Edexcel IAL - Further Pure Mathematics 2- 5.2 Reduction by Substitution- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 2- 5.2 Reduction by Substitution -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 2- 5.2 Reduction by Substitution -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 5.2 Reduction by Substitution
Differential Equations Reducible by Substitution
Some differential equations do not at first appear to be separable or linear, but by making a suitable substitution they can be transformed into one of these standard forms.
The idea is to recognise a repeated combination of variables and replace it with a new variable so that the equation becomes easier to solve.
1. Homogeneous Type
An equation of the form
\( \dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right) \)
can be reduced using the substitution
\( y = vx \)
Then
\( \dfrac{dy}{dx} = v + x\dfrac{dv}{dx} \)
which usually leads to a separable equation.
2. Bernoulli Type
An equation of the form
\( \dfrac{dy}{dx} + P(x)y = Q(x)y^n \)
is reducible to a linear equation by using the substitution
\( v = y^{1-n} \)
3. Special Product or Sum Substitutions
If an equation contains combinations such as \( xy \), \( x+y \) or \( x^2+y^2 \), a substitution like
\( v = xy \), \( v = x+y \)
can simplify the equation.
Example
Solve \( \dfrac{dy}{dx} = \dfrac{y}{x} \).
▶️ Answer / Explanation
\( \dfrac{1}{y}dy = \dfrac{1}{x}dx \)
\( \ln y = \ln x + C \Rightarrow y = Cx \)
Example
Solve \( \dfrac{dy}{dx} = \dfrac{x+y}{x} \).
▶️ Answer / Explanation
Let \( y = vx \), then \( \dfrac{dy}{dx} = v + x\dfrac{dv}{dx} \)
\( v + x\dfrac{dv}{dx} = 1 + v \Rightarrow x\dfrac{dv}{dx} = 1 \)
\( v = \ln x + C \Rightarrow y = x(\ln x + C) \)
Example
Solve \( \dfrac{dy}{dx} + y = y^2 \).
▶️ Answer / Explanation
This is Bernoulli type.
Let \( v = y^{-1} \).
Then it becomes a linear equation in \( v \), which can be solved using an integrating factor.