Edexcel IAL - Further Pure Mathematics 2- 4.3 Differential Equations by Substitution- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 2- 4.3 Differential Equations by Substitution -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 2- 4.3 Differential Equations by Substitution -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 4.3 Differential Equations by Substitution
Differential Equations Reducible by Substitution
Some differential equations are not immediately separable or linear, but they can be turned into one of these standard forms by making a suitable substitution.
Once the substitution is made, the new equation can be solved using:
- Separable variables, or
- The linear integrating factor method.
Idea of Substitution
If an equation involves a repeated combination of variables, such as \( xy \), \( x+y \), or \( y/x \), we define a new variable.
For example:
Let \( v = xy \) or \( v = \dfrac{y}{x} \)
Then differentiate \( v \) with respect to \( x \) and rewrite the equation in terms of \( v \) and \( x \).
Homogeneous Differential Equations
An equation of the form
\( \dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right) \)
can be solved using the substitution:
\( y = vx \)
Then:
\( \dfrac{dy}{dx} = v + x\dfrac{dv}{dx} \)
This usually leads to a separable equation.
Bernoulli Type
An equation of the form
\( \dfrac{dy}{dx} + Py = Qy^n \)
can be made linear using the substitution
\( v = y^{1-n} \)
Example
Solve \( \dfrac{dy}{dx} = \dfrac{y}{x} \).
▶️ Answer / Explanation
This is separable:
\( \dfrac{1}{y}dy = \dfrac{1}{x}dx \)
\( \ln y = \ln x + C \Rightarrow y = Cx \)
Example
Solve \( \dfrac{dy}{dx} = \dfrac{x+y}{x} \).
▶️ Answer / Explanation
Rewrite: \( \dfrac{dy}{dx} = 1 + \dfrac{y}{x} \)
Let \( y = vx \), so \( \dfrac{dy}{dx} = v + x\dfrac{dv}{dx} \)
\( v + x\dfrac{dv}{dx} = 1 + v \Rightarrow x\dfrac{dv}{dx} = 1 \)
\( \dfrac{dv}{dx} = \dfrac{1}{x} \Rightarrow v = \ln x + C \)
So \( y = x(\ln x + C) \)
Example
Solve \( \dfrac{dy}{dx} + y = y^2 \).
▶️ Answer / Explanation
This is Bernoulli type with \( n=2 \).
Let \( v = y^{-1} \).
\( \dfrac{dv}{dx} = -y^{-2}\dfrac{dy}{dx} \)
Substitute to obtain a linear equation in \( v \), then solve.