# Further Maths - Reducible Differential Equations

> Source: https://ollybritton.com/notes/a-level/further-maths/topics/reducible-differential-equations/ · Updated: 2025-03-29 · Tags: school, further-maths, futher-pure-1, differential-equations

## Summary
This topic is about making tricky first order or second order differential equations easier by using a substitution that transforms them into something that we know how to solve from Core Pure 2.

For example, consider the following differential equation:

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

None of the toolkit we've learned so far works here:

* It isn't possible to separate the $x$s and $y$s into $\frac{1}{g(y)}\frac{\text{d}y}{\text{d}x} = f(x)$ to integrate both sides.
* It isn't possible to get into the form $\frac{\text{d}y}{\text{d}x} + P(x)y = Q(x)$

Instead, it's necessary to make the substitution $y = xz$ which transforms the differential equation into:

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

This differential equation can be solved via separating the variables to give:

$$
z^2 = Ax - 1
$$

Reversing the substitution,

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

$$
y^2 = x^2(Ax - 1)
$$

Another example is the differential equation

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

Again, this isn't in the form where we can use separation of variables or an integrating factor. Therefore we make the substitution $z = 1/y$ and as if by magic it becomes:

$$
\frac{\text{d}z}{\text{d}x} - xz = -x
$$

This can be solved via integrating factor and gives:

$$
z = 1 + ce^{x^2/2}
$$

Reversing the substitution:

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

## Flashcards
### 2022-01-19
##### What's the reducible differential equations topic about??
Transforming complicated differential equations into simpler ones using a substitution.

##### If $z = \frac{y}{x}$, or $y = xz$ what is $\frac{\text{d}y}{\text{d}x}$??
$$
\frac{\text{d}y}{\text{d}x} = z + x\frac{\text{d}z}{\text{d}x}
$$

##### What's the first stage in doing a first-order reducible differential equations question??
Working out how to substitute the derivative by differentiating.

##### If a reducible differential equations question asks you to substitute $z = \frac{y}{x}$ then what is the first step??
Rearranging to

$$
y = zt
$$

Then differentiating

$$
\frac{\text{d}y}{\frac{d}t} = z + t\frac{\text{d}z}{\text{d}t}
$$

##### If $z = \frac{1}{y^2}$, or $y = z^{-1/2}$ then what is $\frac{\text{d}y}{\text{d}x}$??
$$
\frac{\text{d}y}{\text{d}x} = -\frac{1}{2} z^{-3/2} \frac{\text{d}z}{\text{d}x}
$$

### 2022-01-20
##### What is the aim of any reducible differential equations question??
Making sure that all instances of the variable (including derivatives) have been replaced.

##### If you're asked to do a substitution for removing $x$ using a function of $z$ in a second differential equation, what would $\frac{\text{d}y}{\text{d}x}$ need to become??
$$
\frac{\text{d}y}{\text{d}z}
$$

##### If you're asked to do a substitution for removing $x$ using a function of $z$ in a second differential equation, what would $\frac{\text{d}^2y}{\text{d}x^2}$ need to become??
$$
\frac{\text{d}^2y}{\text{d}z^2}
$$

##### Imagine you've been given a reducible second-order differential equation where $x = e^u$ and you need a solution for $y$. How could you work out $\frac{\text{d}y}{\text{d}x}$ given that $y$ never appears in the substitution??
Use the chain rule but backwards

$$
\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y}{\text{d}u} \frac{\text{d}z}{\text{d}u}
$$

##### When completing a second-order reducible differential equations question, what is it useful to do at every step where you come up with a formula for a derivative (i.e. $\frac{\text{d}u}{\text{d}x} = \frac{1}{x}$ or $\frac{\text{d}y}{\text{d}x} = \frac{1}{x} \frac{\text{d}y}{\text{d}u}$??
Box the result so that its easy to refer back to later.

##### You're completing a second-order reducible differential equations question and finding a nice form for $\frac{\text{d}^2y}{\text{d}x^2}$ in terms of $\frac{\text{d}y}{\text{d}u}$ and $\frac{\text{d}^2 y}{\text{d}u^2}$ is hard. What could you try instead to attempt it another way??
Starting off by finding a form for

$$
\frac{\text{d}^2y}{\text{d}u^2} = \frac{\text{d}}{\text{d}u}\left[ \frac{\text{d}y}{\text{d}u} \right]
$$

##### What's another way of writing $\frac{\text{d}x}{\text{d}t}$ when you're using a substitution involving a variable $z$??
$$
\frac{\text{d}x}{\text{d}t} = \frac{\text{d}x}{\text{d}z} \times \frac{\text{d}z}{\text{d}t}
$$

##### What is $\frac{\text{d}}{\text{d}u} (\frac{\text{d}y}{\text{d}x})$??
$$
\frac{\text{d}^2 y}{\text{d}x^2} \times \frac{\text{d}x}{\text{d}u}
$$

### 2022-01-21
##### What do you have to remember to do at the end of every reducible differential equations question??
Reverse the substitution.

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