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

Exam code: 9FM0

Author

Mark Curtis

Last updated

6 January 2026

Reducing second-order differential equations

What does reducing second-order differential equations mean?

A hard second-order differential equation can be reduced (transformed)
- into an easier second-order differential equation
  - of the form $a y'' + b y' + c y = . . .$
- using a given transformation
The easier differential equation can then be solved
- by finding
  - the complementary function
  - and the particular integral
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 $\frac{d^{2} y}{d x^{2}} = f (x, y, \frac{d y}{d x})$ 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. $\frac{d^{2} x}{d t^{2}} + x = t^{2}$ 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 follow these steps:
STEP 1
Find $\frac{d y}{d x}$ in terms of $\frac{d z}{d x}$ using the chain rule
- $\frac{d y}{d x} = \frac{d y}{d z} \times \frac{d z}{d x}$
- 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 written in $(x, z)$ only
STEP 2
Find $\frac{d^{2} y}{d x^{2}}$ by taking $\frac{d}{d x}$ of both sides of $\frac{d y}{d x} = . . .$ from Step 1
- Use implicit differentiation where necessary
  - $\frac{d}{d x} (. . .) = \frac{d}{d z} (. . .) \times \frac{d z}{d x}$
- Make sure the final result is written in $(x, z)$ only
STEP 3
Substitute the first derivative and second derivative into the differential equation
- Use the transformation to make sure the final differential equation is written in $(x, z)$ only
  - i.e. no $y$ terms
  - See the worked example below

Worked Example

Use the transformation $y = \frac{1}{z}$ to find the general solution of the differential equation

$\frac{d^{2} y}{d x^{2}} - \frac{2}{y} {(\frac{d y}{d x})}^{2} - 5 \frac{d y}{d x} + 6 y = - e^{x} y^{2}$

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 $y = \frac{1}{z} = z^{- 1}$

$\frac{d y}{d z} = - z^{- 2} = - \frac{1}{z^{2}}$

Substitute this into the chain rule

$\frac{d y}{d x} = - \frac{1}{z^{2}} \frac{d z}{d x}$

Now differentiate both sides of this derivative by $\frac{d}{d x}$

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (- \frac{1}{z^{2}} \frac{d z}{d x})$

Simplify the left-hand side and use the product rule on the right-hand side

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

Use implicit differentiation on the part $\begin{array}{rcl} \frac{d}{d x} (- z^{- 2}) \end{array}$

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

Now substitute $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ into the original differential equation

$(\begin{array}{rcl} \frac{2}{z^{3}} {(\frac{d z}{d x})}^{2} - \frac{1}{z^{2}} \frac{d^{2} z}{d x^{2}} \end{array}) - \frac{2}{y} {(- \frac{1}{z^{2}} \frac{d z}{d x})}^{2} - 5 (- \frac{1}{z^{2}} \frac{d z}{d x}) + 6 y = - e^{x} y^{2}$

This is not yet in the form $(x, z)$ , as there are still $y$ terms on both sides of the equation

Use $y = \frac{1}{z}$ to convert the remaining $y$ terms into $z$ terms, then simplify

$\begin{array}{rcl} \frac{2}{z^{3}} {(\frac{d z}{d x})}^{2} - \frac{1}{z^{2}} \frac{d^{2} z}{d x^{2}} - 2 z {(- \frac{1}{z^{2}} \frac{d z}{d x})}^{2} - 5 (- \frac{1}{z^{2}} \frac{d z}{d x}) + \frac{6}{z} & = & - \frac{e^{x}}{z^{2}} \\ \frac{2}{z^{3}} {(\frac{d z}{d x})}^{2} - \frac{1}{z^{2}} \frac{d^{2} z}{d x^{2}} - 2 z \times \frac{1}{z^{4}} {(\frac{d z}{d x})}^{2} + \frac{5}{z^{2}} \frac{d z}{d x} + \frac{6}{z} & = & - \frac{e^{x}}{z^{2}} \\ \frac{2}{z^{3}} {(\frac{d z}{d x})}^{2} - \frac{1}{z^{2}} \frac{d^{2} z}{d x^{2}} - \frac{2}{z^{3}} {(\frac{d z}{d x})}^{2} + \frac{5}{z^{2}} \frac{d z}{d x} + \frac{6}{z} & = & - \frac{e^{x}}{z^{2}} \\ - \frac{1}{z^{2}} \frac{d^{2} z}{d x^{2}} + \frac{5}{z^{2}} \frac{d z}{d x} + \frac{6}{z} & = & - \frac{e^{x}}{z^{2}} \end{array}$

Multiply both sides by $- z^{2}$

$\begin{array}{rcl} \frac{d^{2} z}{d x^{2}} - 5 \frac{d z}{d x} - 6 z & = & e^{x} \end{array}$

Solve the auxiliary equation

$\begin{array}{rcl} m^{2} - 5 m - 6 & = & 0 \\ (m - 6) (m + 1) & = & 0 \\ m & = & 6 or m = - 1 \end{array}$

Write out the complementary function

$A e^{6 x} + B e^{- x}$

Find the particular integral, of the form $z = λ e^{x}$

$\begin{array}{rcl} λ e^{x} - 5 {λe}^{x} - 6 {λe}^{x} & \equiv & e^{x} \\ - 10 λ & = & 1 \\ λ & = & - \frac{1}{10} \end{array}$

Find the general solution for $z$ in terms of $x$

$z = A e^{6 x} + B e^{- x} - \frac{1}{10} e^{x}$

Use the transformation $y = \frac{1}{z}$ to find the general solution for $y$ in terms of $x$

$y = \frac{1}{A e^{6 x} + B e^{- x} - \frac{1}{10} e^{x}}$

This is the general solution, but it can also be rearranged (with new arbitrary constants) to $y = \frac{10}{C e^{6 x} + D e^{- x} - e^{x}}$

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 follow these steps:
STEP 1
Find $\frac{d y}{d x}$ in terms of $\frac{d y}{d t}$ using the chain rule
- $\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}$
- It sometimes helps to also use that $\frac{d x}{d t} = \frac{1}{\frac{d t}{d x}}$
- as long as the end result is written in $(t, y)$ only
STEP 2
Find $\frac{d^{2} y}{d x^{2}}$ by taking $\frac{d}{d x}$ of both sides of $\frac{d y}{d x} = . . .$ from Step 1
- Use implicit differentiation on the right-hand side
- In particular, use the result that
  - $\frac{d}{d x} (\frac{d y}{d t}) = \frac{d}{d t} (\frac{d y}{d t}) \times \frac{d t}{d x}$
- Make sure the final result is written in $(t, y)$ only
STEP 3
Substitute the first derivative and second derivative into the differential equation
- Use the transformation to make sure the final differential equation is written in $(t, y)$ only
  - i.e. no $x$ terms
  - See the worked example below

Worked Example

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

$x^{2} \frac{d^{2} y}{d x^{2}} + 5 x \frac{d y}{d x} + 4 y = \ln x$

where $y = \frac{3}{4}$ and $\frac{d y}{d x} = - \frac{3}{4}$ 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 $x = e^{t}$ it is easier to use that $\frac{d t}{d x} = \frac{1}{\frac{d x}{d t}}$

$\begin{array}{rcl} \frac{d t}{d x} & = & \frac{1}{\frac{d x}{d t}} \\ = & \frac{1}{e^{t}} \end{array}$

Substitute this into the chain rule

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

Now differentiate both sides of this derivative by $\frac{d}{d x}$

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

Both terms on the right-hand side are in terms of $t$ so you can use implicit differentiation before the product rule

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

Simplify the left-hand side and use the product rule on the right-hand side

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

To find $\frac{d t}{d x}$ from $x = e^{t}$ use that $\frac{d t}{d x} = \frac{1}{\frac{d x}{d t}}$

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

Now substitute $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ into the original differential equation

$x^{2} (\begin{array}{rcl} - e^{- 2 t} \frac{d y}{d t} + e^{- 2 t} \frac{d^{2} y}{d t^{2}} \end{array}) + 5 x (\begin{array}{rcl} e^{- t} \frac{d y}{d t} \end{array}) + 4 y = \ln x$

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

Use $x = e^{t}$ to convert the remaining $x$ terms into $t$ terms, then simplify

$\begin{array}{rcl} {(e^{t})}^{2} (- e^{- 2 t} \frac{d y}{d t} + e^{- 2 t} \frac{d^{2} y}{d t^{2}}) + 5 (e^{t}) (e^{- t} \frac{d y}{d t}) + 4 y & = & \ln (e^{t}) \\ e^{2 t} (- e^{- 2 t} \frac{d y}{d t} + e^{- 2 t} \frac{d^{2} y}{d t^{2}}) + 5 e^{t} (e^{- t} \frac{d y}{d t}) + 4 y & = & t \\ - \frac{d y}{d t} + \frac{d^{2} y}{d t^{2}} + 5 \frac{d y}{d t} + 4 y & = & t \\ \frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} + 4 y & = & t \end{array}$

Solve the auxiliary equation

$\begin{array}{rcl} m^{2} + 4 m + 4 & = & 0 \\ {(m + 2)}^{2} & = & 0 \\ m & = & - 2 repeated \end{array}$

Write out the complementary function

$(A + B t) e^{- 2 t}$

Find the particular integral, of the form $y = λ + μ t$

$\begin{array}{rcl} 4 μ + 4 λ + 4 μ t & \equiv & t \\ 4 μ & = & 1 \Rightarrow μ = \frac{1}{4} \\ μ + λ & = & 0 \Rightarrow λ = - \frac{1}{4} \end{array}$

Find the general solution for $y$ in terms of $t$

$y = (A + B t) e^{- 2 t} - \frac{1}{4} + \frac{1}{4} t$

Use the transformation $x = e^{t}$ (i.e. $t = \ln x$ ) to find the general solution for $y$ in terms of $t$

$y = (A + B \ln x) e^{- 2 \ln x} - \frac{1}{4} + \frac{1}{4} \ln x y = (A + B \ln x) e^{\ln (x^{- 2})} - \frac{1}{4} + \frac{1}{4} \ln x y = (A + B \ln x) x^{- 2} - \frac{1}{4} + \frac{1}{4} \ln x y = A x^{- 2} + B x^{- 2} \ln x - \frac{1}{4} + \frac{1}{4} \ln x$

This is the general solution so to find the particular solution first use that $y = \frac{3}{4}$ when $x = 1$

$\begin{array}{rcl} \frac{3}{4} & = & A + B \times 0 - \frac{1}{4} + \frac{1}{4} \times 0 \\ 1 & = & A \end{array}$

Then use that $\frac{d y}{d x} = - \frac{3}{4}$ when $x = 1$ , which requires differentiating the general solution (with $A = 1$ ) first

$\begin{array}{rcl} y & = & x^{- 2} + B x^{- 2} \ln x - \frac{1}{4} + \frac{1}{4} \ln x \\ \frac{d y}{d x} & = & - 2 x^{- 3} + B (- 2 x^{- 3} \ln x + x^{- 2} \times \frac{1}{x}) - 0 + \frac{1}{4} \times \frac{1}{x} \\ \frac{d y}{d x} & = & - 2 x^{- 3} + B (- 2 x^{- 3} \ln x + x^{- 3}) + \frac{1}{4 x} \end{array}$

Substitute in $\frac{d y}{d x} = - \frac{3}{4}$ when $x = 1$

$\begin{array}{rcl} - \frac{3}{4} & = & - 2 + B (- 2 \times 0 + 1) + \frac{1}{4} \\ - 1 & = & - 2 + B \\ 1 & = & B \end{array}$

Substitute $A = 1$ and $B = 1$ into the general solution to get the particular solution

$y = x^{- 2} + x^{- 2} \ln x - \frac{1}{4} + \frac{1}{4} \ln x$

$y = \frac{1}{x^{2}} + \frac{\ln x}{x^{2}} - \frac{1}{4} + \frac{1}{4} \ln x$

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 general solution of the differential equation

$x^{2} \frac{d^{2} y}{d x^{2}} + 2 (x^{2} - x) \frac{d y}{d x} + (5 x^{2} - 2 x + 2) y = x^{3} \sin x$

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

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

Now differentiate both sides of this derivative by $\frac{d}{d x}$

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

Simplify the left-hand side and find the derivative in $x$ of the two terms on the right-hand side

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

Use the product rule for the last term, then simplify

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

Now substitute $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ into the original differential equation

$x^{2} (\begin{array}{rcl} 2 \frac{d z}{d x} + x \frac{d^{2} z}{d x^{2}} \end{array}) + 2 (x^{2} - x) (\begin{array}{rcl} z + x \frac{d z}{d x} \end{array}) + (5 x^{2} - 2 x + 2) y = x^{3} \sin x$

This is not yet in the form $(x, z)$ , as there is still a $y$ term on the left-hand side

Use $y = x z$ to convert the remaining $y$ term int a $z$ term, then simplify

$\begin{array}{rcl} x^{2} (2 \frac{d z}{d x} + x \frac{d^{2} z}{d x^{2}}) + 2 (x^{2} - x) (z + x \frac{d z}{d x}) + (5 x^{2} - 2 x + 2) x z & = & x^{3} \sin x \\ 2 x^{2} \frac{d z}{d x} + x^{3} \frac{d^{2} z}{d x^{2}} + 2 x^{2} z - 2 x z + 2 x^{3} \frac{d z}{d x} - 2 x^{2} \frac{d z}{d x} + 5 x^{3} z - 2 x^{2} z + 2 x z & = & x^{3} \sin x \\ x^{3} \frac{d^{2} z}{d x^{2}} + 2 x^{3} \frac{d z}{d x} + 5 x^{3} z & = & x^{3} \sin x \end{array}$

Divide both sides by $x^{3}$

$\begin{array}{rcl} \frac{d^{2} z}{d x^{2}} + 2 \frac{d z}{d x} + 5 z & = & \sin x \end{array}$

Solve the auxiliary equation

$\begin{array}{rcl} m^{2} + 2 m + 5 & = & 0 \\ m & = & \frac{- 2 \pm \sqrt{2^{2} - 4 \times 1 \times 5}}{2} \\ m & = & \frac{- 2 \pm \sqrt{- 16}}{2} \\ m & = & \frac{- 2 \pm 4 i}{2} \\ m & = & - 1 \pm 2 i \end{array}$

Write out the complementary function

$e^{- x} (A \cos 2 x + B \sin 2 x)$

Find the particular integral, of the form $z = λ \cos x + μ \sin x$

$(- λ \cos x - μ \sin x) + 2 (- λ \sin x + μ \cos x) + 5 (λ \cos x + μ \sin x) \equiv \sin x - λ + 2 μ + 5 λ = 0 \Rightarrow λ = - \frac{1}{2} μ - μ - 2 λ + 5 μ = 1 \Rightarrow μ = \frac{1}{5} and λ = - \frac{1}{10}$

Find the general solution for $z$ in terms of $x$

$z = e^{- x} (A \cos 2 x + B \sin 2 x) - \frac{1}{10} \cos x + \frac{1}{5} \sin x$

Use the transformation $y = x z$ , i.e. $z = \frac{y}{x}$ , to find the general solution for $y$ in terms of $x$

$y = x (e^{- x} (A \cos 2 x + B \sin 2 x) - \frac{1}{10} \cos x + \frac{1}{5} \sin x)$

$y = x e^{- x} (A \cos 2 x + B \sin 2 x) - \frac{1}{10} x \cos x + \frac{1}{5} x \sin x$

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

Reducing second-order differential equations

What does reducing second-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!

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Author: Mark Curtis