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Module 9 · L13 of 20 ~35 min ⚡ +95 XP available

Separation of Variables — Worked Examples

You know how to separate variables — now you'll see the full solution pathway in action. In this lesson you'll work through three carefully chosen differential equations, applying initial conditions to pin down the unique solution curve each time.

Today's hook — A population satisfies $\dfrac{dN}{dt} = 0.3N$ with $N(0) = 500$. Before working through this lesson, write down what you expect the solution to look like. Will it grow without bound or level off?

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Consider $\dfrac{dy}{dx} = 2y$ with $y(0) = 3$. Without solving formally — what do you expect the solution $y(x)$ to look like? Does it grow, decay, or stay constant? Write your reasoning below.

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The four-step separation method

+5 XP to read

Every separable differential equation is solved using the same four moves. Master these and you can handle any separable ODE the HSC sets.

Separate — rearrange so all $y$ terms (including $dy$) are on one side and all $x$ terms (including $dx$) are on the other.
Integrate both sides, adding one constant $C$.
Solve for $y$ explicitly if possible (make $y$ the subject).
Apply the initial condition to determine $C$.

$\displaystyle \frac{dy}{dx} = f(x)\,g(y) \;\Rightarrow\; \int \frac{dy}{g(y)} = \int f(x)\,dx$

One constant only

Even though you integrate both sides, write only one constant $C$ — the two constants combine into one.

Absolute value trap

$\int \dfrac{dy}{y} = \ln|y| + C$. Drop the absolute value only after applying the initial condition to confirm $y > 0$ (or $y < 0$).

Check by substituting back

After finding $y$, differentiate and substitute into the original ODE to verify. This catches sign errors and missing constants.

What you'll master

Know

Key facts

The four-step separation method
$\int \dfrac{dy}{y} = \ln|y| + C$; absorb $e^C$ into a new constant $A$
Initial conditions uniquely determine the constant of integration

Understand

Concepts

Why separating variables is a valid algebraic manoeuvre
How an initial condition selects one curve from the general solution family
The role of $|y|$ when integrating $1/y$

Can do

Skills

Fully solve a separable ODE including initial condition
Handle the $\ln|y|$ step correctly, transitioning to $y = Ae^{kx}$
Verify a solution by differentiation and back-substitution

Key terms

Separable ODEA differential equation that can be written as $\dfrac{dy}{dx} = f(x)\,g(y)$, with $x$ and $y$ functions on opposite sides.

General solutionThe full family of solution curves, containing an arbitrary constant $C$. Represents infinitely many curves.

Particular solutionThe unique curve obtained after applying an initial (or boundary) condition to determine $C$.

Initial conditionA known value $y(x_0) = y_0$ that pins down the particular solution from the general family.

Arbitrary constantThe constant $C$ (or $A = e^C$) introduced by indefinite integration. Its value is fixed by the initial condition.

Back-substitution checkVerifying a solution by differentiating $y$ and confirming it satisfies the original differential equation.

The separation algebra in full

core concept

Given a separable ODE $\dfrac{dy}{dx} = f(x)\,g(y)$, the separation step rearranges to:

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

Integrating both sides:

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

This is valid because $\dfrac{dy}{dx}\,dx = dy$ (chain rule applied to differentials). A common special case is $g(y) = y$, giving:

$$\int \frac{dy}{y} = \ln|y| + C \quad \Rightarrow \quad y = Ae^{kx} \text{ after exponentiating}$$

where $A = \pm e^C$ is the combined constant. Since $A$ can take any non-zero real value, we often simply write $A$ as an arbitrary constant (positive if context dictates, e.g. population).

Key insight. The initial condition $y(x_0) = y_0$ turns a general solution into a particular one. Substituting $x = x_0$ and $y = y_0$ into the general solution and solving for $C$ (or $A$) gives the unique answer.

Separation algebra: $\frac{dy}{dx}=f(x)g(y)\Rightarrow\frac{1}{g(y)}dy=f(x)dx\Rightarrow\int\frac{1}{g(y)}dy=\int f(x)dx+C$.

Pause — copy the complete separation algebra sequence: $\frac{dy}{dx}=f(x)g(y)\Rightarrow\frac{1}{g(y)}dy=f(x)dx\Rightarrow\int\frac{1}{g(y)}dy=\int f(x)dx$ into your book.

Quick check: When solving $\dfrac{dy}{dx} = 3y$ by separation of variables, which integral appears on the left-hand side?

Reading a worked example

core concept

We just saw the separation algebra: $\frac{1}{g(y)}\,dy=f(x)\,dx$, so $\int\frac{1}{g(y)}\,dy=\int f(x)\,dx$, and then both sides are integrated and $y$ is isolated. That raises a question: when you follow worked examples in a textbook or HSC paper, what features of each line tell you which algebraic move was made? This card answers it → look for (a) the separation step (rearranging the fraction), (b) the integral signs appearing, (c) the $+C$ appearing, (d) the exponential or inverse operation isolating $y$.

In the three worked examples that follow, reveal each step individually and try to predict the next line before clicking "Reveal next step". This active engagement is what builds fluency — passive reading does not.

Cover up lines below, predict the next step, then reveal.
If you got it wrong, re-read the method card (card 02) before continuing.
Pay particular attention to how the initial condition is applied in step 4 of each example.

NESA outcome ME12-4 — uses calculus in the solution of applied problems, including differential equations. Separation of variables is examined every year in HSC Ext 1.

In the three worked examples that follow, reveal each step individually and try to predict the next line before clicking "Reveal next step". This active engagement is what builds fluency — passive reading does not.

Pause — copy the four-line worked-example reading guide: identify the (a) separation move, (b) integration step, (c) constant of integration, (d) solve-for-$y$ step in any solution into your book.

Did you get this? True or false: when separating variables, you write two separate constants of integration — one for each integral.

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

PROBLEM 1 · SIMPLE EXPONENTIAL

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

Separate: $\dfrac{dy}{y} = 2\,dx$

Divide both sides by $y$ (assuming $y \neq 0$) and multiply both sides by $dx$.

PROBLEM 2 · MIXED VARIABLES

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 4$.

Separate: $y\,dy = x\,dx$

Multiply both sides by $y$ and by $dx$ to separate the variables.

PROBLEM 3 · DECAY MODEL

A substance decays according to $\dfrac{dM}{dt} = -0.5M$. If $M(0) = 200$ grams, find $M(t)$ and determine when $M = 50$.

Separate and integrate: $\dfrac{dM}{M} = -0.5\,dt \;\Rightarrow\; \ln|M| = -0.5t + C$

Divide by $M$ and integrate; one constant $C$.

Fill the gap: Solving $\dfrac{dy}{dx} = 5y$ with $y(0) = 2$ gives the particular solution $y = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the constant of integration

Writing $\ln|y| = kx$ instead of $\ln|y| = kx + C$ is a common slip. Without $C$ you cannot apply the initial condition, so you'll always get the trivial solution $A=1$. Always include $C$ immediately after integrating.

Trap 02

Mishandling $\ln|y|$ → $y$

After $\ln|y| = kx + C$, exponentiating gives $|y| = e^{kx+C} = e^C \cdot e^{kx}$, so $y = \pm e^C \cdot e^{kx}$. Writing $y = e^{kx + C}$ and leaving the $\pm$ unresolved loses a mark. Name $A = \pm e^C$ then determine its sign from the initial condition.

Trap 03

Choosing the wrong square root

When solving $y^2 = x^2 + K$, take $y = +\sqrt{x^2+K}$ only if the initial condition gives $y_0 > 0$. If $y_0 < 0$, take the negative root. The initial condition tells you which branch of the general solution you are on.

Did you get this? True or false: when solving $\dfrac{dy}{dx} = ky$, the general solution is $y = e^{kx + C}$, which is equivalent to $y = Ae^{kx}$ where $A = e^C$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Solve $\dfrac{dy}{dx} = -3y$ with $y(0) = 5$. Write the particular solution.

Solve $\dfrac{dy}{dx} = \dfrac{x+1}{y}$ with $y(0) = 3$.

A population grows at $\dfrac{dP}{dt} = 0.02P$ with $P(0) = 1000$. Find $P(t)$ and find $t$ when $P = 2000$. Give an exact answer.

Verify that $y = 3e^{2x}$ satisfies $\dfrac{dy}{dx} = 2y$ with $y(0)=3$ by differentiating and substituting back.

Solve $\dfrac{dy}{dx} = xy$ with $y(0) = 1$. (Hint: $\int x\,dx = \dfrac{x^2}{2} + C$.)

Odd one out: Three of these are valid particular solutions to separable ODEs. Which one is NOT consistent with the stated initial condition?

Revisit your thinking

Earlier you predicted what $\dfrac{dy}{dx} = 2y$, $y(0) = 3$ would look like.

The particular solution is $y = 3e^{2x}$ — an exponentially growing curve through $(0,3)$. The key insight is that $\dfrac{dy}{dx} = 2y$ means the rate of change is proportional to the current value, which always produces exponential behaviour. Did your prediction capture this?

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Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Solve $\dfrac{dy}{dx} = -2y$ with $y(0) = 6$. (2 marks)

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ApplyBand 43 marks

Q2. Solve $\dfrac{dy}{dx} = \dfrac{x}{2y}$ with $y(0) = 2$. Express $y$ as an explicit function of $x$. (3 marks)

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AnalyseBand 53 marks

Q3. A radioactive substance decays according to $\dfrac{dN}{dt} = -\lambda N$ where $\lambda = \ln 2 / T_{1/2}$ and $T_{1/2} = 10$ years. If $N(0) = 800$, find the exact time when $N = 100$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\dfrac{dy}{dx}=-3y$, $y(0)=5$: separate → $\ln|y|=-3x+C$ → $y=Ae^{-3x}$; $A=5$; $y = 5e^{-3x}$.

2. $y\,dy=(x+1)dx$ → $\dfrac{y^2}{2}=\dfrac{x^2}{2}+x+C$; at $x=0$, $y=3$: $C=\dfrac{9}{2}$; $y^2=x^2+2x+9$; $y=\sqrt{x^2+2x+9}$.

3. $P=1000e^{0.02t}$; $P=2000$ when $e^{0.02t}=2$ → $t=\dfrac{\ln 2}{0.02}=50\ln 2 \approx 34.7$ years.

4. $\dfrac{d}{dx}(3e^{2x})=6e^{2x}=2(3e^{2x})=2y$ ✓ and $y(0)=3$ ✓.

5. $\dfrac{dy}{y}=x\,dx$ → $\ln|y|=\dfrac{x^2}{2}+C$ → $y=Ae^{x^2/2}$; $y(0)=1$ → $A=1$; $y=e^{x^2/2}$.

Q1 (2 marks): Separate: $\dfrac{dy}{y}=-2\,dx$ [1]; integrate: $\ln|y|=-2x+C$; general: $y=Ae^{-2x}$; apply $y(0)=6$: $A=6$; $\mathbf{y=6e^{-2x}}$ [1].

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

Q3 (3 marks): $\lambda=\dfrac{\ln 2}{10}$; $N=800e^{-(\ln 2/10)t}$ [1]; set $100=800e^{-(\ln 2/10)t}$ → $e^{-(\ln 2/10)t}=\dfrac{1}{8}$ [1]; $-\dfrac{t\ln 2}{10}=\ln\dfrac{1}{8}=-3\ln 2$ → $t=30$ years [1].

Boss battle · The Separation Master

earn bronze · silver · gold

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

⚔ Enter the arena

Science Jump · platform challenge

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

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 12 · Separation of Variables Lesson 14 · Exponential Growth and Decay →

Module overview · Extension 1 · Checkpoint 3