# Solving Differential Equations: The Variable Separation Technique

In this post, we look at the technique of separating variables to solve first-order differential equations of single variable functions - very common in standard Mathematics courses of Pre-U curriculums. The technique requires us to separate the so-called differentials along with each of its variables to their own side of the equation.

## A Case of Not Walking The Talk

Let's look at a sample problem for which we'll employ this technique.

Solve the following differential equation and state $$y$$ in terms of $$x$$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = 2xy$

The solution:

Separating the differentials $$\mathrm{d}y$$, $$\mathrm{d}x$$ and the variables, we have:

$\frac{1}{y} \:\mathrm{d}y = 2x \: \mathrm{d}x$

Integrating both sides of the equation:

\begin{aligned} \int \; \frac{1}{y} \:\mathrm{d}y &= \int \; 2x \: \mathrm{d}x \\ \ln y &= x^2 + \mathrm{c} \\ y &= e^{x^2 + \mathrm{c}} \\ y &= e^{\mathrm{c}} e^{x^2} \\ y &= \mathrm{k} e^{x^2} \quad \blacksquare \end{aligned}

$$\dagger$$ where $$\mathrm{c, k }$$ are constants

Whilst the above solution is completely accepted in your standard school/public examinations, I have always been uncomfortable with the separation of the differentials - $$\mathrm{d}y$$ and $$\mathrm{d}x$$. Allow me to explain why.

The derivative $$\frac{\mathrm{d}y}{\mathrm{d}x}$$ is not a quotient or a fraction. It is not the same kind of thing as $$\frac{\Delta y}{\Delta x}$$ or $$\frac{\delta y}{\delta x}$$, for the terms $$\frac{\Delta y}{\Delta x}$$ and $$\frac{\delta y}{\delta x}$$ are actual quotients; $$\Delta y, \Delta x, \delta y$$ and $$\delta x$$ are each numerical values. On the other hand, $$\frac{\mathrm{d}y}{\mathrm{d}x}$$ is not decomposable since $$\mathrm{d}y$$ and $$\mathrm{d}x$$ are not numbers standing as a numerator and a denominator, respectively; we say the derivative $$\frac{\mathrm{d}y}{\mathrm{d}x}$$ is atomic - that which cannot be broken up further$$\ddagger$$. So, you can see why moving the differentials $$\mathrm{d}y$$ and $$\mathrm{d}x$$ around as if they're numbers, might be going against the grain to some folks.

$$\ddagger$$ Although these days we've already managed to even split the proton and neutron

To work out of this conundrum, we will explore a train of thought that's coherent with the notion of what the derivative $$\frac{\mathrm{d}y}{\mathrm{d}x}$$ actually is.

## Another way of thinking

Suppose we have a function $$z = y^2$$ and with that, $$y$$ is dependent on $$x$$ - meaning $$y$$ is some function of $$x$$.

Using the Chain Rule for differentiation, we find $$\frac{\mathrm{d}z}{\mathrm{d}x}$$:

\begin{aligned} z &= y^2 \\ \tag{1} \frac{\mathrm{d}z}{\mathrm{d}x} &= \frac{\mathrm{d}z}{\mathrm{d}y} \cdot \frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}z}{\mathrm{d}x} &= 2y \frac{\mathrm{d}y}{\mathrm{d}x} \end{aligned}

We started with $$z$$ and ended up with $$\frac{\mathrm{d}z}{\mathrm{d}x}$$. So, should we wish to "get back" $$z$$, all we need to do is integrate $$\frac{\mathrm{d}z}{\mathrm{d}x}$$ with respect to $$x$$ because that's the way we "undo" the differention we performed. Integrating both sides of the last equation with respect to $$x$$:

\begin{aligned} \frac{\mathrm{d}z}{\mathrm{d}x} &= 2y \cdot \frac{\mathrm{d}y}{\mathrm{d}x} \\ {\color{blue}\int\: \bigg(}\frac{\mathrm{d}z}{\mathrm{d}x}{\color{blue}\bigg)\: dx} &= {\color{blue}\int \: \bigg(}2y \frac{\mathrm{d}y}{\mathrm{d}x} {\color{blue}\bigg) \: dx} \\ z &= y^2 {\color{gray} + \rm{c}} \end{aligned}

$$\dagger$$ The constant $$\rm{c}$$ has to be included in the result since we integrated

Observe that $$\int\big(2y \frac{\mathrm{d}y}{\mathrm{d}x}\big) \: dx$$ had to give us $$y^2 {\color{gray} + \rm{c}}$$ because we just undid the differentiation.

The expression $$\int\big(2y \frac{\mathrm{d}y}{\mathrm{d}x}\big) \: dx$$ gave us $$y^2 {\color{gray} + \rm{c}}$$. Without any deeper thought, we obtained the expression $$y^2 + c$$ purely by viewing the integration process as an undoing of the differentiation.

Recall the steps of differentiation which led us to an expression in terms of $$y$$ (i.e. $$2y$$) multiplied with the term $$\frac{\mathrm{d}y}{\mathrm{d}x}$$. Because the term $$\frac{\mathrm{d}y}{\mathrm{d}x}$$ is present immediately as a multiplier after a function of $$y$$ due to the use of the chain-rule differentiating with respect to $$x$$, we are justified to think the term $$\frac{\mathrm{d}y}{\mathrm{d}x}$$ simply "rolls back in to its hole and vanishes" when we integrate with respect to $$x$$; further inspection reveals only the term $$2y$$ was integrated, and that too with respect to $$y$$, to yield $$y^2 +\mathrm{c}$$. No manipulation was done on the term $$\frac{\mathrm{d}y}{\mathrm{d}x}$$.

We are therefore permitted to think the actual integration in the above was done only on the function of $$y$$ with reference to $$y$$ alone. We can summarise the above reasoning mathematically:

\begin{aligned} \int \: 2y\frac{\mathrm{d}y}{\mathrm{d}x}\:\mathrm{d}x &= y^2 + \rm{c} \\ \int \: 2y\:\mathrm{d}y &= y^2 + \rm{c} \\ \therefore \quad \int \: 2y\frac{\mathrm{d}y}{\mathrm{d}x}\:\mathrm{d}x &= \int \: 2y\:\mathrm{d}y \end{aligned}

Generalizing our foregoing reasoning, we could replace $$2y$$ in the expression $$2y\frac{\mathrm{d}y}{\mathrm{d}x}$$, with any expression/function of $$y$$ as with $$f(y)$$ in $$f(y)\frac{\mathrm{d}y}{\mathrm{d}x}$$. It should be sensible, that even with any general expression of the form $$f(y) \frac{\mathrm{d}y}{\mathrm{d}x}$$, the following is true:

$\boxed{ \int \: f(y)\frac{\mathrm{d}y}{\mathrm{d}x}\:\mathrm{d}x = \int \: f(y)\:\mathrm{d}y } \tag{2}$

At this juncture, there's another false notion that's often bandied about. It is the idea that the differentials of $$\mathrm{d}x$$ actually cancel each other as below:

$\int \: f(y)\frac{\mathrm{d}y}{\color{red}\cancel{\mathrm{d}x}}\enspace{\color{red}\cancel{\mathrm{d}x}} = \int \: f(y)\:\mathrm{d}y$

While we could arrive at the same result of $$(2)$$ by adopting this thought process, it does not make it acceptable in formal mathematics. Thanks to Leibniz's notation however, and I'm saying this as a commendation, we are able to get away with this expedient idea of mutual cancellation of differentials in problem solving.

## Reworded Solution

Let's use the above knowledge to rework the solution of our sample problem:

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= 2xy \\ \end{aligned} We separate only the variables: \begin{aligned} \frac{1}{y} \:\frac{\mathrm{d}y}{\mathrm{d}x} &= 2x \\ \end{aligned} Integrating both sides with respect to $$x$$: \begin{aligned} \int \bigg(\frac{1}{y} \frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\; \mathrm{d}x&= \int (2x) \; \mathrm{d}x \\ \end{aligned} Exploiting the result in (2): \begin{aligned} \int \frac{1}{y}\;\mathrm{d}y &= \int 2x \; \mathrm{d}x \\ \ln y &= x^2 + c \\ y &= e^{x^2 + c} \\ y &= e^ce^{x^2} \\ y &= \mathrm{k} e^{x^2} \quad \blacksquare \end{aligned}

And there you have it - a solution reasoned out without separating $$\mathrm{d}y$$ and $$\mathrm{d}x$$.

## Going Formal

Here we look at a formal derivation of our previous result.

Consider a variable $$z$$, such that $$z = f(y)$$. According to the Chain-Rule:

$\frac{\mathrm{d}}{\mathrm{d}x} [{\color{blue}z}]= \frac{\mathrm{d}}{\mathrm{d}y} [{\color{blue}z}]\cdot \frac{\mathrm{d}y}{\mathrm{d}x} \\$

If we let $${\color{blue}z} = {\color{blue}\int f(y) \mathrm{d}y}$$:

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \bigg[{\color{blue} \int f(y) \: \mathrm{d}y}\bigg] &= \frac{\mathrm{d}}{\mathrm{d}y} \bigg[{\color{blue}\int f(y) \: \mathrm{d}y}\bigg] \cdot \frac{\mathrm{d}y}{\mathrm{d}x} \\ {\color{green}\frac{\mathrm{d}}{\mathrm{d}x} \bigg[\int f(y) \: \mathrm{d}y\bigg]} &= {\color{green}f(y) \cdot \frac{\mathrm{d}y}{\mathrm{d}x}} \\ \end{aligned}

Integrating both sides with respect to $$x$$:

\begin{aligned} {\color{red}\int}\;{\color{green}\frac{\mathrm{d}}{\mathrm{d}x} \bigg[\int f(y) \: \mathrm{d}y\bigg]} \;{\color{red}\mathrm{d}x} &= {\color{red}\int}\; {\color{green}f(y) \cdot \frac{\mathrm{d}y}{\mathrm{d}x}} \:{\color{red}\mathrm{d}x} \\ \int f(y) \: \mathrm{d}y &= \int f(y) \frac{\mathrm{d}y}{\mathrm{d}x} \: \mathrm{d}x \end{aligned}

$\therefore \quad \boxed{\int f(y) \frac{\mathrm{d}y}{\mathrm{d}x} \: \mathrm{d}x = \int f(y) \: \mathrm{d}y}$