IB Maths AA Analysis and Approaches How to Solve: IB AA HL – Differential Equations (Separable, Integrating Factor, Homogeneous)

How to Solve: IB AA HL – Differential Equations (Separable, Integrating Factor, Homogeneous)

Complete Guide

Introduction

"Mastering differential equations in IB AA HL unlocks 10–15% of your Paper 2 marks—and real-world models in physics (radioactive decay), chemistry (reaction rates), biology (population growth), and economics (supply/demand). One separable equation can be the difference between a 5 and a 7."

WHAT YOU NEED TO KNOW FIRST

1. Basic integration (power rule, natural log, substitution).
2. Chain rule & product rule (for differentiating).
3. Logarithm properties (e.g., ln(a) – ln(b) = ln(a/b)).

KEY TERMS & FORMULAS

1. Separable Equations

Definition: A differential equation that can be written as: dy/dx = f(x) · g(y) Method: Separate variables, integrate both sides.

MEMORISE THIS: - If dy/dx = f(x)g(y), rewrite as ∫(1/g(y)) dy = ∫f(x) dx.

2. Integrating Factor (Linear First-Order)

Definition: Used for equations of the form: dy/dx + P(x)y = Q(x)

MEMORISE THIS: - Integrating Factor (IF): e^∫P(x) dx - Solution: y = (1/IF) · ∫(IF · Q(x)) dx

3. Homogeneous Equations

Definition: A differential equation where dy/dx = f(y/x). Method: Substitute v = y/x (so y = vx, dy/dx = v + x dv/dx).

MEMORISE THIS: - Rewrite dy/dx = f(y/x) as v + x dv/dx = f(v). - Solve for v, then substitute back y = vx.

STEP-BY-STEP METHOD

1. Separable Equations

Step 1: Write the equation in the form dy/dx = f(x)g(y). Step 2: Separate variables: ∫(1/g(y)) dy = ∫f(x) dx. Step 3: Integrate both sides. Step 4: Solve for y (include + C). Step 5: Apply initial conditions (if given).

2. Integrating Factor (Linear First-Order)

Step 1: Write the equation in the form dy/dx + P(x)y = Q(x). Step 2: Find the integrating factor: IF = e^∫P(x) dx. Step 3: Multiply both sides by IF. Step 4: Recognise the left side as (IF · y)'. Step 5: Integrate both sides: IF · y = ∫(IF · Q(x)) dx + C. Step 6: Solve for y.

3. Homogeneous Equations

Step 1: Check if dy/dx = f(y/x). Step 2: Substitute v = y/x (so y = vx, dy/dx = v + x dv/dx). Step 3: Rewrite the equation in terms of v and x. Step 4: Separate variables and integrate. Step 5: Substitute back v = y/x to find y.

WORKED EXAMPLES

Example 1 – Basic (Separable)

Problem: Solve dy/dx = 2xy, given y(0) = 3.

Step 1: Separate variables: ∫(1/y) dy = ∫2x dx

Step 2: Integrate: ln|y| = x² + C

Step 3: Solve for y: y = e^(x² + C) = Ae^(x²) (where A = e^C)

Step 4: Apply initial condition (y(0) = 3): 3 = Ae^0 → A = 3

Final Answer: y = 3e^(x²)

What we did and why: - Recognised it was separable. - Integrated both sides, solved for y, and applied the initial condition.

Example 2 – Medium (Integrating Factor)

Problem: Solve dy/dx + 2y = e^(-x).

Step 1: Identify P(x) = 2, Q(x) = e^(-x).

Step 2: Find integrating factor: IF = e^∫2 dx = e^(2x)

Step 3: Multiply both sides by IF: e^(2x) dy/dx + 2e^(2x) y = e^(x)

Step 4: Recognise left side as (e^(2x) y)': (e^(2x) y)' = e^(x)

Step 5: Integrate both sides: e^(2x) y = ∫e^(x) dx = e^(x) + C

Step 6: Solve for y: y = e^(-x) + Ce^(-2x)

What we did and why: - Used the integrating factor method because the equation was linear. - Multiplied by IF, integrated, and solved for y.

Example 3 – Exam-Style (Homogeneous)

Problem: Solve dy/dx = (x² + y²)/(xy).

Step 1: Check if homogeneous: dy/dx = (x² + y²)/(xy) = x/y + y/x = f(y/x)

Step 2: Substitute v = y/x (so y = vx, dy/dx = v + x dv/dx): v + x dv/dx = 1/v + v

Step 3: Simplify: x dv/dx = 1/v

Step 4: Separate variables: ∫v dv = ∫(1/x) dx

Step 5: Integrate: (v²)/2 = ln|x| + C

Step 6: Substitute back v = y/x: (y²)/(2x²) = ln|x| + C

Final Answer: y² = 2x²(ln|x| + C)

What we did and why: - Recognised it was homogeneous, substituted v = y/x, and solved.

COMMON MISTAKES

1. MISTAKE: Forgetting + C when integrating.
   WHY IT HAPPENS: Students rush and forget constants.
   CORRECT APPROACH: Always include + C after integration.

2. MISTAKE: Not separating variables fully (e.g., leaving dy and dx mixed).
   WHY IT HAPPENS: Misapplying the separation method.
   CORRECT APPROACH: Ensure dy is with y terms and dx with x terms.

3. MISTAKE: Incorrect integrating factor (e.g., forgetting the e^).
   WHY IT HAPPENS: Confusing the formula.
   CORRECT APPROACH: IF = e^∫P(x) dx, not just ∫P(x) dx.

4. MISTAKE: Not substituting back v = y/x in homogeneous equations.
   WHY IT HAPPENS: Forgetting the substitution step.
   CORRECT APPROACH: Always replace v with y/x at the end.

5. MISTAKE: Misapplying initial conditions (e.g., plugging in x=0 before solving for y).
   WHY IT HAPPENS: Skipping steps.
   CORRECT APPROACH: Solve for y first, then apply initial conditions.

EXAM TRAPS

1. TRAP: The equation looks separable but isn’t (e.g., dy/dx = x + y).
   HOW TO SPOT IT: If you can’t separate x and y cleanly, it’s not separable.
   HOW TO AVOID IT: Check if it’s linear or homogeneous first.

2. TRAP: The integrating factor is e^(-x), but students forget the negative sign.
   HOW TO SPOT IT: If P(x) is negative (e.g., dy/dx - y = x), the IF will have a negative exponent.
   HOW TO AVOID IT: Always write IF = e^∫P(x) dx carefully.

3. TRAP: The homogeneous substitution v = y/x is used, but students forget dy/dx = v + x dv/dx.
   HOW TO SPOT IT: If the equation is dy/dx = f(y/x), substitution is needed.
   HOW TO AVOID IT: Memorise dy/dx = v + x dv/dx when y = vx.

1-MINUTE RECAP

"Here’s the night-before cheat sheet: 1. Separable? Split dy/dx = f(x)g(y) into ∫(1/g(y)) dy = ∫f(x) dx. 2. Linear? Use IF = e^∫P(x) dx, multiply, integrate, solve for y. 3. Homogeneous? Substitute v = y/x, rewrite, separate, integrate, substitute back. 4. Always include + C and check initial conditions. 5. Watch for traps: Non-separable equations, negative exponents in IF, and forgetting dy/dx = v + x dv/dx. Now go ace that exam!