Separable Equations and Lost Equilibria

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The idea

An equation is separable when $y'=g(t)h(y)$ . Away from zeros of $h$ , integrate $dy/h(y)=g(t)dt$ . The implicit result can be more useful than solving explicitly. An initial value fixes the integration constant.

Visual guide

VISUAL GUIDESeparation can reveal finite-time blow-up

y = 2/(1 − 2t)
Blow-up time t = 1/2

For y′ = y² with y(0) = 2, the solution 2/(1 − 2t) grows without bound as t approaches 1/2 from below. The plotted asymptote marks the end of this initial-value solution’s interval.

Method and assumptions

First list all roots of $h(y)=0$ : they give constant solutions that division would remove. Then integrate, apply the initial data, solve if convenient and locate any poles or domain boundaries. Verify the result in the original equation.

Worked example: decay

For $y'=-3y$ with $y(0)=4$ , integration gives $\ln|y|=-3t+C$ and hence $y=4e^{-3t}$ . The separate equilibrium $y=0$ is also valid for zero initial data.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Separating y′=y(1−y) by dividing by y(1−y) risks losing which solutions?

Only y≡0

All nonconstant solutions

y≡0 and y≡1

Hint 1 · Find a starting point

Division excludes zeros of the divisor.

Hint 2 · Take the next step

Check each excluded constant directly in the original equation.

Show the reasoning

Answer: y≡0 and y≡1

Both constants make y′=0 and y(1−y)=0, so both equilibrium solutions must be retained.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a finite interval

For $y'=y^2$ , $y(0)=2$ , $-1/y=t+C$ gives $y=2/(1-2t)$ . The maximal interval containing $0$ ends at $1/2$ . The equilibrium zero was excluded during division and must be recorded separately.

Interpreting the result

Absolute values in logarithms allow either sign before initial data select a branch. Integration constants cannot repair a lost equilibrium after an invalid division at that equilibrium.

Practice

Solve $y'=ty$ , $y(0)=3$ .
List equilibria of $y'=y(1-y)$ .
Solve $y'=-y^2$ , $y(0)=1$ for $t\ge0$ .

Show worked solutions

$\ln|y|=t^2/2+C$ , so $y=3e^{t^2/2}$ .
$y=0$ and $y=1$ .
$y=1/(1+t)$ ; it remains positive and tends to zero.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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