Separable ODEs — technique, implicit solutions: Visual walkthrough

We are going to derive, from nothing, $\int \frac{d y}{h ( y )} = \int g (x) d x + C .$ But right now that line is just marks on paper. Let us build every mark.

Step 1 — What is $\frac{d y}{d x}$ , really? (a slope on a picture)

WHAT. We have a mystery curve $y (x)$ : for every input $x$ on the horizontal axis, it hands back a height $y$ on the vertical axis. The symbol $\frac{d y}{d x}$ is the slope of that curve — how steeply the height climbs as you nudge $x$ a tiny bit to the right.

WHY. An ODE is a rule that tells you the slope at every point, and asks you to reconstruct the curve. Before we can "solve" anything we must see that $\frac{d y}{d x}$ is a geometric object: the steepness of a line just barely kissing the curve.

PICTURE. In the figure, the black curve is $y (x)$ . Pick a point on it. The lavender line touches at exactly that point — its steepness (rise $d y$ over run $d x$ ) is $\frac{d y}{d x}$ .

Figure — Separable ODEs — technique, implicit solutions

Step 2 — What "separable" looks like: a slope field that factors

WHAT. A separable ODE says the slope at a point $(x, y)$ is a product: $slope here \frac{d y}{d x} = depends only on how far right g (x) \times depends only on how high h (y) .$

WHY. Most ODEs mix $x$ and $y$ hopelessly (like $x + y$ — a sum, which won't split). The magic of a product $g (x) h (y)$ is that later we can shepherd every $x$ to one side and every $y$ to the other. The factoring is exactly what makes separation possible.

PICTURE. At every point we draw a tiny slope-tick. In a separable field, the tick's steepness is "an $x$ -ingredient times a $y$ -ingredient." The mint ticks below all share the same height $y$ , so their steepness differs only through $g (x)$ ; the coral ticks all share the same $x$ , so they differ only through $h (y)$ .

Step 3 — Divide by $h (y)$ : herding $y$ to the left

WHAT. Divide both sides by $h (y)$ (we must assume $h (y) \neq = 0$ — hold that thought for Step 7): $pure- y weight \frac{1}{h ( y )} the slope \frac{d y}{d x} = pure x g (x) .$

WHY. We want the left side to contain only $y$ -stuff (and the slope machinery), and the right side to contain only $x$ -stuff. Dividing by $h (y)$ sweeps the last $y$ off the right side onto the left. Now the two variables live on opposite banks of the river.

PICTURE. Imagine two bins. Before dividing, the $y$ -ingredient $h (y)$ sat on the right, tangled with $g (x)$ . After dividing, it flips to the left as $1/ h (y)$ . Left bin: only $y$ . Right bin: only $x$ .

Step 4 — Integrate both sides over $x$ : summing the slopes

WHAT. Apply $\int (\dots) d x$ to both sides: $\int left bin \frac{1}{h ( y )} \frac{d y}{d x} d x = \int right bin g (x) d x .$

WHY. The ODE fixes the slope at every $x$ . Integration is the tool that stitches infinitely many tiny slopes into an actual curve — it is the exact inverse of "take the slope." Since both sides are equal functions of $x$ , their integrals over $x$ are equal (up to one constant). This is the only legal move that turns slope information back into height information.

PICTURE. The right side is literally the shaded area under $g (x)$ : chop the $x$ -axis into thin strips, each strip's area is $g (x) d x$ , and $\int$ adds them up. That accumulated area is what pins down the curve.

Step 5 — The chain rule, run backwards: collapsing the left side

WHAT. Let $H$ be an antiderivative of $\frac{1}{h}$ , meaning $H^{'} (u) = \frac{1}{h ( u )}$ . The chain rule says $\frac{d}{d x} H (y (x)) = = 1/ h (y) H^{'} (y) \cdot inner slope \frac{d y}{d x} = \frac{1}{h ( y )} \frac{d y}{d x} .$ So the messy left integrand is nothing but the derivative of $H (y (x))$ . Therefore $\int \frac{1}{h ( y )} \frac{d y}{d x} d x = \int \frac{d y}{h ( y )} .$

WHY. This is the honest reason the " $d y / d x$ is a fraction" shortcut works. We never actually split a fraction; we recognised a chain-rule pattern and undid it. The whole $d x$ -integral quietly re-becomes a $d y$ -integral.

PICTURE. Read the chain rule as a two-gear machine: the outer gear $H$ turns at rate $H^{'} (y) = 1/ h (y)$ , the inner gear $y$ turns at rate $d y / d x$ . Their product is how fast $H (y (x))$ spins as $x$ moves. Reversing it (integrating) lands you back on $H (y)$ , an area measured in the $y$ -direction.

Step 6 — The master formula, and its picture as two areas

WHAT. Put Steps 4 and 5 together and attach one constant: $\int \frac{d y}{h ( y )} = \int g (x) d x + C$

Term by term:

$\int \frac{d y}{h ( y )}$ ::: accumulate the $y$ -weight $1/ h (y)$ as $y$ changes — an area in the $y$ -world, call it $H (y)$ .
$\int g (x) d x$ ::: accumulate $g (x)$ as $x$ changes — an area in the $x$ -world, call it $G (x)$ .
$C$ ::: one number setting the height offset between the two worlds.

WHY only one $C$ ? Each indefinite integral is fixed only up to a constant, but you can bundle both leftover constants into a single $C$ by moving one across the equals sign. Two constants would be redundant.

PICTURE. Left area equals right area plus a fixed gap $C$ . Choosing $C$ (via a starting point like $y (0) = 3$ ) is just sliding one area up or down until the two balance at your known point.

Step 7 — The forbidden move: $h (y) = 0$ and the lost horizontal solutions

WHAT. In Step 3 we divided by $h (y)$ — illegal wherever $h (y) = 0$ . If $h (y_{0}) = 0$ for some constant $y_{0}$ , then the flat line $y \equiv y_{0}$ has slope $0$ , and $\frac{d y}{d x} = 0 = g (x) \cdot h (y_{0}) = g (x) \cdot 0 = 0. ✓$ So $y \equiv y_{0}$ is a genuine solution — but the division erased it.

WHY. Dividing by zero is undefined, so the algebra silently threw those flat solutions overboard. The master formula only captures the curved family; the flat equilibrium solutions must be added back by hand.

PICTURE. For the logistic $h (y) = y (1 - y)$ , the roots are $y = 0$ and $y = 1$ : two flat coral lines. The lavender S-curves from the formula flow between them but never include them. You must report the flat lines separately.

Step 8 — Two endings: explicit vs implicit (why you sometimes stop early)

WHAT. After integrating you hold a relation $H (y) = G (x) + C$ .

If you can algebraically isolate $y$ , you get an explicit solution $y = (stuff in x)$ .
If isolating $y$ needs a $\pm$ or a mess, leave it as an implicit solution $F (x, y) = C$ — still perfectly valid.

WHY. Solving for $y$ is a bonus, not a requirement. The relation already defines the curve; whether you can rearrange it is a separate algebra question. Forcing an isolation can introduce sign errors.

PICTURE. Two curves side by side: one that a rearrangement flattens into a clean $y = f (x)$ (explicit), and one — like $y^{2} + y - x^{2} = C$ — that stays a knotted contour you read off as a level set (implicit).

The one-picture summary

WHAT. Everything at once: a product slope-field (Step 2) → herd $y$ left (Step 3) → integrate each bank (Steps 4–5) → equal areas offset by $C$ (Step 6), with the flat equilibrium lines (Step 7) drawn on top and a label for whether the ending is explicit or implicit (Step 8).

Recall Feynman retelling — the whole walkthrough in plain words

An ODE is a rule that whispers the steepness of a hidden curve at every spot. When that steepness happens to be "an $x$ -flavour times a $y$ -flavour," we can sort the mess: shove every $y$ -ingredient to the left, every $x$ -ingredient to the right. Now each side is a pile of tiny steepnesses. Integration is the glue that stacks those tiny steps back into real heights — and thanks to the chain rule run in reverse, the left pile quietly measures area in the $y$ -direction while the right pile measures area in the $x$ -direction. Set the two areas equal, allow one fudge number $C$ for the vertical offset, and you're done — almost. Dividing by the $y$ -flavour secretly deleted any flat line where that flavour was zero, so we hunt down those roots and staple the flat solutions back on. Finally, if the leftover equation untangles into " $y =$ something," great — explicit; if not, we leave it knotted as $F (x, y) = C$ — implicit, and just as true.

Connections

Separable ODEs — Technique & Implicit Solutions (parent)
First-Order ODEs — Overview
Initial Value Problems (choosing $C$ = sliding the area gap)
Partial Fractions (splitting $1/ h (y)$ before integrating)
The Logistic Equation (the equilibrium picture in Step 7)
Exact ODEs (implicit solutions, generalized)
Substitution Method — Homogeneous ODEs
Integrating Factor & Linear ODEs