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

Numerical Solutions of Second-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 second-order differential equations

What is the central difference method for second derivatives?

The central difference method for second derivatives is a numerical method using the iterative formula
- ${(\frac{d^{2} y}{d x^{2}})}_{n} \approx \frac{(y_{n + 1} - 2 y_{n} + y_{n - 1})}{h^{2}}$
- to approximate the particular solution of a second-order differential equation
  - $\frac{d^{2} y}{d x^{2}} = f (x, y)$
  - with initial conditions $y = y_{0}$ and $\frac{d y}{d x} = {(\frac{d y}{d x})}_{0}$ at $x = x_{0}$
- using step sizes of $h$
  - where decreasing $h$ increases the accuracy
You need two starting points, $y_{0}$ and $y_{1}$ , from which you can find $y_{2}$
- $y_{0}$ is given in the question
- $y_{1}$ is found using the forward difference method ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ for first derivatives
  - i.e. rearranging ${(\frac{d y}{d x})}_{0} \approx \frac{y_{1} - y_{0}}{h}$ to give $y_{1}$
The formula comes from writing the second derivative as the rate of change of the first derivative
- ${(\frac{d^{2} y}{d x^{2}})}_{n} \approx \frac{{(\frac{d y}{d x})}_{n} - {(\frac{d y}{d x})}_{n - 1}}{h}$
- then substituting in the forward difference formula for first derivatives and simplifying
  - ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ and ${(\frac{d y}{d x})}_{n - 1} \approx \frac{y_{n} - y_{n - 1}}{h}$

Examiner Tips and Tricks

You will be given the formulas ${(\frac{d^{2} y}{d x^{2}})}_{n} \approx \frac{(y_{n + 1} - 2 y_{n} + y_{n - 1})}{h^{2}}$ and ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ in the exam question.

Worked Example

A differential equation is given by

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

where $y = 1$ and $\frac{d y}{d x} = 0$ when $x = 0$ .

(a) Use the approximation formula ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ once to estimate $y$ at $x = 0.1$

(b) Use the approximation formula ${(\frac{d^{2} y}{d x^{2}})}_{n} \approx \frac{(y_{n + 1} - 2 y_{n} + y_{n - 1})}{h^{2}}$ once to estimate $y$ at $x = 0.2$

Answer:

(a)

Find the step size $h$ required to go from $x = 0$ to $x = 0.1$ in one iteration

$h = 0.1$

Substitute $n = 0$ , $y_{0} = 1$ , ${(\frac{d y}{d x})}_{0} = 0$ and $h = 0.1$ into the formula given

$0 = \frac{y_{1} - 1}{0.1}$

Solve to find $y_{1}$

$\begin{array}{rcl} y_{1} & = & 0.1 \times 0 + 1 \\ y_{1} & = & 1 \end{array}$

$y \approx 1$ at $x = 0.1$

(b)

From the question

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

From part (a)

$x_{1} = 0.1$ and $y_{1} = 1$

Substitute $n = 1$ into the approximation formula given

${(\frac{d^{2} y}{d x^{2}})}_{1} \approx \frac{(y_{2} - 2 y_{1} + y_{0})}{h^{2}}$

Also substitute in $h = 0.1$ , $y_{0} = 1$ and $y_{1} = 1$

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

Find ${(\frac{d^{2} y}{d x^{2}})}_{1}$ by substituting $x_{1} = 0.1$ and $y_{1} = 1$ into the differential equation $\frac{d^{2} y}{d x^{2}} = x - y$

$\begin{array}{rcl} {(\frac{d^{2} y}{d x^{2}})}_{1} & = & x_{1} - y_{1} \\ = & 0.1 - 1 \\ = & - 0.9 \end{array}$

Substitute $\begin{array}{rcl} {(\frac{d^{2} y}{d x^{2}})}_{1} & = & - 0.9 \end{array}$ back into the approximation formula and rearrange to find $y_{2}$

$\begin{array}{rcl} - 0.9 & \approx & \frac{(y_{2} - 2 \times 1 + 1)}{0 . 1^{2}} \\ y_{2} & = & 0 . 1^{2} \times (- 0.9) + 2 \times 1 - 1 \\ y_{2} & = & 0.991 \end{array}$

$y \approx 0.991$ at $x = 0.2$

What do I do if there is a dy/dx term in the second-order differential equation?

If the second-order differential equation has a $\frac{d y}{d x}$ term
- i.e. $\frac{d^{2} y}{d x^{2}} = f (x, y, \frac{d y}{d x})$
- then the central difference formula for a first derivative is also needed
  - ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{2 h}$

Examiner Tips and Tricks

You will be given the formulas ${(\frac{d^{2} y}{d x^{2}})}_{n} \approx \frac{(y_{n + 1} - 2 y_{n} + y_{n - 1})}{h^{2}}$ , ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{2 h}$ and ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ in the exam question.

Worked Example

A differential equation is given by

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

where $y = 1$ and $\frac{d y}{d x} = 0$ when $x = 0$ .

(a) Use the approximation formula ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n}}{h}$ once to estimate $y$ at $x = 0.1$

(b) Use the approximation formulas ${(\frac{d^{2} y}{d x^{2}})}_{n} \approx \frac{(y_{n + 1} - 2 y_{n} + y_{n - 1})}{h^{2}}$ and ${(\frac{d y}{d x})}_{n} \approx \frac{y_{n + 1} - y_{n - 1}}{2 h}$ to estimate $y$ at $x = 0.2$ , giving your answer to 5 decimal places.

Answer:

(a)

Find the step size $h$ required to go from $x = 0$ to $x = 0.1$ in one iteration

$h = 0.1$

Substitute $n = 0$ , $y_{0} = 1$ , ${(\frac{d y}{d x})}_{0} = 0$ and $h = 0.1$ into the formula given

$0 = \frac{y_{1} - 1}{0.1}$

Solve to find $y_{1}$

$\begin{array}{rcl} y_{1} & = & 0.1 \times 0 + 1 \\ y_{1} & = & 1 \end{array}$

$y \approx 1$ at $x = 0.1$

(b)

From the question

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

From part (a)

$x_{1} = 0.1$ and $y_{1} = 1$

Substitute $n = 1$ into the first approximation formula given for second derivatives

${(\frac{d^{2} y}{d x^{2}})}_{1} \approx \frac{(y_{2} - 2 y_{1} + y_{0})}{h^{2}}$

Also substitute in $h = 0.1$ , $y_{0} = 1$ and $y_{1} = 1$ (call this equation A)

${(\frac{d^{2} y}{d x^{2}})}_{1} \approx \frac{(y_{2} - 2 \times 1 + 1)}{0 . 1^{2}} \begin{array}{rcl} A \end{array}$

Find ${(\frac{d^{2} y}{d x^{2}})}_{1}$ by substituting $x_{1} = 0.1$ and $y_{1} = 1$ into the differential equation $\frac{d^{2} y}{d x^{2}} = x - y + \frac{d y}{d x}$ (call this equation B)

$\begin{array}{rcl} {(\frac{d^{2} y}{d x^{2}})}_{1} & = & x_{1} - y_{1} + {(\frac{d y}{d x})}_{1} \\ {(\frac{d^{2} y}{d x^{2}})}_{1} & = & 0.1 - 1 + {(\frac{d y}{d x})}_{1} \\ {(\frac{d^{2} y}{d x^{2}})}_{1} & = & - 0.9 + {(\frac{d y}{d x})}_{1} B \end{array}$

This has the term ${(\frac{d y}{d x})}_{1}$ which cannot immediately be found

Use the second approximation formula given for first derivatives,with $n = 1$ , $h = 0.1$ and $y_{0} = 1$

$\begin{array}{rcl} {(\frac{d y}{d x})}_{1} & \approx & \frac{y_{2} - 1}{2 \times 0.1} \end{array}$

Substitute this into equation B

$\begin{array}{rcl} {(\frac{d^{2} y}{d x^{2}})}_{1} & = & - 0.9 + \frac{y_{2} - 1}{2 \times 0.1} \end{array}$

Then substitute this into equation A

$\begin{array}{rcl} - \end{array} \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 9 \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \frac{y_{2} - 1}{2 \times 0.1} \end{array} \approx \frac{(y_{2} - 2 \times 1 + 1)}{0 . 1^{2}}$

Now solve for $y_{2}$

$\begin{array}{rcl} - 0.9 + \frac{y_{2} - 1}{0.2} & = & \frac{y_{2} - 1}{0.01} \\ - 0.009 + 0.05 (y_{2} - 1) & = & y_{2} - 1 \\ 0.05 (y_{2} - 1) & = & y_{2} - 0.991 \\ 0.05 y_{2} - 0.05 & = & y_{2} - 0.991 \\ 0.941 & = & 0.95 y_{2} \\ y_{2} & = & 0.9905263 . . . \end{array}$

$y \approx 0.99053$ to 5 d.p. at $x = 0.2$

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