Separable ODEs — technique, implicit solutions: Foundations

Before you can trust a single line of the parent note, you need every symbol it uses to feel obvious. Below, in build order, is the whole toolkit — plain words, then the picture, then why the topic needs it.

0. What is a function $y (x)$ , and what is a curve?

The picture below is the mental image behind every symbol on this page. When the parent note says "solve for $y$ ," it means: find the curve.

Figure — Separable ODEs — technique, implicit solutions

Why the topic needs it: an ODE never hands you $y$ directly. It hands you a clue about the curve's steepness and asks you to reconstruct the whole curve. So you must first be comfortable that $y$ is a curve you are hunting for.

1. The slope of a curve, and the symbol $\frac{d y}{d x}$

We write this slope as $\frac{d y}{d x}$ , read "dee-y by dee-x." The $d$ means "a tiny change in." So $\frac{d y}{d x}$ literally reads as "a tiny change in $y$ divided by the tiny change in $x$ that caused it" — a ratio of two tiny steps.

Why the topic needs it: an ODE is by definition an equation involving $\frac{d y}{d x}$ . No slope symbol, no ODE.

2. What "ODE" and "first-order" mean

Think of it as a local instruction: "at any point $(x, y)$ you land on, here is how steep you must be leaving it." A slope field draws that instruction as a tiny dash at many points; a solution is any curve that always flows along the dashes.

Why the topic needs it: this is the whole game. The parent note is a recipe for turning that field of tiny instructions back into an actual curve. See First-Order ODEs — Overview for the wider family.

3. Multiplication that factors: the meaning of $g (x) h (y)$

$x^{2} y = g (x) x^{2} \cdot h (y) y$ ✓ splits.
$x + y$ ✗ — a sum glues $x$ and $y$ together; you cannot peel them apart into a product.
$e^{x + y} = e^{x} \cdot e^{y}$ ✓ — because a sum in the exponent becomes a product (law of indices).

Why the topic needs it: "separable" is literally defined by this factoring. If $\frac{d y}{d x} = g (x) h (y)$ , the $x$ -stuff and $y$ -stuff were never truly mixed — they just looked mixed. That is the door the method walks through.

4. The integral sign $\int$ — "undo the slope"

The picture: slope-taking chops a curve into its steepness at each point; integrating tapes those pieces back into a curve. They undo each other, like cutting and gluing.

Why one lonely $+ C$ appears. Shifting a whole curve straight up by any amount does not change its steepness anywhere — a flat road is flat at sea level or on a plateau. So "the function whose slope is $f$ " is only pinned down up to a vertical shift. That unknown shift is the arbitrary constant $C$ .

Why the $ln ∣ y ∣$ (with bars)? The bars mean absolute value — strip the minus sign, keep size only. We need them because $ln$ only accepts positive inputs, yet $y$ might be negative. Writing $ln ∣ y ∣$ covers both signs of $y$ in one stroke, and that hidden sign is exactly what later gets absorbed into the constant.

Why the topic needs it: the method is "separate, then integrate both sides." No integral sign, no finish.

5. The natural log $ln$ and its partner $e^{x}$ — the undo pair

Whenever we integrate a $\frac{1}{y}$ we land on $ln ∣ y ∣$ ; to peel $y$ back out we hit both sides with $e^{(\dots)}$ . That handoff is exactly how Worked Example 1 goes from $ln ∣ y ∣ = \frac{x ^{3}}{3} + C$ to $y = A e^{x^{3} /3}$ .

Why the topic needs it: almost every separable ODE with a $y$ in a denominator produces a $ln$ , and $e^{x}$ is the only tool that unwraps it.

6. Why splitting $\frac{d y}{d x}$ is legal: the chain rule, run backwards

Read that right to left. The messy-looking left side of the method, $\int \frac{1}{h ( y )} \frac{d y}{d x} d x$ , is exactly the chain-rule pattern $H^{'} (y) \frac{d y}{d x}$ with $H^{'} (y) = \frac{1}{h ( y )}$ . So the whole $\frac{d y}{d x} d x$ collapses to a clean $d y$ : $\int \frac{1}{h ( y )} \frac{d y}{d x} d x = \int \frac{d y}{h ( y )} .$ That is the honest reason we're allowed to write $\frac{1}{h ( y )} d y = g (x) d x$ and integrate each side in its own variable.

Why the topic needs it: it's the load-bearing beam. Without it, "treat $\frac{d y}{d x}$ as a fraction" is a magic trick; with it, it's a theorem. This links straight to Exact ODEs, where the same differential-splitting idea generalises.

7. Explicit vs implicit, and the constant $C$

The relation $F (x, y) = C$ is a curve (or family of curves) drawn by the equation itself. Fixing $C$ using a starting fact — a condition like $y (0) = 3$ — selects one curve from the family. That "fix $C$ from a starting fact" step is the heart of Initial Value Problems.

Why the topic needs it: many honest answers refuse to be solved for $y$ . Recognising an implicit answer as finished saves you from chasing an isolation that doesn't exist.

y (x)

and its graph are

A rule sending each

x

to one

y

; its graph is the curve of points

(x, y)

I can explain what

\frac{d y}{d x}

measures

The slope/steepness: tiny rise

d y

over the tiny run

d x

that caused it.

I know what "first-order ODE" means

An equation linking

y

to its first slope only, viewable as a slope field.

I can tell if an expression factors as

g (x) h (y)

It splits into (pure-

x

)

\times

(pure-

y

); sums like

x + y

do NOT.

I know integration is the reverse of taking a slope

\int f d x

finds the function whose slope is

f

, up to a constant.

I can say why exactly one

+ C

appears

Vertical shifts don't change slope, so the answer is fixed only up to one shift

C

I know why

\int \frac{d y}{y} = ln ∣ y ∣ + C

needs the bars

ln

needs positive input;

∣ y ∣

covers both signs, freeing a

\pm

into

C

I can cancel

ln

with

e^{x}

e^{l n y} = y

and

ln (e^{x}) = x

; they undo each other.

I can state the chain rule backwards

\int \frac{1}{h ( y )} \frac{d y}{d x} d x = \int \frac{d y}{h ( y )}

— that justifies the "fraction split."

I can distinguish explicit from implicit solutions

Explicit isolates

y

; implicit leaves

F (x, y) = C

tangled but valid.

I know what an equilibrium solution is

A flat

y \equiv y_{0}

where

h (y_{0}) = 0

; lost when dividing by

h (y)

Connections

Parent: Separable ODEs Technique
First-Order ODEs — Overview
Integrating Factor & Linear ODEs
Exact ODEs
Substitution Method — Homogeneous ODEs
Partial Fractions
Initial Value Problems
The Logistic Equation

Foundations — Separable ODEs — technique, implicit solutions