Separable ODEs — technique, implicit solutions

WHAT is a separable ODE?

WHY this matters: most ODEs are not solvable in closed form. Separable ones are the friendliest case — the workhorse you reach for first. If it separates, you're basically done after two integrals.

HOW the technique works (derive it, don't memorize)

We start from $\frac{dy}{dx} = g(x)\,h(y).$

Step 1 — Divide by $h(y)$ (assuming $h(y)\neq 0$ , remember this!): $\frac{1}{h(y)}\frac{dy}{dx} = g(x).$

Step 2 — Integrate both sides with respect to $x$ : $\int \frac{1}{h(y)}\frac{dy}{dx}\,dx = \int g(x)\,dx.$

Step 3 — Substitute on the left. The left integrand is exactly the chain-rule form: if $H(y)$ is an antiderivative of $1/h(y)$ , then $\frac{d}{dx}H(y(x)) = \frac{1}{h(y)}\frac{dy}{dx}$ . So the left side becomes $\int \frac{dy}{h(y)}$ : $\boxed{\int \frac{dy}{h(y)} = \int g(x)\,dx + C}$

Figure — Separable ODEs — technique, implicit solutions

Implicit vs explicit solutions

After integrating you get an equation relating $y$ and $x$ , like $H(y) = G(x) + C$ . Sometimes you can solve for $y$ (explicit). Often you can't cleanly — that's fine, leave it as an implicit solution.

Worked example Worked Example 2 — implicit answer (can't isolate

$y$ ) Solve $\dfrac{dy}{dx} = \dfrac{2x}{1+2y}$ .

Separate: $(1+2y)\,dy = 2x\,dx$ . Why? The $y$ -piece $1/(1+2y)$ is the reciprocal of the RHS denominator — multiply across. Integrate: $y + y^2 = x^2 + C$ . Why this on the left? $\int(1+2y)\,dy = y + y^2$ . Stop here. Solving $y^2+y-(x^2+C)=0$ for $y$ needs the quadratic formula and a $\pm$ ambiguity, so the clean answer is the implicit form: $y^2 + y - x^2 = C.$ Check (Why trust it?): differentiate implicitly: $2y\,y' + y' - 2x = 0 \Rightarrow y'(2y+1)=2x \Rightarrow y' = \frac{2x}{1+2y}$ . ✓

The trap everyone hits: lost solutions

Common mistake Dividing by

$h(y)$ can delete a solution The wrong-but-tempting move: "Just divide by $h(y)$ and integrate — done." Why it feels right: algebraically it's the natural first step, and it works for the generic curves. What breaks: if $h(y_0)=0$ for some constant $y_0$ , then $y\equiv y_0$ is itself a solution (since $dy/dx=0=g(x)\cdot 0$ ), but dividing by $h(y)$ erased it. These are called equilibrium / singular solutions. The fix: before dividing, list the roots of $h(y)=0$ and check each constant function separately. Example: In $y'=y(1-y)$ , the family $y=\frac{Ae^x}{1+Ae^x}$ misses $y\equiv 0$ and $y\equiv 1$ . Both are genuine solutions you must report.

Recall checkpoints

Recall What makes an ODE separable, and what's the master formula?

RHS must factor as $g(x)h(y)$ . Then $\displaystyle\int\frac{dy}{h(y)} = \int g(x)\,dx + C$ .

Recall Why is treating

$dy/dx$ as a fraction actually valid here? It's shorthand for the substitution rule (chain rule backwards) in $\int \frac{1}{h(y)}\frac{dy}{dx}dx = \int \frac{dy}{h(y)}$ .

Recall What did you possibly lose by dividing by

$h(y)$ , and how do you recover it? The equilibrium solutions $y\equiv y_0$ where $h(y_0)=0$ . Recover them by checking constants separately before dividing.

Recall Explain it to a 12-year-old (Feynman)

Imagine a recipe where all the apple steps and all the orange steps got mixed together. Separating the ODE is like sorting them: put every apple ( $y$ ) instruction on one page and every orange ( $x$ ) instruction on another. Then you "undo" each page by integration (the reverse of slicing), and tape the pages back together with one mystery number $C$ you figure out from a starting fact like "at the beginning there were 3 apples."

Flashcards

When is a first-order ODE separable?

When

dy/dx

factors as

g(x)\,h(y)

— pure-

x

times pure-

y

State the master solution formula for

dy/dx=g(x)h(y)

\int \frac{dy}{h(y)} = \int g(x)\,dx + C

Why may we "split"

dy/dx

like a fraction in a separable ODE?

It's shorthand for the substitution/chain-rule step

\int\frac{1}{h(y)}\frac{dy}{dx}dx=\int\frac{dy}{h(y)}

What is a singular/equilibrium solution and when does it arise?

A constant

y\equiv y_0

with

h(y_0)=0

; lost when you divide by

h(y)

Solve

dy/dx = x^2 y

y = A e^{x^3/3}

Solve

dy/dx=\frac{2x}{1+2y}

(implicit).

y^2+y-x^2=C

Solve the logistic

y'=y(1-y)

y=\frac{Ae^x}{1+Ae^x}

plus equilibria

y=0,\,y=1

How many arbitrary constants in a first-order separable solution?

Exactly one.

Is an implicit solution

F(x,y)=C

acceptable?

Yes — it's a valid solution even if

y

can't be isolated.

What's the SISC mnemonic?

Separate, Integrate both sides, Single constant, Check

h(y)=0

for lost solutions.

Connections

First-Order ODEs — Overview
Integrating Factor & Linear ODEs (the next tool when it's not separable)
Exact ODEs (implicit solutions generalize here)
Substitution Method — Homogeneous ODEs (turns some non-separable into separable)
Partial Fractions (needed in Example 3)
Initial Value Problems (fixing $C$ )
The Logistic Equation (Example 3 in the wild)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, separable ODE ka funda ekdum simple hai: agar $dy/dx$ ko aise likh sako ki ek taraf sirf $x$ wali cheez ho aur dusri taraf sirf $y$ wali cheez ( $dy/dx = g(x)\,h(y)$ ), toh game won. Bas saare $y$ ko left side bhej do, saare $x$ ko right side, aur dono sides ko integrate kar do. Yeh "fraction ki tarah todna" cheating nahi hai — actually yeh chain rule ko ulta chalane ka shortcut hai, isliye bilkul legal hai.

Answer do tarah ka aata hai. Kabhi tum $y=$ kuch... explicit form me nikal lete ho (jaise $y=3e^{x^3/3}$ ). Aur kabhi $y$ ko alag karna possible hi nahi hota — tab implicit solution $F(x,y)=C$ chhod do, woh bhi bilkul valid answer hai. Verify karne ke liye implicit differentiation kar lo, original ODE wapas aa jaani chahiye.

Sabse bada trap: jab tum $h(y)$ se divide karte ho, toh agar $h(y_0)=0$ hai kisi constant ke liye, toh $y\equiv y_0$ wala solution chup-chaap delete ho jaata hai! In ko equilibrium solutions kehte hain. Logistic equation $y'=y(1-y)$ me $y=0$ aur $y=1$ exactly yahi missing solutions hain. Isliye divide karne se pehle hamesha $h(y)=0$ ke roots check karo. Yaad rakhne ke liye mnemonic: SISC — Separate, Integrate, Single constant, Check lost solutions. Bas itna pakka karlo, separable ODE tumhare liye free marks hai.

Separable ODEs — technique, implicit solutions

WHAT is a separable ODE?

HOW the technique works (derive it, don't memorize)

Implicit vs explicit solutions

The trap everyone hits: lost solutions

Recall checkpoints

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Ordinary Differential Equations

Connections