Your weak spots

Insights load after your first practice round.

Module 9 · L8 of 20 ~40 min ⚡ +100 XP available

Introduction to Differential Equations

A differential equation is not an equation you solve for a number — it is an equation you solve for a function. The equation $\dfrac{dy}{dx} = ky$ says "find a function whose derivative is proportional to itself." Exponential functions are the answer. Understanding this shift from "solve for $x$" to "solve for $f$" unlocks the entire second half of Module 9 — differential equations, direction fields, separation of variables, and real-world modelling.

Today's hook — The equation $\dfrac{dy}{dx} = 2x$ looks simple. What kind of function satisfies it? Is the answer unique? Jot your instinct — you'll verify it rigorously after card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Consider the equation $\dfrac{dy}{dx} = 2x$. Without integrating formally — what kind of function do you think satisfies this? Is there just one such function, or many? Write your reasoning.

auto-saved

What is a differential equation?

+5 XP to read

A differential equation is an equation that involves an unknown function and one or more of its derivatives. Unlike an algebraic equation (which is solved for a number), a differential equation is solved for a function.

General form: an equation relating $y$, $\dfrac{dy}{dx}$ (and possibly higher derivatives) and $x$. A solution is any function $y = f(x)$ that satisfies the equation everywhere on some interval.

The general solution contains an arbitrary constant $C$ (a whole family of curves). An initial condition fixes $C$ to give one particular solution.

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

Solution is a function

Solving $\dfrac{dy}{dx} = 2x$ gives $y = x^2 + C$ — infinitely many parabolas, one for each value of $C$. The solution is a family of curves, not a single number.

Initial condition

If we are also told $y(0) = 3$, we substitute to find $C$: $3 = 0^2 + C$, so $C = 3$. The particular solution is $y = x^2 + 3$.

Order of the DE

The order of a DE is the order of the highest derivative present. $\dfrac{dy}{dx} = 2x$ is first-order; $\dfrac{d^2y}{dx^2} + y = 0$ is second-order. HSC covers first-order only.

What you'll master

Know

Key facts

A differential equation (DE) involves an unknown function and its derivative(s)
The general solution of a first-order DE contains one arbitrary constant $C$
An initial condition $y(x_0) = y_0$ gives a unique particular solution

Understand

Concepts

Why solutions to DEs are functions (entire curves), not single values
The geometric meaning of the general solution as a family of curves
How an initial condition selects one member from the family

Can do

Skills

Recognise and classify differential equations (order, type)
Verify that a given function is a solution of a DE by substitution
Apply an initial condition to find a particular solution

Key terms

Differential equation (DE)An equation that contains an unknown function $y(x)$ and one or more of its derivatives. Its solution is a function (or family of functions).

General solutionThe complete family of solutions, containing an arbitrary constant $C$. It represents infinitely many curves — one for each value of $C$.

Particular solutionA single specific solution obtained by substituting an initial condition into the general solution to find the value of $C$.

Initial conditionA known value $y(x_0) = y_0$ that fixes the arbitrary constant, giving a unique particular solution.

OrderThe order of the highest derivative in the DE. A first-order DE involves only $\dfrac{dy}{dx}$. HSC Extension 1 focuses on first-order DEs.

Verifying a solutionSubstituting $y = f(x)$ (and its derivative) back into the DE to confirm both sides are equal for all $x$ in the domain.

Solving and verifying differential equations

core concept

The simplest differential equations have the form $\dfrac{dy}{dx} = f(x)$. These are solved directly by integration:

$$y = \int f(x)\,dx + C$$

The general solution always includes $+ C$ (unless you know an initial condition). To find a particular solution:

Integrate to get the general solution $y = F(x) + C$.
Substitute the initial condition $y(x_0) = y_0$ to solve for $C$.
Write the particular solution with the specific value of $C$.

Hook answer: $\dfrac{dy}{dx} = 2x \Rightarrow y = \displaystyle\int 2x\,dx = x^2 + C$. The general solution is the family of parabolas $y = x^2 + C$. Given $y(0) = 5$: $5 = 0 + C$, so $C = 5$ and the particular solution is $y = x^2 + 5$.

Verifying by substitution. To verify $y = x^2 + 5$ satisfies $\dfrac{dy}{dx} = 2x$: differentiate to get $\dfrac{dy}{dx} = 2x$ ✓. Both sides match. This substitution method is also used to verify any proposed solution.

The simplest differential equations have the form $\dfrac{dy}{dx} = f(x)$. These are solved directly by integration:

Pause — copy the solve-and-verify procedure: integrate $\frac{dy}{dx}=f(x)$ to get $y=F(x)+C$, then differentiate to confirm $\frac{dy}{dx}=f(x)$ into your book.

Quick check: The general solution of $\dfrac{dy}{dx} = 3x^2$ is:

Recognising and classifying differential equations

core concept

We just saw that a DE of the form $\frac{dy}{dx}=f(x)$ is solved by direct integration: $y=\int f(x)\,dx+C$, and verified by differentiating the solution and checking it matches the original DE. That raises a question: HSC DEs come in many forms — how do you classify each type before selecting a solution method? This card answers it → check the right-hand side: function of $x$ only (direct integration), function of $y$ only (autonomous), separable $f(x)g(y)$, or growth/logistic.

DEs appear in many forms at HSC level. Here are the key types you will encounter:

Type 1: $\dfrac{dy}{dx} = f(x)$ — the derivative is a function of $x$ only. Solve by direct integration.
Type 2: $\dfrac{dy}{dx} = g(y)$ — the derivative is a function of $y$ only. Requires separation of variables (Lesson 11).
Type 3: $\dfrac{dy}{dx} = f(x)g(y)$ — the derivative is a product. Requires full separation of variables (Lesson 12).

In this lesson we focus on Type 1 and on the fundamental concepts of solution families and initial conditions.

Examples of DEs:

$\dfrac{dy}{dx} = 2x + 1$ (Type 1 — integrate directly)
$\dfrac{dy}{dx} = y$ (Type 2 — proportional growth, solution $y = Ce^x$)
$\dfrac{dy}{dx} = xy$ (Type 3 — solution $y = Ce^{x^2/2}$)

Real-world connection. Newton's law of cooling is $\dfrac{dT}{dt} = -k(T - T_{\text{env}})$, where $T$ is temperature, $t$ is time, and $k > 0$ is a constant. This is a Type 2 DE — its solution (covered in Lesson 18) describes exponential cooling.

DEs appear in many forms at HSC level. Here are the key types you will encounter:

Pause — copy the four DE classification types with examples: (1) $\frac{dy}{dx}=f(x)$; (2) $\frac{dy}{dx}=g(y)$ autonomous; (3) $\frac{dy}{dx}=f(x)g(y)$ separable; (4) logistic into your book.

Did you get this? True or false: the DE $\dfrac{dy}{dx} = \cos x$ can be solved by direct integration to give $y = \sin x + C$.

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

PROBLEM 1 · FIND THE PARTICULAR SOLUTION

Given $\dfrac{dy}{dx} = 4x - 3$ and $y(1) = 2$, find the particular solution.

Integrate: $y = \displaystyle\int (4x - 3)\,dx = 2x^2 - 3x + C$

Direct integration of the right-hand side; don't forget $+C$.

PROBLEM 2 · VERIFY A SOLUTION

Show that $y = 3e^{2x}$ is a solution of the DE $\dfrac{dy}{dx} = 2y$.

Differentiate: $\dfrac{dy}{dx} = 3 \cdot 2e^{2x} = 6e^{2x}$.

Apply the chain rule to $y = 3e^{2x}$.

PROBLEM 3 · GENERAL SOLUTION + INITIAL CONDITION

Find the particular solution of $\dfrac{dy}{dx} = \cos x + e^x$ given $y(0) = 4$.

General solution: $y = \displaystyle\int (\cos x + e^x)\,dx = \sin x + e^x + C$

Integrate term by term: $\int\cos x\,dx = \sin x$; $\int e^x\,dx = e^x$. Include $+C$.

Fill the gap: The general solution of $\dfrac{dy}{dx} = 5x^4$ is $y =$ $+ C$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the constant of integration

Writing $y = x^2$ instead of $y = x^2 + C$ for $\dfrac{dy}{dx} = 2x$ is the single most common DE error. The $+ C$ is essential — without it, you have lost infinitely many valid solutions. Always include it unless you have an initial condition to eliminate it.

Trap 02

Confusing "solving for a number" with "solving for a function"

A DE does not have a unique numerical answer. The answer is a function (or family of functions). If you find yourself writing a number as the solution, something has gone wrong — unless you were given an initial condition and asked for the value of $y$ at a specific point.

Trap 03

Applying the initial condition before integrating

You must integrate first (to get the general solution including $C$), then substitute the initial condition. If you substitute before integrating, you will find $C = 0$ in all cases — which is usually wrong and misses the entire point of the initial condition.

Did you get this? True or false: to find the particular solution of a DE, you should integrate first to get the general solution (including $+C$), then substitute the initial condition.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find the general solution of $\dfrac{dy}{dx} = 6x^2 - 4x + 1$.

Given $\dfrac{dy}{dx} = e^{2x}$ and $y(0) = 1$, find the particular solution.

Verify that $y = \sin(2x) + 4$ is a solution of $\dfrac{dy}{dx} = 2\cos(2x)$.

Classify each DE as Type 1, 2, or 3: (a) $\dfrac{dy}{dx} = x^3$ (b) $\dfrac{dy}{dx} = y^2$ (c) $\dfrac{dy}{dx} = xy^2$.

Find the particular solution of $\dfrac{dy}{dx} = \sin x$ given $y(\pi) = 0$.

Odd one out: Three of these are correct statements about differential equations. Which one is NOT?

Revisit your thinking

Earlier you predicted what kind of function satisfies $\dfrac{dy}{dx} = 2x$ and whether the solution is unique.

The answer: integrating gives $y = x^2 + C$ — a whole family of parabolas, one for each value of $C$. Without an initial condition, there are infinitely many solutions. With one (e.g. $y(0) = 5$), you get exactly one: $y = x^2 + 5$. This is the fundamental difference between DEs and algebraic equations.

auto-saved

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. Find the particular solution of $\dfrac{dy}{dx} = 6x^2 - 4x + 1$ given $y(0) = 3$. (2 marks)

auto-saved

ApplyBand 42 marks

Q2. Show that $y = 5e^{-3x}$ is a solution of $\dfrac{dy}{dx} = -3y$. (2 marks)

auto-saved

AnalyseBand 53 marks

Q3. Explain the difference between a general solution and a particular solution of a differential equation. Your answer should include why the constant of integration is essential and what an initial condition does. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $y = 2x^3 - 2x^2 + x + C$

2. General: $y = \tfrac{1}{2}e^{2x} + C$. $y(0)=1$: $1 = \tfrac{1}{2}+C$, $C = \tfrac{1}{2}$. Particular: $y = \tfrac{1}{2}e^{2x} + \tfrac{1}{2}$.

3. $\frac{dy}{dx} = 2\cos(2x)$. RHS $= 2\cos(2x)$. Equal ✓ (note: $4$ is a constant and drops out on differentiation).

4. (a) Type 1 — RHS is $f(x)$ only. (b) Type 2 — RHS is $g(y)$ only. (c) Type 3 — RHS is $f(x) \cdot g(y)$.

5. General: $y = -\cos x + C$. $y(\pi)=0$: $0 = -\cos\pi + C = 1 + C$, $C = -1$. Particular: $y = -\cos x - 1$.

Q1 (2 marks): $y = 2x^3 - 2x^2 + x + C$ [1]. $y(0)=3$: $3 = C$. Particular: $y = 2x^3 - 2x^2 + x + 3$ [1].

Q2 (2 marks): $\dfrac{dy}{dx} = 5(-3)e^{-3x} = -15e^{-3x}$ [1]. $-3y = -3(5e^{-3x}) = -15e^{-3x}$. Both sides equal $-15e^{-3x}$ ✓ [1].

Q3 (3 marks): The general solution $y = F(x) + C$ contains an arbitrary constant representing the entire family of solution curves — one for each value of $C$ [1]. Without an initial condition, there are infinitely many solutions; the constant $C$ cannot be omitted because integration always introduces it [1]. An initial condition $y(x_0) = y_0$ is a known point that the curve must pass through; substituting it determines the unique value of $C$, giving the particular solution — a single specific curve [1].

Boss battle · The DE Detective

earn bronze · silver · gold

Five timed questions on differential equations — recognising, solving, and verifying. 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 differential equation questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 7 · Volumes Between Curves Lesson 9 · Direction Fields →

Module overview · Extension 1 · Checkpoint 2