Edexcel IAL - Further Pure Mathematics 2- 4.1 Separable First Order Differential Equations- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 2- 4.1 Separable First Order Differential Equations -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 2- 4.1 Separable First Order Differential Equations -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 4.1 Separable First Order Differential Equations
First Order Differential Equations with Separable Variables
A first order differential equation involves the first derivative \( \dfrac{dy}{dx} \) of a function.
A differential equation is called separable if it can be written in the form
\( \dfrac{dy}{dx} = f(x)\,g(y) \)
This means that all terms involving \( y \) can be separated from those involving \( x \).
Solving a Separable Differential Equation
Starting with
\( \dfrac{dy}{dx} = f(x)g(y) \)
we rearrange:
\( \dfrac{1}{g(y)}\,dy = f(x)\,dx \)
Then integrate both sides:
\( \displaystyle \int \dfrac{1}{g(y)}\,dy = \int f(x)\,dx \)
This gives the general solution.
Using given values (initial conditions) produces a particular solution.
Formation of the Differential Equation
Sometimes the differential equation is formed from a statement describing how two variables change together.
For example:
- “The rate of increase of \( y \) is proportional to \( y \)” gives \( \dfrac{dy}{dx} = ky \).
- “The rate of decrease of \( y \) is proportional to \( x \)” gives \( \dfrac{dy}{dx} = -kx \).
Family of Curves
The constant of integration \( C \) produces a family of curves.
Different values of \( C \) give different solution curves on the same axes.
Example
Solve \( \dfrac{dy}{dx} = 3x y \).
▶️ Answer / Explanation
Separate variables: \( \dfrac{1}{y}dy = 3x\,dx \)
Integrate: \( \ln |y| = \dfrac{3x^2}{2} + C \)
So: \( y = Ce^{\tfrac{3x^2}{2}} \)
Example
Solve \( \dfrac{dy}{dx} = x(1+y^2) \), given that \( y=0 \) when \( x=0 \).
▶️ Answer / Explanation
\( \dfrac{dy}{1+y^2} = x\,dx \)
Integrate: \( \tan^{-1}y = \dfrac{x^2}{2} + C \)
Use \( y=0 \) when \( x=0 \): \( C=0 \)
So \( y = \tan\left(\dfrac{x^2}{2}\right) \)
Example
The rate of increase of \( y \) is proportional to \( xy \). Given \( y=2 \) when \( x=1 \), find \( y \) in terms of \( x \).
▶️ Answer / Explanation
\( \dfrac{dy}{dx} = kxy \)
Separate: \( \dfrac{1}{y}dy = kx\,dx \)
Integrate: \( \ln y = \dfrac{kx^2}{2} + C \)
Using \( y=2 \) when \( x=1 \) gives \( C = \ln2 – \dfrac{k}{2} \)