Further Maths - Reducible Differential Equations
Pearson Edexcel Further Mathematics 2022
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
What the topic is and the first-order substitutions
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
\[\frac{\text{d}y}{\text{d}t} = z + t\frac{\text{d}z}{\text{d}t}\]If $z = \frac{1}{y^2}$, or $y = z^{-1/2}$ then what is $\frac{\text{d}y}{\text{d}x}$?
Second-order reducible equations
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 it’s 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 $\frac{\text{d}}{\text{d}u} (\frac{\text{d}y}{\text{d}x})$?
Finishing the question
What do you have to remember to do at the end of every reducible differential equations question?
Reverse the substitution.