A LevelFurther MathsEdexcelFurther Pure 1Revision NotesFurther Numerical MethodsFurther Numerical MethodsNumerical Solutions of First-Order Differential Equations

Numerical Solutions of First-Order Differential Equations (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 22 January 2026

Numerical solutions of first-order differential equations

What is the forward difference method?

The forward difference method is a numerical method using the iterative formula
- ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$
- to approximate the particular solution of a first-order differential equation
  - $\frac{d y}{d x} = f (x, y)$
  - with boundary condition $y = y_{0}$ when $x = x_{0}$
- using step sizes of $h$
  - where decreasing $h$ increases the accuracy
For each iteration
- the method uses the gradient at that point to create a straight-line segment
  - which gives an estimate of the next $y$ value
- This is also known as Euler's method

Graph show the curve of an exact solution, followed by an approximation using line segments. Both start from the point (x0, y0) and the first line segment goes to (x1, y1) with a gradient of (dy/dx) at x0. The second line segment goes to (x2, y2) from (x1, y1) with a gradient of (dy/dx) at x1, and so on.

Examiner Tips and Tricks

You will be given the formula ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ in the exam question, though you may need to change the variables, e.g. $\frac{d θ}{d t}$ .

How do I use the forward difference method?

To use the forward difference method, see the worked example below

Worked Example

The velocity of a drone, $v$ ms^-1, varies with time, $t$ seconds, according to the model

$\frac{d v}{d t} = t - 3 v + 1$

The initial velocity of the drone is 2 ms^-1.

Use two iterations of the approximation formula ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ to estimate the velocity of the drone after 1 second.

Answer:

Find the step size, $h$ , to get from $t = 0$ to $t = 1$ in 2 iterations

$h = 0.5$

Write out the approximation formula using $v$ , $t$ and $h = 0.5$

${(\frac{d v}{d t})}_{n} \approx \frac{v_{n + 1} - v_{n}}{0.5}$

Start the first iteration by substituting $n = 0$ into the approximation formula

${(\frac{d v}{d t})}_{0} \approx \frac{v_{1} - v_{0}}{0.5}$

Identify the initial conditions, $t_{0}$ and $v_{0}$ , from the question

$t_{0} = 0$ and $v_{0} = 2$

Find ${(\frac{d v}{d t})}_{0}$ by substituting $t_{0} = 0$ and $v_{0} = 2$ into $\frac{d v}{d t} = t - 3 v + 1$

$\begin{array}{rcl} {(\frac{d v}{d t})}_{0} & = & t_{0} - 3 v_{0} + 1 \\ = & 0 - 3 \times 2 + 1 \\ = & - 5 \end{array}$

Substitute ${(\frac{d v}{d t})}_{0} = - 5$ and $v_{0} = 2$ into the approximation formula and rearrange to find $v_{1}$

$\begin{array}{rcl} - 5 & \approx & \frac{v_{1} - 2}{0.5} \\ v_{1} & = & - 5 \times 0.5 + 2 \\ v_{1} & = & - 0.5 \end{array}$

The first iteration is now complete

Start the second iteration by substituting $n = 1$ into the approximation formula

${(\frac{d v}{d t})}_{1} \approx \frac{v_{2} - v_{1}}{0.5}$

Note that the time $t$ has increased by 0.5

$\begin{array}{rcl} t_{1} & = & t_{0} + 0.5 \\ = & 0 + 0.5 \\ = & 0.5 \end{array}$

Find ${(\frac{d v}{d t})}_{1}$ by substituting $t_{1} = 0.5$ and $v_{1} = - 0.5$ into $\frac{d v}{d t} = t - 3 v + 1$

$\begin{array}{rcl} {(\frac{d v}{d t})}_{1} & = & t_{1} - 3 v_{1} + 1 \\ = & 0.5 - 3 \times (- 0.5) + 1 \\ = & 3 \end{array}$

Substitute ${(\frac{d v}{d t})}_{1} = 3$ and $v_{1} = - 0.5$ into the approximation formula and rearrange to find $v_{2}$

$\begin{array}{rcl} 3 & \approx & \frac{v_{2} - (- 0.5)}{0.5} \\ v_{2} & = & 3 \times 0.5 - 0.5 \\ v_{2} & = & 1 \end{array}$

This is the estimate of $v$ when $t = 1$

The velocity after 1 second is approximately 1 ms^-1

What is the central difference method?

The central difference method is a numerical method using the iterative formula
- ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{2 h}$
- to approximate the particular solution of a first-order differential equation
  - $\frac{d y}{d x} = f (x, y)$
  - with boundary condition $y = y_{0}$ when $x = x_{0}$
- using step sizes of $h$
  - where decreasing $h$ increases the accuracy
For each iteration
- the method finds the average gradient from a point before and a point after to estimate the next value of $y$
This means you need two starting points, $y_{0}$ and $y_{1}$ , before you can work out $y_{2}$
- $y_{0}$ is given in the question
- $y_{1}$ is found using the forward difference method ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$
  - i.e. rearranging ${(\frac{d y}{d x})}_{0} \approx \frac{y_{1} - y_{0}}{h}$ to give $y_{1}$

Examiner Tips and Tricks

You will be given the formulas ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{2 h}$ and ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ in the exam question, though you may need to change the variables, e.g. $\frac{d θ}{d t}$ .

How do I use the central difference method?

To use the central difference method, see the worked example below

Worked Example

The nitrate concentration in a river, $y$ mg L^-1, changes with the distance downstream, $x$ km, according to the model

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

At the point farthest upstream, $x = 0$ , the nitrate concentration is 1 mg L^-1.

(a) Use one iteration of the approximation formula ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ to estimate the nitrate concentration 0.5 km downstream.

(b) Hence use the approximation formula ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{2 h}$ to estimate the nitrate concentration 1 km downstream.

Answer:

(a)

This is asking for the forward difference method

Find the step size, $h$ , to get from $x = 0$ to $x = 0.5$ in 1 iteration

$h = 0.5$

Write out the approximation formula using $h = 0.5$

${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{0.5}$

Start the first iteration by substituting $n = 0$ into the approximation formula

${(\frac{d y}{d x})}_{0} \approx \frac{y_{1} - y_{0}}{0.5}$

Identify the initial conditions, $x_{0}$ and $y_{0}$ , from the question

$x_{0} = 0$ and $y_{0} = 1$

Find ${(\frac{d y}{d x})}_{0}$ by substituting $x_{0} = 0$ and $y_{0} = 1$ into $\frac{d y}{d x} = x - \frac{3 y}{2}$

$\begin{array}{rcl} {(\frac{d y}{d x})}_{0} & = & x_{0} - \frac{3 y_{0}}{2} \\ = & 0 - \frac{3 \times 1}{2} \\ = & - 1.5 \end{array}$

Substitute ${(\frac{d y}{d x})}_{0} = - 1.5$ and $y_{0} = 1$ into the approximation formula and rearrange to find $y_{1}$

$\begin{array}{rcl} - 1.5 & \approx & \frac{y_{1} - 1}{0.5} \\ y_{1} & = & 0.5 \times (- 1.5) + 1 \\ y_{1} & = & 0.25 \end{array}$

The nitrate concentration 0.5 km downstream is approximately 0.25 mg L^-1

(b)

This is asking for the central difference method, which requires knowing $y_{0}$ and $y_{1}$ first

Find $y_{0}$ at $x_{0} = 0$ from the question

$y_{0} = 1$

Find $y_{1}$ at $x_{1} = 0.5$ from the answer in part (a)

$y_{1} \approx 0.25$

Now use the central difference formula with $n = 1$

${(\frac{d y}{d x})}_{1} \approx \frac{y_{2} - y_{0}}{2 \times 0.5}$

Find ${(\frac{d y}{d x})}_{1}$ by substituting $x_{1} = 0.5$ and $y_{1} \approx 0.25$ into $\frac{d y}{d x} = x - \frac{3 y}{2}$

$\begin{array}{rcl} {(\frac{d y}{d x})}_{1} & = & x_{1} - \frac{3 y_{1}}{2} \\ = & 0.5 - 3 \times \frac{0.25}{2} \\ = & 0.125 \end{array}$

Substitute ${(\frac{d y}{d x})}_{1} = 0.125$ and $y_{0} = 1$ into the approximation formula and rearrange to find $y_{2}$

$\begin{array}{rcl} 0.125 & \approx & \frac{y_{2} - 1}{2 \times 0.5} \\ y_{2} & = & 0.125 \times 2 \times 0.5 + 1 \\ y_{2} & = & 1.125 \end{array}$

The nitrate concentration 1 km downstream is approximately 1.125 mg L^-1

What else could I be asked about difference methods?

You need to know that as you decrease the step size, $h$ ,
- the central difference method gives values closer to the true solution than the forward difference method
  - i.e. its errors are smaller
You need to know how to find the percentage error of your estimate if given the actual value
- using the formula
  - $percentage error = \frac{| estimate - actual |}{actual} \times 100 %$
If the estimate is less than the actual value
- it is an underestimate
If the estimate is greater than the actual value
- it is an overestimate

Examiner Tips and Tricks

To compare an estimate to an actual answer, you either need to be given the actual answer or you need to solve the differential equation algebraically.

Worked Example

The velocity of a drone, $v$ ms^-1, varies with time, $t$ seconds, according to the model

$\frac{d v}{d t} = t - 3 v + 1$

The initial velocity of the drone is 2 ms^-1.

Two iterations of the approximation formula ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ give an estimate of the velocity after 1 second as 1 ms^-1.

By solving the differential equation,

(a) determine whether the estimate is an overestimate or an underestimate,

(b) find the percentage error of this estimate.

Answer:

(a)

This differential equation can be solved using the integrating factor method

First rearrange into the correct form, $\frac{d v}{d t} + p (t) v = q (t)$

$\frac{d v}{d t} + 3 v = t + 1$

Find the integrating factor

$e^{\int 3 d t} = e^{3 t}$

Rewrite the differential equation

$\frac{d}{d t} (v e^{3 t}) = (t + 1) e^{3 t}$

Integrate both sides

$v e^{3 t} = \int (t + 1) e^{3 t} d t$

Use integration by parts on the right-hand side

$v e^{3 t} = \frac{1}{3} t e^{3 t} + \frac{2}{9} e^{3 t} + c$

Make $v$ the subject to get the general solution

$v = \frac{1}{3} t + \frac{2}{9} + c e^{- 3 t}$

Find $c$ using $v = 2$ when $t = 0$

$2 = \frac{2}{9} + c c = \frac{16}{9}$

Write out the particular solution

$v = \frac{1}{3} t + \frac{2}{9} + \frac{16}{9} e^{- 3 t}$

Now, to compare after 1 second, substitute in $t = 1$ to get the actual answer

$\begin{array}{rcl} v & = & \frac{1}{3} + \frac{2}{9} + \frac{16}{9} e^{- 3} \\ = & 0.6440658 . . . \end{array}$

Compare the estimate of $v = 1$ to the actual solution

The estimate of 1 ms^-1 is an overestimate of the actual value, 0.644 ms^-1to 3 s.f.

(b)

Use $\frac{| estimate - actual |}{actual} \times 100 %$ to find the percentage error of the estimate

$\frac{| 1 - \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 6440658 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} . \end{array} |}{\begin{array}{rcl} 0.6440658 . . . \end{array}} \times 100 = 55.26 . . . . %$

Round to 3 s.f.

The percentage error is 55.3%

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Numerical Solutions of First-Order Differential Equations (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Numerical solutions of first-order differential equations

What is the forward difference method?

How do I use the forward difference method?

What is the central difference method?

How do I use the central difference method?

What else could I be asked about difference methods?

Unlock more, it's free!

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

Author: Mark Curtis