Series Solutions to Differential Equations (Edexcel A Level Further Maths): Revision Note

Exam code: 9FM0

Author

Mark Curtis

Last updated

6 January 2026

Series solutions to differential equations

How do I find a series solution to a differential equation?

For complicated differential equations that cannot be solved, an approximate solution can be found using Taylor series
- e.g. solve $\frac{d y}{d x} = x \ln y + y^{2}$ where $y = b$ when $x = a$
The series solution is given by
- $y (x) = y (a) + (x - a) y' (a) + \frac{{(x - a)}^{2}}{2!} y'' (a) + . . . + \frac{{(x - a)}^{r}}{r!} y^{(r)} (a) + . . .$
  - where $y (a) = b$ , $y' (a) = {(\frac{d y}{d x})}_{x = a}, y'' (a) = {(\frac{d^{2} y}{d x^{2}})}_{x = a}$ , etc

Examiner Tips and Tricks

You will be given the Taylor series expansion below within the exam question itself. For differential equation solutions it helps to swap the $f$ to a $y$ .

$f (x) = f (a) + (x - a) f' (a) + \frac{{(x - a)}^{2}}{2!} f'' (a) + . . . + \frac{{(x - a)}^{r}}{r!} f^{(r)} (a) + . . .$

How do I calculate the derivatives at x=a?

To calculate $y' (a)$
- substitute in $x = a$ and $y = b$ to the differential equation
- rearrange to make $\frac{d y}{d x}$ the subject
To calculate $y'' (a)$
- differentiate both sides of the differential equation with respect to $x$
  - this may involve implicit differentiation
- substitute in $x = a$ and $y = b$ to both sides
- then rearrange to make $\frac{d^{2} y}{d x^{2}}$ the subject
This process is repeated to find $\frac{d^{3} y}{d x^{3}}$ etc.

How do I find higher-order derivatives?

The differentiation involved gets harder and harder for higher-order derivatives
- All derivatives must be with respect to $x$
  - which means functions of $y$ require implicit differentiation
  - which could include implicit chain, product or quotient rules
- There are also derivatives of derivatives that can be simplified
  - e.g. $\begin{array}{rcl} \frac{d}{d x} (\frac{d^{2} y}{d x^{2}}) & = & \frac{d^{3} y}{d x^{3}} \end{array}$
The table below shows some common examples

Term	Derivative, $\frac{d}{d x} (. . .)$	Result
$x^{2}$	$\frac{d}{d x} (x^{2})$	$2 x$
$y$	$\frac{d}{d x} (y)$	$\frac{d y}{d x}$
$y^{2}$	$\frac{d}{d x} (y^{2}) = \frac{d}{d y} (y^{2}) \times \frac{d y}{d x}$	$2 y \frac{d y}{d x}$
$\frac{d y}{d x}$	$\frac{d}{d x} (\frac{d y}{d x})$	$\frac{d^{2} y}{d x^{2}}$
$\frac{d^{2} y}{d x^{2}}$	$\frac{d}{d x} (\frac{d^{2} y}{d x^{2}})$	$\frac{d^{3} y}{d x^{3}}$
$x y$	$\frac{d}{d x} (x) y + x \frac{d}{d x} (y)$	$y + x \frac{d y}{d x}$
$e^{2 y} \frac{d y}{d x}$	$\frac{d}{d x} (e^{2 y}) \frac{d y}{d x} + e^{2 y} \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d y} (e^{2 y}) \frac{d y}{d x} \times \frac{d y}{d x} + e^{2 y} \frac{d}{d x} (\frac{d y}{d x})$	$2 e^{2 y} {(\frac{d y}{d x})}^{2} + e^{2 y} \frac{d^{2} y}{d x^{2}}$
${(\frac{d y}{d x})}^{5}$	$5 {(\frac{d y}{d x})}^{4} \times \frac{d}{d x} (\frac{d y}{d x})$	$5 {(\frac{d y}{d x})}^{4} \frac{d^{2} y}{d x^{2}}$

Examiner Tips and Tricks

Make sure you know the difference between powers of derivatives and higher-order derivatives.

E.g. ${(\frac{d y}{d x})}^{2}$ is the first derivative of $y$ with respect to $x$ , squared
$\frac{d^{2} y}{d x^{2}}$ is the second derivative of $y$ with respect to $x$
They are not the same thing!

Worked Example

The Taylor series expansion of $f (x)$ about $x = a$ is given by

$f (x) = f (a) + (x - a) f' (a) + \frac{{(x - a)}^{2}}{2!} f'' (a) + . . . + \frac{{(x - a)}^{r}}{r!} f^{(r)} (a) + . . .$

A differential equation is given by

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

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

Determine a series solution for $y$ , in ascending powers of $(x - 2)$ , up to and including the term in ${(x - 2)}^{2}$ , giving each coefficient in simplest form.

Answer:

Write out the relevant terms of the Taylor series formula, changing $f$ into $y$ and substituting in $a = 2$

$y (x) = y (2) + (x - 2) y' (2) + \frac{{(x - 2)}^{2}}{2!} y'' (2) + . . .$

Find $y (2)$ , the value of $y$ when $x = 2$ (it is given in the question)

$y (2) = 1$

Next find $y' (2)$

First substitute $x = 2$ and $y = 1$ into the original differential equation

$1^{3} \frac{d y}{d x} = \ln 1 + 2^{2}$

Then make $\frac{d y}{d x}$ the subject (this gives the value of $y' (2)$ )

$\begin{array}{rcl} \frac{d y}{d x} & = & 0 + 4 \\ y' (2) & = & 4 \end{array}$

To find $y'' (2)$ , don't differentiate the version with numbers substituted in

Instead, go back to the original differential equation and differentiate both sides with respect to $x$

$\begin{array}{rcl} \frac{d}{d x} (y^{3} \frac{d y}{d x}) & = & \frac{d}{d x} (\ln y + x^{2}) \end{array}$

It is easier here to start on the right-hand side, which can be separated as follows

$\begin{array}{rcl} \frac{d}{d x} (\ln y) + \frac{d}{d x} (x^{2}) \end{array}$

To calculate $\begin{array}{rcl} \frac{d}{d x} (\ln y) \end{array}$ , use implicit differentiation

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

To calculate $\frac{d}{d x} (x^{2})$ , differentiate $x^{2}$ with respect to $x$

$\frac{d}{d x} (x^{2}) = 2 x$

Now look at the left-hand side, which is the derivative of a product of terms

$\begin{array}{rcl} \frac{d}{d x} (y^{3} \frac{d y}{d x}) \end{array}$

Apply the product rule, showing clearly the derivatives with respect to $x$

$\begin{array}{rcl} \frac{d}{d x} (y^{3}) \frac{d y}{d x} + y^{3} \frac{d}{d x} (\frac{d y}{d x}) \end{array}$

Use implicit differentiation to simplify the part $\begin{array}{rcl} \frac{d}{d x} (y^{3}) \end{array}$

$\begin{array}{rcl} \frac{d}{d x} (y^{3}) & = & \frac{d}{d y} (y^{3}) \times \frac{d y}{d x} \\ = & 3 y^{2} \frac{d y}{d x} \end{array}$

Also simplify the derivative of a derivative, $\begin{array}{rcl} \frac{d}{d x} (\frac{d y}{d x}) \end{array}$

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

Substituting these parts back into the left-hand side

$\begin{array}{rcl} 3 y^{2} \frac{d y}{d x} \times \frac{d y}{d x} + y^{3} \frac{d^{2} y}{d x^{2}} \end{array}$

Simplify the first term using that $\frac{d y}{d x} \times \frac{d y}{d x} = {(\frac{d y}{d x})}^{2}$

$\begin{array}{rcl} 3 y^{2} {(\frac{d y}{d x})}^{2} + y^{3} \frac{d^{2} y}{d x^{2}} \end{array}$

Put the left-hand side and right-hand side back together

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

To find $y'' (2)$ , substitute $x = 2$ and $y = 1$ into the equation above and make $\begin{array}{rcl} \frac{d^{2} y}{d x^{2}} \end{array}$ the subject

${(\frac{d y}{d x})}_{x = 2} = y' (2) = 4$ , as found above

$\begin{array}{rcl} 3 \times 1^{2} {(4)}^{2} + 1^{3} \frac{d^{2} y}{d x^{2}} & = & \frac{1}{1} \times 4 + 2 \times 2 \\ 48 + \frac{d^{2} y}{d x^{2}} & = & 4 + 4 \\ \frac{d^{2} y}{d x^{2}} & = & - 40 \\ y'' (2) & = & - 40 \end{array}$

Finally, substitute $y (2) = 1$ , $y' (2) = 4$ and $y'' (2) = - 40$ into the Taylor series

$\begin{array}{rcl} y (x) & = & y (2) + (x - 2) y' (2) + \frac{{(x - 2)}^{2}}{2!} y'' (2) + . . . \\ = & 1 + (x - 2) \times 4 + \frac{{(x - 2)}^{2}}{2 \times 1} \times (- 40) + . . . \\ = & 1 + 4 (x - 2) - 20 {(x - 2)}^{2} + . . . \end{array}$

Write out the series solution for $y$ , up to and including the term in ${(x - 2)}^{2}$

$\begin{array}{rcl} y & = & 1 + 4 (x - 2) - 20 {(x - 2)}^{2} \end{array}$

Examiner Tips and Tricks

Don't forget to put $y =$ at the start of your series solution, as this is an approximation to the exact curve $y$ that satisfies the differential equation.

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Series Solutions to Differential Equations (Edexcel A Level Further Maths): Revision Note

Series solutions to differential equations

How do I find a series solution to a differential equation?

How do I calculate the derivatives at x=a?

How do I find higher-order derivatives?

Unlock more, it's free!

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Author: Mark Curtis