Differential Equations
Modelling real-world change by relating a function to its derivatives.
### Introduction to Differential Equations
A differential equation is a powerful mathematical tool that describes the relationship between a function and its derivatives. In essence, it's an equation that involves an unknown function and its rates of change. For the Cambridge A-Level Mathematics (9709) syllabus, we focus on first-order differential equations. These are equations that involve only the first derivative of a function, typically written as `dy/dx`.
Why are they important? Differential equations are the language of change. They are used to model dynamic systems in physics, biology, economics, and engineering. Examples include modelling population growth, the cooling of a hot object, the decay of radioactive material, or the flow of electricity in a circuit. Our focus is on learning how to set up these equations from a given context and solve them using a specific technique.
### Formulating First-Order Differential Equations
Before solving a differential equation, you must often formulate it from a problem described in words. This involves translating a statement about a rate of change into a mathematical equation. The key is to identify phrases that indicate proportionality.
Example: The rate at which a liquid cools is proportional to the difference between its temperature, `θ`, and the constant room temperature, `S`. This is Newton's Law of Cooling and can be formulated as:
`dθ/dt = -k(θ - S)`
The negative sign is crucial as it indicates that the temperature is decreasing (cooling).
### Solving by Separation of Variables
The primary method for solving first-order differential equations in this syllabus is the method of separating variables. This technique can be used when the equation can be written in the form `dy/dx = f(x)g(y)`, where the expression for the derivative is a product of a function of `x` and a function of `y`.
The process follows these clear steps:
`h(y) dy = f(x) dx`
`∫ h(y) dy = ∫ f(x) dx`
### General and Particular Solutions
The general solution is not a single answer but a family of them. To find a specific solution, we need more information, known as an initial condition or a boundary condition. This is a given point `(x₀, y₀)` that lies on the curve of the specific solution we are looking for.
To find the particular solution:
Worked Example:
Find the particular solution of the differential equation `dy/dx = (y + 1) / x` given that `y = 2` when `x = 1`.
* Step 1: Separate the variables.
`1 / (y + 1) dy = 1 / x dx`
* Step 2: Integrate both sides.
`∫ 1 / (y + 1) dy = ∫ 1 / x dx`
* Step 3: Find the general solution.
`ln|y + 1| = ln|x| + C`
This is the general solution. It is often useful to simplify this logarithmic form. Let `C = ln(A)`, where `A` is another constant. Then:
`ln|y + 1| = ln|x| + ln(A) = ln|Ax|`
`y + 1 = Ax`
`y = Ax - 1`
* Step 4: Find the particular solution using the condition (1, 2).
Substitute `x = 1` and `y = 2` into the general solution `y = Ax - 1`:
`2 = A(1) - 1`
`3 = A`
* Step 5: State the final answer.
Substitute `A = 3` back into the general solution:
`y = 3x - 1`
This is the particular solution that satisfies the given differential equation and passes through the point (1, 2).
Key Points to Remember
- 1A **first-order differential equation** relates a variable `y` to its first derivative `dy/dx`.
- 2**Formulating** an equation involves translating word problems about rates of change into mathematical expressions, often using a constant of proportionality, `k`.
- 3The core method of solving is **separation of variables**: grouping all `y` terms with `dy` and all `x` terms with `dx` on opposite sides of the equation.
- 4After separating, **integrate** both sides of the equation with respect to their respective variables.
- 5Always add a single **constant of integration**, `C`, to find the **general solution**, which represents a family of functions.
- 6Use given **initial conditions** (a specific point `(x, y)`) to find the value of `C` and determine the unique **particular solution**.
- 7Logarithmic results from integration, like `ln|y| = ...`, can often be simplified into an exponential form `y = Ae^...`.
- 8Common applications include population growth (`dP/dt = kP`) and Newton's Law of Cooling (`dθ/dt = -k(θ - S)`).
Pakistan Example
Modelling Water Evaporation from Tarbela Dam
The rate of water loss due to evaporation from a reservoir like Tarbela Dam can be modelled as being directly proportional to the current volume of water, `V`. This gives the differential equation `dV/dt = -kV`, where `k` is the evaporation constant and `t` is time in days. If the dam holds 10 billion cubic metres of water and the constant `k` is found to be 0.001, students can solve this equation by separating variables: `(1/V)dV = -0.001 dt`. Integrating and using the initial condition `V(0) = 10`, they can find the particular solution `V(t) = 10e^(-0.001t)`. This allows them to predict the volume of water remaining after a certain number of days, connecting differential equations to Pakistan's critical water resource management.
Quick Revision Infographic
Mathematics — Quick Revision
Differential Equations
Key Concepts
Formulas to Know
Logarithmic results from integration, like `ln|y| = ...`, can often be simplified into an exponential form `y = Ae^...`.P/dt = kP`) and Newton's Law of Cooling (`dθ/dt = -k(θ - S)`).Modelling Water Evaporation from Tarbela Dam
The rate of water loss due to evaporation from a reservoir like Tarbela Dam can be modelled as being directly proportional to the current volume of water, `V`. This gives the differential equation `dV/dt = -kV`, where `k` is the evaporation constant and `t` is time in days. If the dam holds 10 billion cubic metres of water and the constant `k` is found to be 0.001, students can solve this equation by separating variables: `(1/V)dV = -0.001 dt`. Integrating and using the initial condition `V(0) = 10`, they can find the particular solution `V(t) = 10e^(-0.001t)`. This allows them to predict the volume of water remaining after a certain number of days, connecting differential equations to Pakistan's critical water resource management.