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Module 9 · L12 of 20 ~40 min ⚡ +95 XP available

Separation of Variables

When $\dfrac{dy}{dx} = f(x)\,g(y)$, the right side is a product of a function of $x$ and a function of $y$. The separation trick collects all $y$-terms on the left and all $x$-terms on the right, then integrates both sides independently. This is the most powerful first-order DE technique you will meet in the HSC — and it works for a huge variety of real-world models.

Today's hook — The DE $\dfrac{dy}{dx} = xy$ has both $x$ and $y$ on the right side. Before reading on, can you think of a way to rewrite it so that all $y$-terms are on one side and all $x$-terms are on the other? Jot your idea below.

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Recall — your gut answer first

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The DE $\dfrac{dy}{dx} = xy$ involves both $x$ and $y$ on the right side. Before using the formal method — try rearranging this equation so that the left side involves only $y$ and the right side involves only $x$. What do you get?

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The key move: separate then integrate

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When $\dfrac{dy}{dx} = f(x)\,g(y)$, we separate the variables by collecting $y$-terms on one side and $x$-terms on the other, then integrate independently:

Step 1 — Rearrange: $\dfrac{1}{g(y)}\,dy = f(x)\,dx$.

Step 2 — Integrate both sides:

$\displaystyle\int \dfrac{1}{g(y)}\,dy = \int f(x)\,dx + C$

One constant $C$ is sufficient (the arbitrary constants from each side combine into one).

$\displaystyle\int \frac{1}{g(y)}\,dy = \int f(x)\,dx + C$

One $C$ only

Both integrals produce a constant, but they merge into a single $C$ on one side. Never write $C_1 + C_2$ as your final answer — combine them first.

Check $g(y) = 0$ first

Equilibrium solutions occur where $g(y) = 0$. The separation step divides by $g(y)$, which is invalid at those points — so identify them first.

NESA outcome ME12-4

Separation of variables is explicitly required for HSC Ext 1. It generalises both $\tfrac{dy}{dx} = f(x)$ (Lesson 10) and $\tfrac{dy}{dx} = g(y)$ (Lesson 11).

What you'll master

Know

Key facts

$\dfrac{dy}{dx} = f(x)\,g(y)$ is separable if the right side factors
Rearrange to $\dfrac{1}{g(y)}\,dy = f(x)\,dx$, then integrate both sides
One constant $C$ remains; use an initial condition to find it

Understand

Concepts

Why multiplying through by $dx$ (treating Leibniz notation heuristically) is valid here
The connection to Lessons 10 and 11 as special cases of this technique
When a DE is not separable (e.g. $\dfrac{dy}{dx} = x + y$)

Can do

Skills

Identify whether a DE is separable
Solve separable DEs by separating and integrating
Apply an initial condition to obtain the particular solution

Key terms

Separable DEA first-order DE that can be written as $\dfrac{dy}{dx} = f(x)\,g(y)$ — the right side factors into a function of $x$ times a function of $y$.

Separation of variablesThe technique of rearranging a separable DE so that $y$-terms are on the left and $x$-terms are on the right, then integrating independently.

Implicit solutionA solution where $y$ is not expressed explicitly as $y = F(x)$, but is defined implicitly by a relation such as $\ln|y| = x^2 + C$.

Explicit solutionA solution rearranged to give $y$ directly as a function of $x$, e.g. $y = e^{x^2+C} = Ae^{x^2}$.

Initial conditionA value such as $y(0) = 3$ substituted after integrating to determine the constant $C$ uniquely.

Non-separable DEA DE whose right side cannot be factored into $f(x) \cdot g(y)$. Example: $\dfrac{dy}{dx} = x + y$ is not separable without further manipulation.

Separation of variables — the method

core concept

A DE is separable if it can be written as $\dfrac{dy}{dx} = f(x)\,g(y)$. The method has four steps:

Identify equilibria. Solve $g(y) = 0$; these are constant solutions.
Separate. Rearrange to $\dfrac{1}{g(y)}\,dy = f(x)\,dx$ (heuristically: "multiply by $dx$, divide by $g(y)$").
Integrate. $\displaystyle\int \dfrac{1}{g(y)}\,dy = \int f(x)\,dx + C$.
Solve for $y$. If possible, rearrange to an explicit form $y = F(x)$. Apply the initial condition to find $C$.

$$\frac{dy}{dx} = f(x)\,g(y) \;\Longrightarrow\; \int \frac{1}{g(y)}\,dy = \int f(x)\,dx + C$$

Example: Solve $\dfrac{dy}{dx} = xy$ with $y(0) = 2$.

Step 1: Equilibrium at $y = 0$.

Step 2: Separate: $\dfrac{1}{y}\,dy = x\,dx$.

Step 3: $\ln|y| = \dfrac{x^2}{2} + C$.

Step 4: $|y| = e^{x^2/2 + C} = Ae^{x^2/2}$. Apply $y(0) = 2$: $A = 2$.

Particular solution: $y = 2e^{x^2/2}$.

Connection to Lessons 10 and 11. If $g(y) = 1$, you get $\dfrac{dy}{dx} = f(x)$ (Lesson 10 — integrate directly). If $f(x) = 1$, you get $\dfrac{dy}{dx} = g(y)$ (Lesson 11 — inversion method). Separation of variables is the full general technique that includes both as special cases.

Separation of variables: $\frac{dy}{dx}=f(x)g(y)\Rightarrow\int\frac{dy}{g(y)}=\int f(x)\,dx$; integrate both sides; solve for $y$.

Pause — copy the four-step separation method with the Leibniz notation move $\frac{dy}{g(y)}=f(x)\,dx$ written out explicitly into your book.

Quick check: Which of the following is the correct separated form of $\dfrac{dy}{dx} = x^2 y$?

Recognising separable DEs

core concept

We just saw the four-step separation method: write $\frac{dy}{dx}=f(x)g(y)$, rearrange to $\frac{dy}{g(y)}=f(x)\,dx$, integrate both sides, solve for $y$. That raises a question: how do you quickly test whether a DE is separable before committing to this method — and what are the three common forms that pass the test? This card answers it → check whether the right-hand side factorises as a product of a function of $x$ alone and a function of $y$ alone.

Not every DE is separable. Before applying the method, check whether the right side can be written as a product $f(x)\cdot g(y)$:

Separable$\dfrac{dy}{dx} = xy$ → $f(x) = x,\;g(y) = y$ ✓

Not separable$\dfrac{dy}{dx} = x + y$ — cannot be written as a product ✗

Separable$\dfrac{dy}{dx} = \dfrac{\sin x}{y^2}$ → $f(x)=\sin x,\;g(y)=y^2$ ✓

Not separable$\dfrac{dy}{dx} = x^2 + y^2$ — sum, not product ✗

Separable$\dfrac{dy}{dx} = e^{x+y} = e^x \cdot e^y$ → factor as product ✓

Special case$\dfrac{dy}{dx} = g(y)$ → Lesson 11 (set $f(x) = 1$)

Trick for $e^{x+y}$. Use the exponent law $e^{x+y} = e^x \cdot e^y$ to split into a product. This is a common exam question that tests algebraic fluency before the integration even starts.

Not every DE is separable. Before applying the method, check whether the right side can be written as a product $f(x)\cdot g(y)$:

Pause — copy the separability test: the DE $\frac{dy}{dx}=h(x,y)$ is separable iff $h=f(x)g(y)$; include three examples (separable) and one non-example into your book.

Did you get this? True or false: $\dfrac{dy}{dx} = e^{x+y}$ is separable.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · STANDARD PRODUCT

Solve $\dfrac{dy}{dx} = 2xy$ with $y(0) = 3$.

Equilibrium: $g(y) = y = 0 \Rightarrow y = 0$. Separate (for $y \neq 0$): $\dfrac{1}{y}\,dy = 2x\,dx$.

Identify the equilibrium, then rearrange so $y$ terms are left and $x$ terms are right.

PROBLEM 2 · TRIG NUMERATOR

Find the general solution of $\dfrac{dy}{dx} = \dfrac{\cos x}{y^2}$.

Rewrite as $\dfrac{dy}{dx} = (\cos x)\cdot\dfrac{1}{y^2}$. Here $f(x) = \cos x$, $g(y) = y^{-2}$.

Identify the product structure. Equilibrium of $1/y^2$ would require $y \to \infty$, so there is no finite equilibrium — no constant solution except by inspection.

PROBLEM 3 · EXPONENTIAL SPLIT

Solve $\dfrac{dy}{dx} = e^{x+y}$ with $y(0) = 0$.

Factor: $e^{x+y} = e^x \cdot e^y$. So $f(x) = e^x$, $g(y) = e^y$. Equilibrium: $e^y = 0$ has no solution (never zero), so no equilibrium.

Use $e^{a+b} = e^a \cdot e^b$ to make the product structure visible.

Fill the gap: After separating $\dfrac{dy}{dx} = 3x^2 y$, integrating the left side gives $\ln|y|$ and integrating the right side gives $x^{$$} + C$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Adding instead of multiplying to recognise separability

$\dfrac{dy}{dx} = x + y$ is not separable — it is a sum, not a product. $\dfrac{dy}{dx} = xy$ is separable. Always check the algebraic structure before attempting to separate.

Trap 02

Writing two constants $C_1$ and $C_2$

Both sides of the separated equation produce a constant when integrated. But the difference (or combination) of two arbitrary constants is still an arbitrary constant. Always consolidate to a single $C$ on one side before applying an initial condition.

Trap 03

Skipping the equilibrium check

Dividing by $g(y)$ in the separation step is invalid when $g(y) = 0$. Any constant $y = c$ satisfying $g(c) = 0$ is a solution that your integration will miss entirely. State equilibria separately.

Did you get this? True or false: the DE $\dfrac{dy}{dx} = x^2 + y^2$ is separable.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Decide whether each DE is separable and explain why: (a) $\dfrac{dy}{dx} = xy^2$, (b) $\dfrac{dy}{dx} = x + y$, (c) $\dfrac{dy}{dx} = \dfrac{y}{x}$.

Find the general solution of $\dfrac{dy}{dx} = \dfrac{x}{y}$.

Solve $\dfrac{dy}{dx} = -xy$ with $y(0) = 5$.

Find the particular solution of $\dfrac{dy}{dx} = \dfrac{\sin x}{y}$ with $y\!\left(\dfrac{\pi}{2}\right) = 2$.

A quantity $Q$ decays according to $\dfrac{dQ}{dt} = -kQ^2$ with $Q(0) = 1$. Find $Q(t)$ and determine $Q(1)$ if $k = 0.5$.

Odd one out: Three of these DEs are separable. Which one is NOT separable?

Revisit your thinking

Earlier you tried rearranging $\dfrac{dy}{dx} = xy$ so that $y$-terms and $x$-terms are separated.

The formal move is: divide both sides by $y$ and multiply by $dx$ to get $\dfrac{1}{y}\,dy = x\,dx$. Then integrate: $\ln|y| = \dfrac{x^2}{2} + C$, giving $y = Ae^{x^2/2}$. The "trick" you may have spotted is that treating $\dfrac{dy}{dx}$ like a fraction (even though it isn't literally one) gives the right result — the Chain Rule justifies it rigorously.

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Multiple choice

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Short answer

ApplyBand 32 marks

Q1. Find the general solution of $\dfrac{dy}{dx} = x y^2$. (2 marks)

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ApplyBand 43 marks

Q2. Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 2$. Give the answer in the form $y = f(x)$. (3 marks)

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AnalyseBand 54 marks

Q3. A biologist models the growth of a bacterial population by $\dfrac{dP}{dt} = 0.2P(10 - P)$. State the equilibrium values of $P$. For $0 < P < 10$, separate and integrate to find the general solution. (You may leave it in implicit form.) (4 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. (a) separable; (b) not separable; (c) separable ($f(x)=1/x$, $g(y)=y$) · 2. $y^2 = x^2 + C$ · 3. $y = 5e^{-x^2/2}$ · 4. $y^2 = -2\cos x + C$; IC: $4 = 0 + C$ so $y^2 = -2\cos x + 4$, $y = \sqrt{4-2\cos x}$ · 5. $Q = \dfrac{1}{kt+1}$; $Q(1) = \dfrac{1}{1.5} \approx 0.67$

Q1 (2 marks): Equilibrium $y = 0$ [½]. Separate: $y^{-2}\,dy = x\,dx$ [½]. Integrate: $-\dfrac{1}{y} = \dfrac{x^2}{2} + C$ [½]. General solution: $y = -\dfrac{1}{\frac{x^2}{2}+C} = \dfrac{-2}{x^2+K}$ (where $K = 2C$) [½].

Q2 (3 marks): $y\,dy = x\,dx$ [1]. $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$ [1]. Apply $y(0) = 2$: $\dfrac{4}{2} = 0 + C$, so $C = 2$. $y^2 = x^2 + 4$, hence $y = \sqrt{x^2+4}$ (positive since $y(0) = 2 > 0$) [1].

Q3 (4 marks): Equilibria $P = 0$ and $P = 10$ [1]. Separate: $\dfrac{dP}{P(10-P)} = 0.2\,dt$ [½]. Partial fractions: $\dfrac{1}{P(10-P)} = \dfrac{1}{10}\!\left(\dfrac{1}{P} + \dfrac{1}{10-P}\right)$ [1]. Integrate: $\dfrac{1}{10}(\ln P - \ln(10-P)) = 0.2t + C$ [1]. Simplify: $\ln\dfrac{P}{10-P} = 2t + K$, i.e. $\dfrac{P}{10-P} = Ae^{2t}$ [½].

Boss battle · The Variable Separator

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Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering separation of variables questions. Lighter alternative to the boss.

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← Lesson 11 · Solving dy/dx = g(y) Lesson 13 · Separation of Variables — Worked Examples →

Module overview · Extension 1 · Checkpoint 3