IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~10 min read

Introduction to First-Order Differential Equations

A differential equation (DE) is an equation that contains one or more derivatives. A first-order DE contains a first derivative (dy/dx) but no second or higher derivatives. The simplest type — dy/dx = f(x) — you’ve already met: just integrate the right-hand side. But most DEs in this chapter have dy/dx = f(x, y) — depending on BOTH variables — and direct integration alone is impossible. This chapter introduces four analytical techniques (separation of variables, integrating factor, homogeneous substitution, Euler’s method numerical approximation) for solving them. This note sets up the language and re-cements the direct-integration case.

📘 What you need to know

Differential equation: an equation containing one or more derivatives. E.g., dy/dx = 12xy² is a DE; y² + 5y = 2x is NOT (no derivative).
First-order DE: highest derivative is the first derivative. E.g., dy/dx = sin x is first-order; d²y/dx² − 5 dy/dx + 7y = 0 is NOT first-order (contains d²y/dx²).
“Order” vs “degree”: order = highest derivative; degree = highest power of that derivative. AA HL focuses on first-ORDER (and usually first-DEGREE) equations only.
General solution vs particular solution: integrating gives a general solution containing +c. Using an initial or boundary condition (e.g., y(0) = 5) determines c, giving a particular solution.
Direct integration works iff RHS = f(x) only: if dy/dx depends only on x (and not on y), simply integrate both sides to find y. If RHS depends on y too, new methods are needed.
Four analytical methods in this chapter (each is a separate note): separation of variables, integrating factor, homogeneous substitution, Euler’s method (numerical approximation when analytical solution is hard).
Verifying a solution: substitute the proposed function into the DE; if LHS = RHS for all x, it’s a valid solution. This is the standard “show that…” setup.
Variables aren’t always x, y: many DE problems use t (time) as the independent variable. dy/dt, dP/dt, dN/dt are common in modelling. Read the variables from the question.

Classifying DEs and the direct-integration case

Simplest first-order DE — direct integration dydx = f(x) ⟹ y = ∫f(x) dx + c

When dy/dx depends only on x, integrate to recover y. This is the case you’ve already met in Topic 5 integration. The four advanced methods coming up handle the harder case where dy/dx also depends on y.

The chapter covers four analytical techniques for first-order DEs, plus Euler’s method for numerical approximations. This note is on direct integration (top-left, highlighted green) — the simplest case where dy/dx depends ONLY on x. The other three (preview boxes) handle the harder cases where dy/dx also depends on y; each gets its own note.

The reason direct integration is special: ∫f(x) dx requires only x to be the variable of integration. If the RHS contains y too — e.g., dy/dx = xy — you can’t integrate the RHS directly because y is itself a function of x (and that function is exactly what you’re trying to find!). The advanced methods all work around this circular dependency in clever ways.

General solution vs particular solution

General solution

contains “+ c“

Solution to the DE with the constant of integration left unspecified. Represents an infinite family of curves, one for each value of c.

Particular solution

c is determined

Solution where c has been pinned down using a boundary or initial condition (e.g., y(0) = 5). Represents one specific curve from the family.

🧭 Recipe — direct-integration first-order DEs

Confirm it’s a first-order DE: the highest derivative is the first derivative (dy/dx, ds/dt, etc.). If second derivatives appear, this method doesn’t apply.
Check if RHS depends only on the independent variable: if dy/dx = f(x) (no y on the right), direct integration applies. If y appears on the right, use one of the other techniques (next notes).
Integrate both sides with respect to the independent variable. The LHS becomes y (the antiderivative of dy/dx is y itself). The RHS is ∫f(x) dx.
Add the constant of integration +c. This gives the GENERAL solution — a family of curves parameterized by c.
Apply boundary/initial conditions if given: substitute the known (x, y) pair to solve for c, giving the PARTICULAR solution. If no condition is given, leave +c in the answer.

Worked examples

WE 1

Classification — which of the following are first-order DEs?

For each equation, determine whether it is a differential equation, and if so, whether it is a first-order DE:
(a) dy/dx + 3y = sin x (b) y² + 5y = 2x (c) d²y/dx² − 4 dy/dx + 3y = 0 (d) dr/dθ = r²

(a) dy/dx + 3y = sin x Contains dy/dx — IS a DE Highest derivative is dy/dx (first order) — IS first-order DE ✓ (b) y² + 5y = 2x No derivatives — NOT a DE (it’s a polynomial equation) (c) d²y/dx² − 4 dy/dx + 3y = 0 Contains d²y/dx² (second derivative) — IS a DE Highest derivative is d²y/dx² (second order) — NOT first-order ✗ (d) dr/dθ = r² Contains dr/dθ (first derivative w.r.t. θ) — IS a DE Highest derivative is first — IS first-order DE ✓ First-order DEs: (a) and (d) “first-order” is about the HIGHEST DERIVATIVE order, not about polynomial degree. (d) has r² (a square), but it’s still first-order because the highest derivative is dr/dθ.

WE 2

Verify a function is a solution

Show that y = 2e^3x is a solution to the differential equation dy/dx = 3y.

Step 1 — differentiate the proposed solution y = 2 e^(3x) dy/dx = 2 · 3 · e^(3x) = 6 e^(3x) (chain rule) Step 2 — compute the RHS of the DE using the proposed y RHS = 3y = 3 · 2 e^(3x) = 6 e^(3x) Step 3 — compare LHS and RHS LHS = dy/dx = 6 e^(3x) RHS = 3y = 6 e^(3x) LHS = RHS ✓ y = 2 e^(3x) satisfies dy/dx = 3y, as required. verification questions (“show that… is a solution”) DON’T require you to solve the DE from scratch — just check by substitution. This is faster than solving and is the standard exam pattern.

WE 3

General solution by direct integration

Find the general solution of the differential equation dy/dx = 6x² − 4x + 3.

Step 1 — confirm direct integration applies RHS = 6x² – 4x + 3 depends only on x (no y) — ✓ direct integration works Step 2 — integrate both sides ∫ dy/dx dx = ∫ (6x² – 4x + 3) dx y = 6 · x³/3 – 4 · x²/2 + 3x + c y = 2x³ – 2x² + 3x + c y = 2x³ − 2x² + 3x + c (general solution) verify: dy/dx = 6x² – 4x + 3 ✓. The +c is essential — without it, you’ve found ONE particular solution rather than the general family. The general solution is an infinite family of cubics, all with the same shape but shifted vertically by c.

WE 4

Particular solution using an initial condition

Find the particular solution to the differential equation ds/dt = 4t − 6 given that s = 5 when t = 0.

Step 1 — direct integration applies (RHS depends only on t) ∫ ds/dt dt = ∫ (4t – 6) dt s = 2t² – 6t + c (general solution) Step 2 — apply the initial condition s(0) = 5 5 = 2(0)² – 6(0) + c 5 = 0 – 0 + c c = 5 Step 3 — substitute c back into the general solution s = 2t² – 6t + 5 s = 2t² − 6t + 5 (particular solution) verify: ds/dt = 4t – 6 ✓ and s(0) = 0 – 0 + 5 = 5 ✓. The initial condition pins down a SPECIFIC curve from the infinite family — this is the “particular” solution. Without the IC, all you get is the general family with +c.

WE 5

Recognizing when direct integration FAILS

Explain why the differential equation dy/dx = e^x · y² cannot be solved by direct integration. State which technique would apply instead.

Step 1 — examine the RHS RHS = e^x · y² This depends on BOTH x (through e^x) AND y (through y²) Step 2 — why direct integration fails Direct integration: y = ∫ (RHS) dx But ∫ e^x · y² dx requires y to be a known function of x y is exactly what we’re trying to FIND — circular! We cannot integrate the RHS without already knowing y(x) Step 3 — identify the appropriate technique RHS = e^x · y² = (function of x) · (function of y) = g(x) · h(y) This matches the “separation of variables” pattern Use separation of variables (next note) — split the equation as (1/y²) dy = e^x dx and integrate both sides. this conceptual question is exam-relevant: identifying WHICH technique applies is half the battle. When you see d/dx = (something with both x AND y), reach for the toolkit: separation of variables, integrating factor, or homogeneous substitution — never direct integration alone.

WE 6

Real-world application — particle kinematics

A particle moves so that its velocity v (m/s) at time t (seconds) satisfies the differential equation dv/dt = 6 − 0.5t. The particle starts from rest at t = 0.
(a) Find v as a function of t.
(b) Find the time at which the particle reaches its maximum velocity, and find this maximum velocity.

(a) Step 1 — direct integration RHS = 6 – 0.5t depends only on t — direct integration applies v = ∫ (6 – 0.5t) dt = 6t – 0.25t² + c (a) Step 2 — apply initial condition v(0) = 0 (starts from rest) 0 = 6(0) – 0.25(0)² + c ⟹ c = 0 v(t) = 6t – 0.25t² = 6t – t²/4 (a) v = 6t − t²4 m/s (b) Step 1 — maximum velocity occurs when dv/dt = 0 dv/dt = 6 – 0.5t = 0 (from the original DE) 0.5t = 6 ⟹ t = 12 seconds (b) Step 2 — compute v at t = 12 v(12) = 6(12) – (12)²/4 = 72 – 36 = 36 m/s (b) Maximum velocity = 36 m/s at t = 12 seconds classic application: DE gives velocity from acceleration. Maximum occurs when the derivative of velocity (= acceleration = original RHS) equals zero — that’s where the velocity transitions from increasing to decreasing. The +c was 0 here because the particle starts from rest.

💡 Top tips

“Order” refers to the highest derivative, NOT the polynomial degree. dy/dx = y⁵ is FIRST-ORDER (the derivative is dy/dx, not d²y/dx²) even though y⁵ is a fifth power.
If RHS = f(x) only, use direct integration — the technique you already know from Topic 5. Don’t reach for fancier tools when the simple one works.
Always include +c in the general solution. If an initial/boundary condition is given, substitute it to find c; the final answer is the particular solution.
Verify by differentiating your solution. d/dx of your y should reproduce the original RHS — this catches arithmetic errors.
Variables aren’t always x and y: modelling problems use t, P, N, T, v, s, etc. Match the differential to the independent variable in the problem.

⚠ Common mistakes

Confusing order with degree: d²y/dx² + 3y = 0 is SECOND-order (because of d²y/dx²), not first. Order = highest derivative; degree = highest power.
Forgetting +c: writing y = 2x³ – 2x² + 3x without the +c loses a mark because that’s ONE solution out of infinitely many, not the general solution.
Trying direct integration when RHS contains y: ∫(e^x · y²) dx is undefined without knowing y(x). Use separation of variables (or another method) instead.
Not using the initial condition: solving for c from y(0) = 5 is essential when the question asks for a particular solution. Leaving the answer with +c is wrong if a condition was given.
Mixing variables: integrating dy/dt with respect to x is meaningless. Match the integration variable to the independent variable in the derivative.

Up next: Separation of Variables. The first technique for DEs where dy/dx depends on both x and y. If the RHS factors as g(x) · h(y) (a function of x times a function of y), you can rearrange so all y‘s are on one side and all x‘s on the other, then integrate both sides separately. The technique handles a huge class of DEs — especially in population modelling and other “rate of change is proportional to …” problems.

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