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 $t<0$, has zero slope at 0, and increases for $t>0$, 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>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 maximal interval around 0 ends where $t+t^2/2=\pm\pi/2$. 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:

$\int_{y_{0}}^{y (t)} \frac{1}{b ( u )} d u = \int_{t_{0}}^{t} a (s) d s .$

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.

Implicit can be finished

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

$\int_{1}^{y (t)} (1 + u^{2}) d u = \int_{0}^{t} s d s,$

$y + \frac{y ^{3}}{3} = \frac{t ^{2}}{2} + \frac{4}{3} .$

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 $y>0$, 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

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