_2}, \qquad c_2 = \frac{r_1\alpha - \beta}{r_1 - r_2}.$$ Deriving it each time is fine; recognizing it saves a minute.
Long-run behavior
The largest root dominates. Both roots negative → every solution decays to 0. Any positive root → typical solutions blow up. A zero root leaves a constant term behind.
Working backwards from a given general solution: the exponents are the roots, so rebuild $(r - r_1)(r - r_2)$ and read off $b$ and $c$.
Definitions and theorems
Worked example
Solve $y'' + y' - 6y = 0$, $y(0) = 1$, $y'(0) = -8$.
- Characteristic equation: $r^2 + r - 6 = 0$.
- Factor: $(r+3)(r-2) = 0 \Rightarrow r = -3, 2$.
- General solution: $y = c_1e^{-3t} + c_2e^{2t}$.
- $y(0) = c_1 + c_2 = 1$; $y'(0) = -3c_1 + 2c_2 = -8$.
- Subtract: $5c_1 = 10 \Rightarrow c_1 = 2$, so $c_2 = -1$. Thus $y = 2e^{-3t} - e^{2t}$, and $y \to -\infty$ because of the $e^{2t}$ term.
Common mistakes
- Dropping the leading coefficient. For $2y'' + 5y' - 3y = 0$ the characteristic equation is $2r^2 + 5r - 3 = 0$, not $r^2 + 5r - 3$.
- Differentiating wrong when applying $y'(0)$. $y' = r_1c_1e^{r_1t} + r_2c_2e^{r_2t}$ — each term keeps its own $r$.
- Assuming decay because one root is negative. One positive root is enough to blow up.
- Fractional roots. When $a \neq 1$ the roots are often fractions; keep them exact rather than decimalizing early.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.