Review / Part I — Explicit & Qualitative Methods / Ch 2

Autonomous Equations

Ch 2 — Part I — Explicit & Qualitative Methods

An ODE of the form \(y' = f(y)\) — right-hand side depending only on \(y\), not on \(t\) — is called autonomous. These are the workhorses of qualitative analysis.

Time-translation invariance

If \(y(t)\) is a solution, so is \(y(t - t_0)\) for any shift \(t_0\). Slope fields are columns of identical vertical patterns.

Stationary solutions

A constant function \(y(t) \equiv c\) with \(f(c) = 0\) is called a stationary (or equilibrium) solution.

Separability for free

Every autonomous ODE is separable: \(\dfrac{dy}{f(y)} = dt\), integrable away from zeros of \(f\).

Practice Problems

Problem 1easy
Find all stationary solutions of \(y' = y(y-2)(y+1)\).
Hint
Solve \(f(y) = 0\).
Solution

The product is zero exactly at \(y = 0, 2, -1\).

Answer: \(y = -1, 0, 2\).

Problem 2medium
Solve \(y' = y^2\) with \(y(0) = 1\) using separation.
Hint
\(\int y^{-2} dy = \int dt\).
Solution

\(-y^{-1} = t + C\). From \(y(0) = 1\), \(C = -1\), so \(y(t) = 1/(1-t)\). Note the blow-up at \(t = 1\).

Answer: \(y(t) = 1/(1-t)\), valid for \(t < 1\).

Problem 3medium
Give an autonomous ODE whose only stable equilibrium is \(y = 5\).
Hint
Try \(y' = -(y-5)^3\) or \(y' = 5 - y\).
Solution

For \(y' = 5-y\), the only equilibrium is \(y=5\), and \(f'(5) = -1 < 0\), so stable.

Answer: \(y' = 5 - y\) works.

Problem 4easy
If \(y_1(t)\) solves the autonomous ODE and \(y_1(0) = 3\), how do you get the solution with \(y(7) = 3\)?
Hint
Time translation.
Solution

Set \(y(t) = y_1(t - 7)\); then \(y(7) = y_1(0) = 3\) and the ODE is unchanged.

Answer: \(y(t) = y_1(t-7)\).