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IB Mathematics AI AHL Solving by separation of variables MAI Study Notes - New Syllabus

IB Mathematics AI AHL Solving by separation of variables MAI Study Notes

LEARNING OBJECTIVE

  • Setting up a model/differential equation from a context. 

Key Concepts: 

  • separation of variables. 

MAI HL and SL Notes – All topics

SETTING UP A DIFFERENTIAL EQUATION

Setting Up a Differential Equation

In many real-life situations, quantities change over time or with respect to another variable. A differential equation is an equation that relates a function with its derivative and is commonly used to model such changing systems.

Key steps in modeling a situation using a differential equation:

  1. Identify the dependent variable: This is the quantity that is changing (e.g. population, temperature, amount of drug, water level).
  2. Identify the independent variable: This is the variable with respect to which the change occurs (usually time \( t \)).
  3. Describe the relationship: Use the problem context to express how the rate of change of the quantity depends on the quantity itself or other factors.
  4. Translate the relationship into a differential equation.

Common Types of Contexts:

  • Exponential Growth/Decay: “The rate of change of a population is proportional to the size of the population.”
  • Cooling/Heating: “The rate of cooling is proportional to the temperature difference between an object and its surroundings.”
  • Economics: “The rate of change of profit depends on cost and revenue functions.”
  • Motion problems: “Acceleration is proportional to velocity or displacement.”

Example

Context: A tank contains 100 liters of pure water. A solution containing 0.2 grams of salt per liter flows into the tank at a rate of 5 liters per minute. The mixture is kept uniform and flows out at the same rate. Set up the differential equation that models the amount of salt in the tank over time.

▶️Answer/Explanation

 Let \( S(t) \) be the amount of salt (in grams) in the tank at time \( t \) (in minutes).

Salt enters at a rate:

\( \text{inflow rate} = 0.2 \times 5 = 1 \text{ gram/min} \)

Salt leaves at a rate:

\( \frac{S(t)}{100} \times 5 = \frac{5S(t)}{100} = \frac{S(t)}{20} \text{ grams/min} \)

$ \frac{dS}{dt} = \text{rate in} – \text{rate out} = 1 – \frac{S(t)}{20}$

This is the differential equation that models the amount of salt in the tank over time.

VARIABLES SEPARABLE METHOD

Variables Separable Method

Concept:

Applicable when a differential equation can be written as:

$ \frac{dy}{dx} = f(x)g(y) $

We separate the variables and integrate:

$ \int \frac{1}{g(y)}\,dy = \int f(x)\,dx $

General Steps:

  1. Separate variables
  2. Integrate both sides
  3. Solve for \( y \), include constant \( C \)

Example

Solve \( \frac{dy}{dx} = 3y \).

▶️Answer/Explanation

 $ \frac{dy}{y} = 3 dx \quad \Rightarrow \quad \ln|y| = 3x + C \quad \Rightarrow \quad y = Ae^{3x} $

 Growth of Algae – Exponential Model

The growth of algae \( G \), at time \( t \), is proportional to the amount present, so:

$ \frac{dG}{dt} = kG $

This is a first-order separable differential equation and models exponential growth. We are to solve it using the separation of variables method.

Example

$ \frac{dG}{dt} = kG $ Find general solution of the differential equation.

▶️Answer/Explanation

We are given: $ \frac{dG}{dt} = kG $

Separate variables: $ \frac{1}{G} \, dG = k \, dt $

Integrate both sides: $ \int \frac{1}{G} \, dG = \int k \, dt \Rightarrow \ln|G| = kt + C $

Exponentiate both sides: $ |G| = e^{kt+C} = Ae^{kt}, \text{ where } A = e^C $ Since \( G > 0 \) (algae population),

we can write the solution as: $ G(t) = Ae^{kt} $ This is the general solution of the differential equation.

Integrating Factor Method

Concept:

Used for linear first-order differential equations of the form:

$ \frac{dy}{dx} + P(x)y = Q(x) $

We multiply both sides by the integrating factor:

$ \mu(x) = e^{\int P(x)\,dx} $

This turns the left-hand side into a product derivative:

$ \frac{d}{dx}[\mu(x)y] = \mu(x)Q(x) $

Then integrate both sides to find \( y \).

Example

Solve \( \frac{dy}{dx} + 2y = x \)

▶️Answer/Explanation

 $ \mu(x) = e^{\int 2 dx} = e^{2x} \quad \Rightarrow \quad \frac{d}{dx}(e^{2x}y) = xe^{2x} $

Integrate both sides: $ e^{2x}y = \int xe^{2x} dx = \frac{xe^{2x}}{2} – \frac{e^{2x}}{4} + C $

$ y = \frac{x}{2} – \frac{1}{4} + Ce^{-2x} $

HOMOGENEOUS DIFFERENTIAL EQUATIONS

Homogeneous Differential Equations

Concept:

A differential equation is homogeneous if it can be expressed in the form:

$ \frac{dy}{dx} = F\left( \frac{y}{x} \right) $

We use substitution:

$ y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} $

Then reduce to a separable equation in \( v \) and \( x \).

Example

Solve \( \frac{dy}{dx} = \frac{x + y}{x} \)

▶️Answer/Explanation

 $ \Rightarrow \frac{dy}{dx} = 1 + \frac{y}{x} \Rightarrow \text{homogeneous} $

Let \( y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} \)

$ v + x \frac{dv}{dx} = 1 + v \Rightarrow x \frac{dv}{dx} = 1 \Rightarrow dv = \frac{1}{x} dx $

Integrate: $ v = \ln|x| + C \Rightarrow \frac{y}{x} = \ln|x| + C \Rightarrow y = x(\ln|x| + C) $

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