# Further Maths - Reducible 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 $\frac{\text{d}y}{\text{d}x}$?

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 $\frac{\text{d}y}{\text{d}x}$?

### 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?

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?

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 $z$?

What is

\[\frac{\text{d}}{\text{d}u} (\frac{\text{d}y}{\text{d}x})\]
?

### 2022-01-21

What do you have to remember to do at the end of every reducible differential equations question?

Reverse the substitution.