Differential Equations | FractionRush A-Level

Welcome to Differential Equations

Differential equations describe how quantities change — they are the mathematical language of the physical world. Population growth, cooling coffee, radioactive decay, drug dosage in medicine, electrical circuits — all modelled by differential equations. Mastering this topic connects pure mathematics directly to physics, biology, economics, and engineering.

Separable: dy/dx = f(x)g(y) → ∫(1/g(y)) dy = ∫f(x) dx | Integrating factor: μ = e^{∫P dx}

Key idea: A differential equation links a function to its rate of change. Solving it means finding the original function — often a family of curves (general solution) or one specific curve (particular solution).

Learning Objectives

Form differential equations from word problems and contextual descriptions
Separate variables and integrate both sides correctly
Find the general solution containing an arbitrary constant C
Apply initial conditions to find the particular solution
Sketch families of solution curves for different values of C
Identify and solve exponential growth and decay models
Use the integrating factor method for dy/dx + P(x)y = Q(x)
Solve boundary value problems in context
Connect the mathematical solution back to the physical context
Verify a proposed solution by differentiation and substitution

Topic Overview

Forming DEs

Translating rate-of-change language into equations

Separation of Variables

Rearranging and integrating both sides independently

General Solution

Family of curves — contains arbitrary constant C

Particular Solution

Applying initial conditions to pin down C

Integrating Factor

Method for first-order linear DEs: dy/dx + Py = Q

Exponential Models

Growth, decay, Newton's cooling, half-life

Learn 1 — Forming Differential Equations

Before solving a differential equation, we must be able to form one from a written description. The key is recognising "rate of change" language.

Rate of Change Language

"y increases at a rate proportional to y"
→ dy/dt = ky (k > 0, growth)

"y decreases at a rate proportional to y"
→ dy/dt = −ky (k > 0, decay)

"Rate of change of y is proportional to the product of x and y"
→ dy/dx = kxy

Newton's Law of Cooling

The rate at which a body cools is proportional to the difference between its temperature and the surrounding temperature T₀.

dT/dt = −k(T − T₀) where k > 0

Note: The minus sign ensures the temperature decreases when T > T₀. k is always stated as a positive constant.

Population Growth

dP/dt = kP (k > 0 for growth, k < 0 for decline)

Radioactive Decay

dM/dt = −kM (k > 0; mass decreases over time)

More Complex Situations

Description	Differential Equation
Rate proportional to square of y	dy/dx = ky²
Rate proportional to √y	dy/dx = k√y
Rate proportional to (L − y) where L is max value	dy/dt = k(L − y)
Rate proportional to y(L − y)	dy/dt = ky(L − y) — logistic growth

Tip: Always define your variables clearly. State what y, t, k represent before writing the DE. In exams, marks are awarded for correct notation.

Learn 2 — Separation of Variables

If a differential equation can be written as dy/dx = f(x) · g(y), we can separate the variables — put all y-terms on one side, all x-terms on the other — then integrate both sides.

The Method

dy/dx = f(x) · g(y) → (1/g(y)) dy = f(x) dx → ∫(1/g(y)) dy = ∫f(x) dx

Step 1: Rewrite as (1/g(y)) dy = f(x) dx
Step 2: Integrate both sides
Step 3: Include + C on one side only
Step 4: Rearrange for y if possible (explicit solution)

Example 1: dy/dx = xy

Separate: (1/y) dy = x dx

Integrate: ∫(1/y) dy = ∫x dx → ln|y| = x²/2 + C

Exponentiate: |y| = e^{x²/2 + C} = e^C · e^x²/2

Write A = ±e^C (arbitrary constant): y = Ae^x²/2

Example 2: dy/dx = (1 + y)/x

Separate: dy/(1+y) = dx/x

Integrate: ln|1+y| = ln|x| + C

Exponentiate: |1+y| = e^C|x|, so y = Ax − 1

Example 3: dy/dx = y² cos x

Separate: dy/y² = cos x dx

Integrate: −1/y = sin x + C

Rearrange: y = −1/(sin x + C)

Always include + C on exactly one side. Adding C to both sides is incorrect — it just combines into one constant anyway.

Tip: After integrating ln|y|, write y = Ae^(expression) where A = ±e^C is an arbitrary constant. A can be positive or negative.

Learn 3 — Particular Solutions

The general solution contains an arbitrary constant C and represents a whole family of curves. A particular solution satisfies a specific initial condition — a given value of y at a given x — which determines C exactly.

The Process

Step 1: Solve the DE to get the general solution (with C)
Step 2: Substitute the initial condition (x₀, y₀)
Step 3: Solve for C
Step 4: Write the particular solution with the specific value of C

Example: Population Model

dP/dt = 0.05P, P(0) = 1000.

General solution: P = Ae^0.05t

Apply P(0) = 1000: 1000 = Ae⁰ = A, so A = 1000

Particular solution: P = 1000e^0.05t

Interpretation: Population starts at 1000 and grows at 5% per unit time.

Example: Cooling Problem

dT/dt = −k(T − 20), T(0) = 90.

Separate: dT/(T−20) = −k dt

Integrate: ln|T−20| = −kt + C

T − 20 = Ae^−kt

Apply T(0) = 90: 90 − 20 = A, so A = 70

Particular solution: T = 20 + 70e^−kt

Families of Solution Curves

For dy/dx = y, the general solution is y = Ce^x. Different values of C give different curves:

C = 1: y = e^x (passes through (0,1))
C = 2: y = 2e^x (passes through (0,2))
C = −1: y = −e^x (passes through (0,−1))
C = 0: y = 0 (the x-axis — a valid solution!)

All curves have the same shape; the initial condition shifts the curve vertically.

Sketching tip: For y = Ce^kx (k > 0), all curves pass through (0, C), increase for C > 0, decrease toward 0 for C < 0. The x-axis is a horizontal asymptote for C = 0.

Learn 4 — Integrating Factor Method

When a first-order linear DE cannot be separated (or separation is impractical), use the integrating factor method. The equation must be in standard form:

dy/dx + P(x) · y = Q(x)

The Integrating Factor

μ(x) = e^{∫P(x) dx}

Note: no + C needed when computing the integrating factor itself.

The Method — Step by Step

Step 1: Rearrange into standard form: dy/dx + P(x)y = Q(x)

Step 2: Compute μ = e^{∫P dx}

Step 3: Multiply both sides by μ:
μ · dy/dx + μ · P(x) · y = μ · Q(x)

Step 4: Recognise the left side as d/dx(μ · y) [product rule!]:
d/dx(μ · y) = μ · Q(x)

Step 5: Integrate both sides: μ · y = ∫μ · Q(x) dx + C

Step 6: Divide by μ to find y.

Example 1: dy/dx + y = e^x

P(x) = 1, Q(x) = e^x

μ = e^{∫1 dx} = e^x

Multiply: e^x dy/dx + e^xy = e^2x → d/dx(e^xy) = e^2x

Integrate: e^xy = e^2x/2 + C

Divide: y = e^x/2 + Ce^−x

Example 2: dy/dx + (2/x)y = x²

P(x) = 2/x, Q(x) = x²

μ = e^∫(2/x)dx = e^{2 ln x} = x²

Multiply: x² dy/dx + 2xy = x⁴ → d/dx(x²y) = x⁴

Integrate: x²y = x⁵/5 + C

Divide: y = x³/5 + C/x²

Common error: The integrating factor is e^{∫P dx} — not ∫P dx itself. Always exponentiate after integrating P.

Key check: After multiplying by μ, the left side MUST equal d/dx(μy). If it doesn't, you've made an error in computing μ. This is an automatic check you can always use.

Learn 5 — Solution Curves, Verification & Exponential Models

Sketching Families of Curves

The general solution y = Ae^kx gives a family of curves parameterised by A.

For k > 0 (growth): All curves increase. Positive A → above x-axis. Negative A → below. A = 0 → y = 0.
For k < 0 (decay): All curves decrease toward y = 0 (horizontal asymptote).
Key feature: No two curves from the same family ever intersect (uniqueness theorem).

Verifying a Solution

To verify that a proposed function y = f(x) satisfies a DE, differentiate y, then substitute both y and dy/dx into the original equation and confirm it holds.

Example: Verify y = (x+1)e^x satisfies dy/dx − y = e^x
dy/dx = e^x + (x+1)e^x = (x+2)e^x
dy/dx − y = (x+2)e^x − (x+1)e^x = e^x ✓

Exponential Growth: y = Ae^kt

dy/dt = ky ⟺ y = Ae^kt

Doubling time: Set y = 2A → 2A = Ae^kt → ln 2 = kt → t = (ln 2)/k

Half-life: Set y = A/2 → 1/2 = e^kt → t = −(ln 2)/k (used when k < 0)

Exponential Decay: y = Ae^−kt

dM/dt = −kM ⟺ M = Ae^−kt (k > 0)

Logistic Growth (Qualitative)

The logistic model dy/dt = ky(L − y) produces S-shaped curves. When y is small, growth is nearly exponential. As y approaches the carrying capacity L, growth slows to zero. This is used for population ecology and spread of disease.

Exam tip: For A-Level 9709 P3, you will not be asked to solve logistic equations — but you may need to interpret or sketch them qualitatively.

Worked Examples

Example 1 — Population Growth

A population grows at a rate proportional to its current size. Initially 500. After 10 years, 800. Find the population after 20 years.

Form DE: dP/dt = kP

General solution: P = Ae^kt

Apply P(0) = 500: A = 500, so P = 500e^kt

Apply P(10) = 800: 800 = 500e^10k → e^10k = 1.6 → k = (ln 1.6)/10 ≈ 0.04700

Find P(20): P = 500e^20k = 500 × (1.6)² = 500 × 2.56 = 1280

M1 Correct DE and separation A1 A = 500 M1 Using P(10) A1 1280

Example 2 — Separation with Initial Condition

Solve dy/dx = xy given y(0) = 2.

Separate: (1/y) dy = x dx

Integrate: ln|y| = x²/2 + C

General solution: y = Ae^x²/2

Apply y(0) = 2: 2 = Ae⁰ = A

Particular solution: y = 2e^x²/2

Example 3 — Rational Separation

Solve dy/dx = (1 − y)/x given y(1) = 3.

Separate: dy/(1−y) = dx/x (note: 1−y ≠ 0)

Integrate: −ln|1−y| = ln|x| + C

Rearrange: ln|1−y| = −ln|x| + C → |1−y| = e^C/|x| → 1−y = A/x

General solution: y = 1 − A/x

Apply y(1) = 3: 3 = 1 − A → A = −2

Particular solution: y = 1 + 2/x

Example 4 — Newton's Cooling

dT/dt = −k(T − 20). Initially T = 90°C. After 5 min, T = 70°C. Find time when T = 30°C.

Separate: dT/(T−20) = −k dt

Integrate: ln|T−20| = −kt + C → T − 20 = Ae^−kt

T(0) = 90: 70 = A, so T = 20 + 70e^−kt

T(5) = 70: 50 = 70e^−5k → e^−5k = 5/7 → k = (ln(7/5))/5 ≈ 0.06729

T = 30: 10 = 70e^−kt → e^−kt = 1/7 → t = (ln 7)/k = (ln 7)/0.06729 ≈ 29.0 min

Example 5 — Integrating Factor: dy/dx + 2y = 4x

Standard form: P = 2, Q = 4x

μ = e^{∫2 dx} = e^2x

Multiply: d/dx(e^2xy) = 4xe^2x

Integrate RHS by parts: ∫4xe^2xdx = 4[xe^2x/2 − e^2x/4] = 2xe^2x − e^2x

So: e^2xy = 2xe^2x − e^2x + C

Divide by e^2x: y = 2x − 1 + Ce^−2x

Example 6 — Integrating Factor: dy/dx + (1/x)y = 2

P = 1/x, Q = 2

μ = e^∫(1/x)dx = e^{ln x} = x

Multiply: d/dx(xy) = 2x

Integrate: xy = x² + C

Divide: y = x + C/x

Example 7 — Verify a Solution

Verify that y = (x+1)e^x satisfies dy/dx − y = e^x.

Differentiate: dy/dx = e^x + (x+1)e^x = (x+2)e^x M1

Compute dy/dx − y: (x+2)e^x − (x+1)e^x = e^x ✓ A1

Since dy/dx − y = e^x, the function satisfies the DE.

Example 8 — Sketch Solution Curves for dy/dx = y

General solution: y = Ce^x. Sketch for C = −1, 0, 1, 2.

C = 0: y = 0 (the x-axis; a valid solution)

C = 1: y = e^x; passes through (0,1), increases steeply

C = 2: y = 2e^x; passes through (0,2), steeper scaling

C = −1: y = −e^x; passes through (0,−1), decreases (negative)

All curves approach 0 as x → −∞. None intersect each other.

Common Mistakes

Mistake 1 — Not Integrating Both Sides

✗ dy/y = x dx → y = x (just moved terms, didn't integrate)

✓ ∫dy/y = ∫x dx → ln|y| = x²/2 + C

Mistake 2 — + C on Both Sides

✗ ln|y| + C₁ = x²/2 + C₂ (two constants — wrong)

✓ ln|y| = x²/2 + C (combine C₁ − C₂ = C on one side only)

Mistake 3 — Integrating Factor Computed Incorrectly

✗ μ = ∫P dx (forgetting to exponentiate)

✓ μ = e^{∫P dx} (always exponentiate the integral of P)

Mistake 4 — Not Recognising d/dx(μy)

✗ After multiplying by μ, integrating each term separately

✓ After multiplying by μ, the left side is always d/dx(μy) by the product rule. Write it that way, then integrate.

Mistake 5 — Sign Error in ln Integration

✗ dy/y = k dx → y = kx + C (confusing integration of 1/y)

✓ ∫dy/y = ∫k dx → ln|y| = kx + C → y = Ae^kx

Mistake 6 — Applying Initial Condition Before Solving

✗ Substituting y₀ into the DE itself to find k, then trying to solve

✓ First find the general solution (with C). Then substitute the initial condition to find C. The constant in the solution, not the DE, is what we determine.

Mistake 7 — Sign Confusion in Exponential Decay

✗ dM/dt = −kM → M = Ae^kt (growth, not decay!)

✓ dM/dt = −kM with k > 0 → M = Ae^−kt (decays toward zero). The k in the DE is positive; the minus sign produces the decay.

Key Formulas

Name	Formula	Notes
Separation of Variables	∫(1/g(y)) dy = ∫f(x) dx	Works when DE has form dy/dx = f(x)g(y)
Integrating Factor	μ = e^{∫P(x) dx}	For dy/dx + P(x)y = Q(x)
IF Solution	μy = ∫μQ dx + C	After multiplying both sides by μ
Exponential Growth	y = Ae^kt	Solves dy/dt = ky (k > 0)
Exponential Decay	y = Ae^−kt	Solves dy/dt = −ky (k > 0)
Newton's Cooling	T = T₀ + (T_init − T₀)e^−kt	T₀ = ambient temperature
Doubling Time	t = (ln 2)/k	From y = Ae^kt, when y = 2A
Half-Life	t_½ = (ln 2)/k	From y = Ae^−kt, when y = A/2
General to Particular	Substitute (x₀, y₀) into general solution to find C	Always solve for C after, not before

Proof Bank

Proof 1 — y = Ae^kt satisfies dy/dt = ky

Claim: If y = Ae^kt, then dy/dt = ky.

Proof:

Start with y = Ae^kt where A and k are constants.

Differentiate with respect to t:

dy/dt = A · k · e^kt = k · (Ae^kt) = k · y

So dy/dt = ky. ∎

Converse (solving the DE): Given dy/dt = ky, separate variables:

∫(1/y) dy = ∫k dt → ln|y| = kt + C → y = e^C · e^kt = Ae^kt

So every solution of dy/dt = ky has this form. ∎

Proof 2 — Derivation of the Integrating Factor Method

Goal: Solve dy/dx + P(x)y = Q(x).

Idea: We seek a function μ(x) such that when we multiply both sides by μ, the left side becomes the derivative of a product.

Multiply by μ: μ dy/dx + μP(x)y = μQ(x)

We want the left side to equal d/dx(μy) = μ dy/dx + y dμ/dx

So we need: y dμ/dx = μP(x)y → dμ/dx = μP(x)

This is a separable DE in μ: dμ/μ = P(x) dx

Integrating: ln|μ| = ∫P(x) dx → μ = e^{∫P(x) dx}

With this μ, the equation becomes: d/dx(μy) = μQ(x)

Integrating both sides: μy = ∫μQ(x) dx + C

Therefore: y = (1/μ)[∫μQ(x) dx + C] ∎

Solution Curves Visualiser — y = A · e^kx

Explore how the growth/decay rate k and initial condition A shape the solution curves of dy/dx = ky.

k = 0.50

k = 0.50 Behaviour: Growth Curves: A = −2, −1, 0, 1, 2 DE: dy/dx = ky

Showing 5 solution curves for k = 0.50. Blue = positive A (growth above x-axis). Red = negative A (growth below). Green = A = 0 (trivial solution y = 0).

Exercise 1 — Forming Differential Equations

Exercise 2 — Separating Variables

Exercise 3 — Particular Solutions

Exercise 4 — Integrating Factor

Exercise 5 — Exponential Models

Practice — 30 Questions

Challenge — 15 Questions

Exam Style Questions — Cambridge 9709 P3

Question 1 [6 marks]

A variable y satisfies the differential equation dy/dx = (y + 1)sin x. Given that y = 0 when x = 0, find y in terms of x.

Mark Scheme:
Separate variables: dy/(y+1) = sin x dx M1
Integrate: ln|y+1| = −cos x + C A1
Apply y(0) = 0: ln 1 = −1 + C → C = 1 M1 A1
ln|y+1| = 1 − cos x A1
y + 1 = e^{1−cos x}, so y = e^{1−cos x} − 1 A1

Question 2 [7 marks]

The population P of bacteria in a colony satisfies dP/dt = kP − 500, where k is a positive constant and t is time in hours. Initially P = 2000. After 2 hours P = 3000. Find P when t = 4 and find the value of t when P = 10000.

Mark Scheme:
Separate: dP/(kP−500) = dt → (1/k)ln|kP−500| = t + C M1 A1
P(0) = 2000: (1/k)ln|2000k−500| = C M1
P(2) = 3000: (1/k)ln|3000k−500| − (1/k)ln|2000k−500| = 2
Solving numerically or analytically gives k ≈ 0.203 A1
P(4): kP − 500 = (2000k−500)e^4k → P ≈ 5000 (exact if k = ln(5/4)/2) M1 A1
P = 10000: t = (1/k)ln(9500k/1500k... ) — follow method A1 ft

Question 3 [6 marks]

Find the general solution of dy/dx + 3y = e^−x. Hence find the particular solution for which y → 0 as x → ∞.

Mark Scheme:
Integrating factor: μ = e^3x M1 A1
d/dx(e^3xy) = e^−x · e^3x = e^2x M1
e^3xy = e^2x/2 + C A1
General solution: y = e^−x/2 + Ce^−3x A1
As x→∞, both terms → 0 for any C. But for y→0 non-trivially: C = 0 gives y = e^−x/2 A1

Question 4 [8 marks]

The temperature T (°C) of a body at time t minutes satisfies dT/dt = −0.04(T − 15). The body starts at 75°C. (a) Find T in terms of t. (b) Find the time for the body to cool to 25°C. (c) Find dT/dt when T = 25°C and explain its physical significance.

Mark Scheme:
(a) Separate: dT/(T−15) = −0.04 dt → ln|T−15| = −0.04t + C M1 A1
T(0) = 75: C = ln 60, T − 15 = 60e^−0.04t, T = 15 + 60e^−0.04t A1
(b) 25 = 15 + 60e^−0.04t → e^−0.04t = 1/6 → t = (ln 6)/0.04 = 44.8 min M1 A1
(c) dT/dt = −0.04(25−15) = −0.04 × 10 = −0.4°C/min M1 A1
Significance: at T = 25°C, the body is cooling at 0.4°C per minute. B1

Question 5 [5 marks]

Verify that y = xe^2x is a solution of dy/dx − 2y = e^2x. Find the general solution of this DE.

Mark Scheme:
Verify: dy/dx = e^2x + 2xe^2x M1
dy/dx − 2y = e^2x + 2xe^2x − 2xe^2x = e^2x ✓ A1
General solution: μ = e^−2x, d/dx(e^−2xy) = 1 M1
e^−2xy = x + C → y = xe^2x + Ce^2x A1 A1

Question 6 [6 marks]

The rate of decay of a substance is proportional to the amount present. Initially 100g. After 20 days, 60g remains. Find: (a) the mass after 50 days, (b) the half-life of the substance.

Mark Scheme:
DE: dM/dt = −kM → M = 100e^−kt M1 A1
M(20) = 60: e^−20k = 0.6 → k = (ln(5/3))/20 ≈ 0.02554 M1 A1
(a) M(50) = 100 × 0.6^50/20 = 100 × 0.6^2.5 = 100 × 0.2786 ≈ 27.9 g A1
(b) Half-life: 0.5 = e^−kt → t = (ln 2)/k = (ln 2 × 20)/(ln(5/3)) ≈ 27.1 days A1

Question 7 [7 marks]

Solve the differential equation (x + 1)dy/dx + y = (x + 1)², given y = 4 when x = 1. Give your answer in the form y = f(x).

Mark Scheme:
Divide by (x+1): dy/dx + y/(x+1) = x+1 → standard form P = 1/(x+1) M1
μ = e^∫1/(x+1)dx = e^ln(x+1) = x+1 A1
d/dx((x+1)y) = (x+1)² M1 A1
(x+1)y = (x+1)³/3 + C A1
Apply y(1) = 4: 2 × 4 = 8/3 + C → C = 8 − 8/3 = 16/3 M1
y = (x+1)²/3 + 16/(3(x+1)) A1

Question 8 [5 marks]

The gradient of a curve at point (x, y) is given by dy/dx = y²e^x. The curve passes through (0, 1). Find the equation of the curve and state any restrictions on x.

Mark Scheme:
Separate: dy/y² = e^x dx M1
Integrate: −1/y = e^x + C A1
Apply (0,1): −1 = 1 + C → C = −2 M1 A1
y = 1/(2 − e^x), valid for e^x < 2, i.e. x < ln 2 A1

Past Paper Questions — Cambridge 9709

9709/32/O/N/20 Q8 [9 marks]

The variables x and y satisfy the differential equation (1 + x²)dy/dx = xy + x. It is given that y = 0 when x = 0. Find y in terms of x.

Mark Scheme:
Factorise RHS: xy + x = x(y + 1) M1
Separate: dy/(y+1) = x/(1+x²) dx M1 A1
Integrate: ln|y+1| = ½ ln(1+x²) + C A1
Apply y(0) = 0: 0 = 0 + C → C = 0 M1 A1
ln|y+1| = ½ ln(1+x²) = ln√(1+x²) A1
y + 1 = √(1+x²) (positive since y(0)=0 requires y+1 > 0) A1
y = √(1+x²) − 1 A1

9709/32/M/J/21 Q9 [9 marks]

The variables x and t satisfy the differential equation dx/dt = (3 − x)(1 + x) for t ≥ 0 and 0 ≤ x < 3. Use partial fractions to find x in terms of t, given that x = 0 when t = 0.

Mark Scheme:
Separate: dx/[(3−x)(1+x)] = dt M1
Partial fractions: 1/[(3−x)(1+x)] = A/(3−x) + B/(1+x); A = 1/4, B = 1/4 M1 A1
Integrate: (1/4)[−ln|3−x| + ln|1+x|] = t + C A1
i.e. (1/4)ln|(1+x)/(3−x)| = t + C A1
Apply x(0) = 0: C = (1/4)ln(1/3) M1 A1
(1/4)ln|(1+x)/(3−x)| = t + (1/4)ln(1/3)
(1+x)/(3−x) = (1/3)e^4t A1
Solve: 3(1+x) = (3−x)e^4t... → x = 3(e^4t−1)/(e^4t+3) A1

9709/31/M/J/22 Q6 [7 marks]

Find the general solution of the differential equation dy/dx + y cot x = cos x, giving y in terms of x.

Mark Scheme:
Standard form: P = cot x, Q = cos x B1
μ = e^{∫cot x dx} = e^{ln|sin x|} = sin x M1 A1
d/dx(y sin x) = cos x · sin x = ½ sin 2x M1 A1
y sin x = −¼ cos 2x + C A1
y = (−¼ cos 2x + C)/sin x = ½ sin x + C cosec x (using cos 2x = 1−2sin²x) A1

9709/32/O/N/22 Q7 [8 marks]

The variables x and y are related by dy/dx = 6e^2xy². It is given that y = 1 when x = 0. (a) Find y in terms of x. (b) State the value of x for which the solution is not valid.

Mark Scheme:
(a) Separate: dy/y² = 6e^2x dx M1
Integrate: −1/y = 3e^2x + C A1
Apply y(0) = 1: −1 = 3 + C → C = −4 M1 A1
−1/y = 3e^2x − 4 → y = 1/(4 − 3e^2x) A1
(b) y undefined when 4 − 3e^2x = 0 → e^2x = 4/3 → x = ½ ln(4/3) M1 A1

9709/33/M/J/23 Q5 [7 marks]

The variables x and y satisfy dy/dx − 2y = e^3x. Find the particular solution for which y = 2 when x = 0.

Mark Scheme:
Standard form: P = −2, Q = e^3x B1
μ = e^−2x M1 A1
d/dx(e^−2xy) = e^3x · e^−2x = e^x M1
e^−2xy = e^x + C A1
y = e^3x + Ce^2x
Apply y(0) = 2: 2 = 1 + C → C = 1 M1
y = e^3x + e^2x A1

Differential Equations A-Level Pure 3