Reducing First-Order Differential Equations (Edexcel A Level Further Maths): Revision Note

Exam code: 9FM0

Author

Mark Curtis

Last updated

6 January 2026

Reducing first-order differential equations

What does reducing first-order differential equations mean?

A hard first-order differential equation can be reduced (transformed) into an easier first-order differential equation
- using a given transformation
The easier differential equation can then be solved
- e.g. using
  - direct integration
  - separation of variables
  - or the integrating factor
The general solution to the easier differential equation can then be transformed back
- to give the general solution to the harder differential equation
- from which you can work out the particular solution
  - using the given boundary conditions

What is the dependent variable and what is the independent variable?

If the solution to the differential equation $\frac{d y}{d x} = f (x, y)$ is $y = g (x)$ , then
- $y$ is the dependent variable
- $x$ is the independent variable
  - as $y$ depends on $x$

Examiner Tips and Tricks

Be careful in modelling questions, as the letters can change, e.g. $\frac{d x}{d t} = x + t^{2}$ has a dependent variable of $x$ and an independent variable of $t$ .

How do I transform the dependent variable?

If you are given a transformation of the dependent variable
- i.e. changing $(x, y)$ into $(x, z)$
- using the transformation
  - $y = h (z)$
  - or $z = h^{- 1} (y)$
- then use the chain rule to rewrite $\frac{d y}{d x}$ in terms of $\frac{d z}{d x}$
  - $\frac{d y}{d x} = \frac{d y}{d z} \times \frac{d z}{d x}$
  - It sometimes also helps to use that $\frac{d y}{d z} = \frac{1}{\frac{d z}{d y}}$
  - as long as the end result is in $(x, z)$ only
  - i.e. no $y$ terms
- then substitute this derivative and the transformation into the differential equation
  - See the worked example below

Worked Example

Use the transformation $y = e^{z}$ to find the particular solution of the differential equation

$\frac{d y}{d x} = x y \ln y$

where $y = 2$ when $x = 0$ .

Answer:

Identify the variables being transformed

$(x, y) \to (x, z)$

This is a transformation of the dependent variable, $y$

Write $\frac{d y}{d x}$ in terms of $\frac{d z}{d x}$ using the chain rule

$\frac{d y}{d x} = \frac{d y}{d z} \times \frac{d z}{d x}$

Find $\frac{d y}{d z}$ from $y = e^{z}$

$\frac{d y}{d z} = e^{z}$

Substitute this into the chain rule

$\frac{d y}{d x} = e^{z} \frac{d z}{d x}$

Now substitute this derivative into the original differential equation

$e^{z} \frac{d z}{d x} = x y \ln y$

This is not yet in the form $(x, z)$ , as there are still $y$ terms on the right-hand side

Use $y = e^{z}$ and $\ln y = \ln (e^{z}) = z$ to convert the remaining $y$ terms into $z$ terms

$\begin{array}{rcl} e^{z} \frac{d z}{d x} & = & x \times e^{z} \times z \\ e^{z} \frac{d z}{d x} & = & x z e^{z} \\ \frac{d z}{d x} & = & x z \end{array}$

Solve this differential equation using separation of variables

Remember to add a constant of integration

$\begin{array}{rcl} \frac{1}{z} \frac{d z}{d x} & = & x \\ \int \frac{1}{z} d z & = & \int x d x \\ \ln | z | & = & \frac{x^{2}}{2} + c \\ z & = & e^{\frac{x^{2}}{2} + c} \\ z & = & e^{\frac{x^{2}}{2}} e^{c} \\ z & = & A e^{\frac{x^{2}}{2}} \end{array}$

Transform the general solution back to the variables $(x, y)$ using $y = e^{z}$ and $\ln y = z$

$\begin{array}{rcl} \ln y & = & A e^{\frac{x^{2}}{2}} \\ y & = & e^{A e^{\frac{x^{2}}{2}}} \end{array}$

Now substitute in the boundary conditions $x = 0$ and $y = 2$ to find $A$

$\begin{array}{rcl} 2 & = & e^{A e^{0}} \\ 2 & = & e^{A} \\ \ln 2 & = & A \end{array}$

Substitute this value of $A$ back into the general solution to get the particular solution

$\begin{array}{rcl} y & = & e^{(\ln 2) e^{\frac{x^{2}}{2}}} \end{array}$

This is the answer, but it can also be simplified to $y = {(e^{\ln 2})}^{e^{\frac{x^{2}}{2}}} = 2^{e^{\frac{x^{2}}{2}}}$

How do I transform the independent variable?

If you are given a transformation of the independent variable
- i.e. changing $(x, y)$ into $(t, y)$
- using the transformation
  - $x = h (t)$
  - or $t = h^{- 1} (x)$
- then use the chain rule to rewrite $\frac{d y}{d x}$ in terms of $\frac{d y}{d t}$
  - $\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}$
  - It sometimes helps to also use that $\frac{d t}{d x} = \frac{1}{\frac{d x}{d t}}$
  - as long as the end result is in $(t, y)$ only
  - i.e. no $x$ terms
- then substitute this derivative and the transformation into the differential equation
  - See the worked example below

Worked Example

Use the transformation $x = \frac{1}{t}$ where $t > 0$ to find the particular solution of the differential equation

$x^{2} \frac{d y}{d x} - x y = - \frac{3}{x}$

where $y = 2$ when $x = 1$ .

Answer:

Identify the variables being transformed

$(x, y) \to (t, y)$

This is a transformation of the independent variable, $x$

Write $\frac{d y}{d x}$ in terms of $\frac{d y}{d t}$ using the chain rule

$\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}$

To find $\frac{d t}{d x}$ from $x = \frac{1}{t} = t^{- 1}$ it is easier to use that $\frac{d t}{d x} = \frac{1}{\frac{d x}{d t}}$

$\begin{array}{rcl} \frac{d t}{d x} & = & \frac{1}{\frac{d x}{d t}} \\ = & \frac{1}{- t^{- 2}} \\ = & \frac{1}{(- \frac{1}{t^{2}})} \\ = & 1 \div \frac{- 1}{t^{2}} \\ = & 1 \times \frac{t^{2}}{- 1} \\ = & - t^{2} \end{array}$

Substitute this into the chain rule

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d y}{d t} \times (- t^{2}) \\ = & - t^{2} \frac{d y}{d t} \end{array}$

Now substitute this derivative into the original differential equation

$x^{2} (- t^{2} \frac{d y}{d t}) - x y = - \frac{3}{x}$

This is not yet in the form $(t, y)$ , as there are still $x$ terms in it

Use $x = \frac{1}{t}$ to convert the remaining $x$ terms into $t$ terms

$\begin{array}{rcl} {(\frac{1}{t})}^{2} (- t^{2}) \frac{d y}{d t} - (\frac{1}{t}) y & = & - \frac{3}{(\frac{1}{t})} \\ - \frac{d y}{d t} - \frac{y}{t} & = & - 3 t \\ \frac{d y}{d t} + \frac{y}{t} & = & 3 t \end{array}$

This is in the correct form $\frac{d y}{d t} + p (t) y = q (t)$ for the integrating factor

the integrating factor is $e^{\int p (t) d t}$
where $\frac{d}{d t} (y e^{\int p (t) d t}) = q (t) e^{\int p (t) d t}$

Find the integrating factor

$e^{\int \frac{1}{t} d t} = e^{\ln t} = t$

Solve the differential equation using the integrating factor

Remember to add a constant of integration

$\begin{array}{rcl} \frac{d}{d t} (y t) & = & 3 t \times t \\ \frac{d}{d t} (y t) & = & 3 t^{2} \\ y t & = & \int 3 t^{2} d t \\ y t & = & t^{3} + c \\ y & = & t^{2} + \frac{c}{t} \end{array}$

Convert the general solution back into the variables $(x, y)$ using $x = \frac{1}{t}$ and $t = \frac{1}{x}$

$\begin{array}{rcl} y & = & {(\frac{1}{x})}^{2} + \frac{c}{(\frac{1}{x})} \\ y & = & \frac{1}{x^{2}} + c x \end{array}$

Find $c$ by substituting in $x = 1$ and $y = 2$

$\begin{array}{rcl} 2 & = & \frac{1}{1^{2}} + c \times 1 \\ 1 & = & c \end{array}$

Substitute $c = 1$ into the general solution to get the particular solution

$y = \frac{1}{x^{2}} + x$

Examiner Tips and Tricks

Don't forget to add a constant of integration when using the integrating factor, otherwise entire terms will be missing in the general solution!

How do I transform with products or quotients of variables?

Transforming the variables $(x, y)$ into $(x, z)$ using
- a product of variables
  - e.g. $y = x^{2} z$
- or a quotient of variables
  - e.g. $y = \frac{z}{x}$
- can be done using the product rule or quotient rule respectively
Some transformations may also involve implicit differentiation
- e.g. $y = x^{2} z^{3}$
  - where $\frac{d y}{d x} = \frac{d}{d x} (x^{2} z^{3}) = 2 x z^{3} + 3 x^{2} z^{2} \frac{d z}{d x}$

Worked Example

Use the transformation $y = x z$ to find the particular solution of the differential equation

$\frac{d y}{d x} = \frac{y}{x} + 5 x^{5}$

where $y = 2$ when $x = 1$ .

Answer:

Identify the variables being transformed

$(x, y) \to (x, z)$

This is a transformation of the dependent variable, $y$

Write $\frac{d y}{d x}$ in terms of $\frac{d z}{d x}$ using the product rule

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d}{d x} (x z) \\ \frac{d y}{d x} & = & 1 \times z + x \times \frac{d z}{d x} \\ \frac{d y}{d x} & = & z + x \frac{d z}{d x} \end{array}$

Substitute this into the original differential equation

$\begin{array}{rcl} z + x \frac{d z}{d x} & = & \frac{y}{x} + 5 x^{5} \end{array}$

This is not yet in the form $(x, z)$ , as there are still $y$ terms on the right-hand side

Use $y = x z$ to convert the remaining $y$ terms into $x$ and $z$ terms then simplify

$\begin{array}{rcl} z + x \frac{d z}{d x} & = & \frac{x z}{x} + 5 x^{5} \\ z + x \frac{d z}{d x} & = & z + 5 x^{5} \\ x \frac{d z}{d x} & = & 5 x^{5} \\ \frac{d z}{d x} & = & 5 x^{4} \end{array}$

Solve this differential equation by direct integration of the right-hand side

Remember to add a constant of integration

$z = x^{5} + c$

Transform the general solution back to the variables $(x, y)$ using $y = x z$ so $z = \frac{y}{x}$

$\begin{array}{rcl} \frac{y}{x} & = & x^{5} + c \\ y & = & x^{6} + c x \end{array}$

Now substitute in the boundary conditions $x = 1$ and $y = 2$ to find $c$

$\begin{array}{rcl} 2 & = & 1^{6} + c \times 1 \\ 1 & = & c \end{array}$

Substitute this value of $c$ back into the general solution to get the particular solution

$\begin{array}{rcl} y & = & x^{6} + x \end{array}$

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Reducing First-Order Differential Equations (Edexcel A Level Further Maths): Revision Note

Reducing first-order differential equations

What does reducing first-order differential equations mean?

What is the dependent variable and what is the independent variable?

How do I transform the dependent variable?

How do I transform the independent variable?

How do I transform with products or quotients of variables?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

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Author: Mark Curtis