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Introduction to ODEs & Direction Fields

Differential Equations · Section 1.1 · generated practice set

Read an autonomous equation without solving it: equilibria, a sign chart, stability, and slopes straight off the direction field.

Practice this set → Fresh numbers on every attempt. No account needed.

Method

Section 1.1 is about reading an ODE without solving it. For an autonomous equation $y' = f(y)$, the whole long-term story is in the sign of $f$.

The three-step read

  1. Find equilibria. Set $f(y) = 0$ and solve. These are the constant solutions.
  2. Sign chart. On each interval between equilibria, pick a test value and record whether $f > 0$ (solutions rise) or $f < 0$ (solutions fall).
  3. Classify. Arrows pointing in from both sides → asymptotically stable. Out from both sides → unstable. In on one side, out on the other → semistable.

The derivative test is the shortcut: $f'(y^*) < 0$ means stable, $f'(y^*) > 0$ means unstable, $f'(y^*) = 0$ means go back to the sign chart.

For a direction field, remember the slope at a point $(t,y)$ is just $f(t,y)$ — you evaluate, you do not integrate.

Definitions and theorems

Equilibrium solution
A constant function $y(t) = y^*$ solves $y' = f(y)$ exactly when $f(y^*) = 0$.
Stability by linearization
If $f(y^*) = 0$ and $f'(y^*) < 0$, then $y^*$ is asymptotically stable; if $f'(y^*) > 0$ it is unstable.

Worked example

Find and classify the equilibria of $y' = y^2 - 4y$.

  1. Set $y^2 - 4y = y(y-4) = 0$, so the equilibria are $y = 0$ and $y = 4$.
  2. $f'(y) = 2y - 4$.
  3. $f'(0) = -4 < 0$, so $y = 0$ is asymptotically stable.
  4. $f'(4) = 4 > 0$, so $y = 4$ is unstable.
  5. Sign check: on $(0,4)$ pick $y = 1$: $f(1) = -3 < 0$, solutions fall toward 0. Above 4, pick $y = 5$: $f(5) = 5 > 0$, solutions rise away. Consistent.

Common mistakes

  1. Solving instead of reading. You do not need the explicit solution to classify equilibria. Reaching for separation of variables here wastes half the exam period.
  2. Sign of $f'$, not sign of $f$. Stability is about how $f$ changes at the equilibrium, not its value there (which is always 0).
  3. Forgetting the equilibrium is a solution. "As $t\to\infty$, $y \to k$" is wrong if the initial value already equals $k$ — then $y \equiv k$ exactly.
  4. Semistable exists. A double root like $y' = (y-3)^2$ gives arrows in one side, out the other. Do not force every equilibrium into stable/unstable.

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.

Open 1.1 →

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