Separable equations

Chapter 4 - Put each variable on its own side

A first-order equation is separable when its rate factors into a time-only part and a state-only part:

$y'=a(t)b(y).$

The method reverses the chain rule. It rewrites the equation so one integral uses $y$ and the other uses $t$ .

The fingerprint and the move

See

$\dfrac{dy}{dt}=a(t)b(y)$

$\dfrac{1}{b(y)}\,dy=a(t)\,dt$

integrate both sides, then use the initial condition

Before dividing by $b(y)$ , solve $b(y)=0$ . Every root may give an equilibrium solution that division would erase.

Equilibrium safeguard

Write "lost-solution check: $b(y)=0$ " above the first line of every separation. This five-second habit prevents one of the most common errors in the course.

Worked example: full support

Solve $y'=ty$ , $y(0)=2$ fully worked

Recognize. The rate is $a(t)b(y)$ with $a(t)=t$ and $b(y)=y$ .
Protect equilibria. $b(y)=0$ gives $y=0$ , an equilibrium of the differential equation. It does not meet $y(0)=2$ , but it belongs to the general solution set.
Separate. On a region where $y\ne0$ , $\dfrac{1}{y}\,dy=t\,dt$ .
Integrate. $\ln|y|=t^2/2+C$ .
Expose the family. $|y|=e^Ce^{t^2/2}$ . Absorb sign and positive magnitude into one nonzero constant: $y=Ce^{t^2/2}$ . Including $C=0$ restores the equilibrium.
Use the condition. $2=C$ , so $\boxed{y=2e^{t^2/2}}$ .
Check. $y'=2te^{t^2/2}=ty$ , and $y(0)=2$ . The curve decreases for $t\lt 0$ , has zero slope at 0, and increases for $t\gt 0$ , exactly as $ty$ predicts for positive $y$ .

Why the absolute value disappears carefully

From $\ln|y|=F(t)+C$ , exponentiation gives $|y|=Ae^{F(t)}$ with $A\gt 0$ . A solution stays either positive or negative on the interval where the derivation holds. Write $y=Ce^{F(t)}$ , where $C\ne0$ can have either sign. Then check whether $C=0$ restores an equilibrium.

Erasing $|\cdot|$ without accounting for sign can delete negative solutions.

Fade the support backward

Solve $y'=(1+t)(1+y^2)$ , $y(0)=0$ .

Given

$\dfrac{dy}{1+y^2}=(1+t)dt$

$\arctan y=t+t^2/2+C$

Complete

Use $y(0)=0$ to find $C=\underline{\hspace{2em}}$ .

Apply tangent: $y=\underline{\hspace{8em}}$ .

Verify

Differentiate your function and name the interval containing 0 on which it remains finite.

Check the completion

$C=0$ , so $y=\tan(t+t^2/2)$ . The argument has minimum $-1/2$ , so it never reaches $-\pi/2$ . Both nearest poles instead occur where $t+t^2/2=\pi/2$ , at $t=-1\pm\sqrt{1+\pi}$ . The maximal interval containing 0 is $(-1-\sqrt{1+\pi},-1+\sqrt{1+\pi})$ . Substitution confirms $y'=(1+t)\sec^2(t+t^2/2)=(1+t)(1+y^2)$ .

Definite integrals keep conditions attached

Instead of adding constants and solving for them later, you can incorporate $y(t_0)=y_0$ directly:

$\int_{y_0}^{y(t)}\frac{1}{b(u)},du=\int_{t_0}^{t}a(s),ds.$

The dummy variables $u$ and $s$ prevent the same symbol from serving as both a limit and an integration variable. This form often reduces constant errors and naturally produces an implicit solution, an equation relating the variables without isolating the state.

Implicit can be finished

For $y'=t/(1+y^2)$ with $y(0)=1$ , definite integration gives

$\int_1^{y(t)}(1+u^2)du=\int_0^t s,ds,$

$y+\frac{y^3}{3}=\frac{t^2}{2}+\frac43.$

Solving the cubic explicitly would make the answer harder to read. The implicit equation already defines the solution locally and is easy to differentiate for verification.

Lost solution check. Separating $y'=y(1-y)$ requires division by $y(1-y)$ . Which solutions must be recorded first?

Error clinic

Incorrect:

y'=t+y \Rightarrow dy/y=t\,dt+dt

Division by $y$ gives $y'/y=t/y+1$ , not $t+1$ . A sum does not separate unless the entire right side factors into a time-only part times a state-only part.

Incomplete:

\int y^{-1}dy=\ln y

The real antiderivative is $\ln|y|$ on intervals avoiding zero. If an initial condition forces $y\gt 0$ , say so; otherwise preserve both signs.

Practice

Practice - core move
4.1 Solve $y'=3t^2/y$ , $y(0)=2$ . State the branch selected by the initial condition and verify it.
Detailed solution
Practice - equilibrium safeguard
4.2 Find all equilibrium and non-equilibrium solutions of $y'=y(y-2)$ . Then solve the initial-value problem $y(0)=1$ .
Detailed solution
Integrated - modeling
4.3 A culture grows at rate $P'=0.6P(1-P/500)$ , with $P(0)=50$ . Separate and solve. Use the equation, not only the formula, to predict the long-run population.
Detailed solution
Challenge - interval
4.4 Solve $y'=1+y^2$ , $y(0)=1$ . Find the maximal open interval containing zero on which the solution exists.
Detailed solution
Practice - extension
4.5 Solve $y'=2ty^2$ , $y(0)=-2$ . Include the lost-solution check, state the maximal interval containing zero, and explain why the solution stays negative even though its derivative changes sign.
Detailed solution
Challenge - transfer
4.6 Solve $y'=y^{3/2}$ , $y(0)=4$ , for real nonnegative $y$ . State the maximal interval containing zero and verify the branch restriction before squaring. Explain why the resulting rational expression is not a continuation of this solution past its pole.
Detailed solution

Spaced retrieval 2

Without notes:

rebuild the three-pass verification from Chapter 1;
state the horizontal-row fingerprint from Chapter 2; and
write the equilibrium safeguard for separation.

I solved the faded example on paper and retrieved the three earlier ideas.

Next: Use an integrating factor to turn a linear equation into one product derivative.

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Differential Equations as a Second Language