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📚 DIFFERENTIAL EQUATIONS

Cambridge AS & A Level Mathematics (Chapter 10)

Sections 10.1 - 10.2: Separating Variables & Forming Equations

🎯 What is a Differential Equation?

Definition: An equation containing derivatives such as dy/dx or d²y/dx² is called a differential equation.

KEY CONCEPT: The solution of a differential equation is a function, not just a number!

Types of Solutions

Solution Type	Description	Example
General Solution	Contains arbitrary constant(s)	y = x³ + C
Particular Solution	Specific solution using initial conditions	y = x³ + 5

First-Order Differential Equation:
Contains only dy/dx (first derivative)

Example: dy/dx = 3x²

🔧 Section 10.1: Separation of Variables

The Basic Method

STEP-BY-STEP PROCEDURE:

1. Separate the variables: Rearrange to get f(y) dy = g(x) dx
2. Integrate both sides: ∫f(y) dy = ∫g(x) dx
3. Solve for the relationship between x and y (general solution)
4. Apply initial conditions to find the constant C (particular solution)
5. Check your answer by differentiating

KEY RULE:
If you integrate f(y) · dy/dx with respect to x, it's the same as integrating f(y) with respect to y:

∫ f(y) · dy/dx dx = ∫ f(y) dy

Example 1: Simple Separation

Problem: Solve dy/dx = xy, given y = 2 when x = 0 (y > 0)

Solution:
Step 1 - Separate:
dy/dx = xy
(1/y) dy = x dx

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

Step 3 - Solve for y:
y = e^((1/2)x² + C) = Ae^((1/2)x²)

Step 4 - Use initial condition:
When x = 0, y = 2:
2 = A · e^0 = A
Therefore: y = 2e^((1/2)x²)

Example 2: Using Initial Conditions

Problem: Solve dy/dx = x/(2y), given y = 3 when x = 0

Solution:
Separate: 2y dy = x dx
Integrate: ∫2y dy = ∫x dx
y² = (1/2)x² + C

Use y = 3 when x = 0:
9 = 0 + C → C = 9
Answer: y² = (1/2)x² + 9

⚠️ Important: Check Excluded Values!

When separating variables, if you divide by an expression in y (like y³ or (y-1)), check afterward whether the excluded values (like y = 0 or y = 1) are also solutions of the original equation.

📝 Section 10.2: Forming Differential Equations

Translating Words into Math

English Phrase	Mathematical Form
Rate of increase of y	dy/dt
Rate of decrease of y	-dy/dt or dy/dt = -k(...)
Proportional to x	= kx
Inversely proportional to x	= k/x
Proportional to x²	= kx²
Proportional to √x	= k√x

Common Models & Their Equations

1. EXPONENTIAL GROWTH
"Rate of increase is proportional to current population"
dP/dt = kP (k > 0)
Solution: P = P₀e^(kt)

2. EXPONENTIAL DECAY
"Rate of decrease is proportional to current amount"
dV/dt = -kV (k > 0)
Solution: V = V₀e^(-kt)
Examples: Radioactive decay, depreciation

3. NEWTON'S LAW OF COOLING
"Rate of change of temperature is proportional to temperature difference"
dT/dt = k(T - Tₛ)
where Tₛ = surrounding temperature
Solution: T = Tₛ + Ae^(kt)

4. LOGISTIC GROWTH
"Population growth with limited resources"
dP/dt = rP(1 - P/K)
where K = carrying capacity
As t → ∞, P → K

Example 3: Velocity Problem

Problem: The velocity of a particle is inversely proportional to its displacement s from origin O. Initially s = 2m. After 1 second, s = 5m.

Form the differential equation:
Velocity = ds/dt
"Inversely proportional to s" means:
ds/dt = k/s

Solve:
s · ds = k · dt
∫s ds = ∫k dt
(1/2)s² = kt + C

Use initial conditions:
When t = 0, s = 2: (1/2)(4) = C → C = 2
When t = 1, s = 5: (1/2)(25) = k + 2 → k = 10.5

Final answer: s² = 21t + 4

Example 4: Population Density

Given:
• At t = 0: P = 2 people/hectare, dP/dt = 0.5
• Rate of increase is inversely proportional to P²

Form equation:
dP/dt = k/P²
0.5 = k/4 → k = 2
dP/dt = 2/P²

Solve:
P² dP = 2 dt
∫P² dP = ∫2 dt
(1/3)P³ = 2t + C

Using P = 2 when t = 0:
(1/3)(8) = C → C = 8/3
P³ = 6t + 8

Example 5: Motorbike Depreciation

Given:
• Initial value: V₀ = 10,000
• Rate of decrease proportional to V
• After 3 years, V = 5,000

Model: dV/dt = -kV
Solution: V = 10,000e^(-kt)

Find k:
When t = 36 months, V = 5,000:
5,000 = 10,000e^(-36k)
e^(-36k) = 0.5
k = -ln(0.5)/36

Value after 2 years (t=24):
V = 10,000e^(24·ln(0.5)/36) ≈ 6,300

📋 Quick Reference Formulas

STANDARD INTEGRALS:
• ∫(1/y) dy = ln|y| + C
• ∫e^y dy = e^y + C
• ∫y^n dy = y^(n+1)/(n+1) + C (n ≠ -1)
• ∫sin y dy = -cos y + C
• ∫cos y dy = sin y + C

LOGARITHM RULES:
• ln(AB) = ln A + ln B
• ln(A/B) = ln A - ln B
• ln(A^n) = n ln A
• e^(ln A) = A
• ln(e^A) = A

✅ Important Tips & Warnings

✓ Always define your variables clearly (e.g., "Let P = population at time t")
✓ Check if modulus signs |y| are needed in ln|y|
✓ Remember: one arbitrary constant for first-order equations
✓ Always use initial conditions if given
✓ Verify your solution by differentiating

⚠️ Common Mistakes:
• Forgetting to separate variables completely
• Missing the constant of integration C
• Not using initial conditions
• Dividing by expressions that could be zero
• Wrong sign on "rate of decrease" (should be negative)

🎓 Practice Strategy

To master this chapter:

1. Technical skill: Practice separating variables and integrating (use partial fractions when needed)
2. Initial conditions: Always apply them to find specific solutions
3. Word problems: Train yourself to translate English phrases into mathematical equations
4. Real-world context: Understand what your solution means physically (population, temperature, etc.)
5. Check solutions: Differentiate your answer to verify it satisfies the original equation

📖 Cambridge International AS & A Level Mathematics
Pure Mathematics 2 & 3 - Chapter 10
Good luck with your studies! 🚀