Separation of Variables

Separation of Variables Meaning

As the name suggests, these are the types of Differential Equations when the variables can be separated explicitly, such a class of Differential equations are very easy to deal with.

In fact, you can simply plan down an algorithm to solve such differential equations. Then, you shall only be dealing with Ordinary Differential Equations of first order.

It has to be said that most of the applications of differential equations are mostly in the realm of Partial Differential Equations, which, for now, remains out of the scope of this article.

Separation of Variables differential equations

A separable differential equation of the \(n^{\text {th }}\) order can be written in the form:

$$\frac{\mathrm{d}^n y}{\mathrm{d} x^n}=f(x) g(y)$$

where \(y\) is differentiable and, therefore, continuous on the appropriately defined domain of the function.

Also note that \(y\) is a function of \(x\), i.e., \(y=f(x)\), which implies that \(x\) is the independent variable here. The equation above might look a bit scary at first since you have a differential equation of order \(n\) where \(n \in \mathbb{N}\). But you are only going to deal with cases where \(n=1\), is the first-order separable differential equation.

Methods of separation of variables

You shall only be looking at DEs of the first order, i.e. when \(n=1\).

The Ordinary Differential Equation you will deal with, is:

$$\frac{\mathrm{d} y}{\mathrm{d} x}=f(x) g(y)$$

To solve this ODE, you will separate the two variables, having like variables on each side, i.e. having \(\mathrm{d} y\), \(h(y)\) on one side and \(\mathrm{d} x, f(x)\) on the other. You get:

$$\frac{\mathrm{d} y}{h(y)}=f(x) \, \mathrm{d} x$$

Now, the variables are separated successfully. The reason for doing it, is to have functions on both sides, which can be integrated easily.

Integrating on both sides:

$$\int \frac{\mathrm{d} y}{h(y)}=\int f(x) \, \mathrm{d}x$$

And there you have it! The differential equation \(y'=f(x)h(y)\) has been solved, you have an explicit solution to the given separable differential equation.

Separation of Variables Examples

Let's look at another example, just slightly different:

Application of Separation of Variables

These ordinary differential equations are not here as just pure mathematical objects that you play with, as discussed firsthand. There is a very solid reason these differential equations and their solutions exist. There is a wide range of stuff to which one can apply these differential equations to. You will look at some of these applications which involve separable variables. The most common of these problems are growth problems of populations, decay rates, Newton's law of cooling, etc.

Model of Exponential Population Growth

An essential application of first-order differential equations of separable form is using it to model the population growth of different groups such as bacteria or the growth rate of rabbits.

Such models are not exactly accurate and often ignore certain outside factors to keep the differential equations simple enough to deal with.

Let's start with a population group of \(P_0\), which is the initial population at the time \(t=0\). And let the population be \(P\) at some arbitrary time \(t=t\).

And suppose that the population is increasing at a constant rate \(k\), and assume that there are no external factors that are reducing the population (for simplicity).

You want to form an ordinary differential equation to model this situation, note that the growth rate at any certain point is proportional to the population at that time:$$ \frac{\mathrm{d} P}{\mathrm{d} t}=kP$$

where \(k\) is the rate of population growth. Now, since the equation is of separable form, you get:

$$\frac{\mathrm{d} P}{P}=k \, \mathrm{d} t$$

Integrating on both sides:

$$\int_{P_0}^P \frac{\mathrm{d} P}{P}=\int_0^t k \, \mathrm{d} t$$

The limits of integration are from \(P=P_0\) to \(P=P\), which the population grows to. And the time goes from \(t=0\) to \(t=t\). Integrating definitely:

$$\begin{aligned} & \left.\ln P\right|_{P_0} ^P=\left.k t\right|_0 ^t \\ \implies & \ln P-\ln P_0=k t \\ \implies & \ln \left(\frac{P}{P_0}\right)=k t \end{aligned}$$

Taking exponential9 on both sides:

$$ \begin{aligned} &\frac{P}{P_0}=e^{k t} \\ \implies \ &P=P_0 e^{kt} \end{aligned}$$

Now, you can model the population of the given group at a time \(t\).

Radioactive Decay

Radioactive is one of the most useful applications of separable differential equations. Suppose you have \(N_0\) amount of radioactive material at the time \(t=0\) and you want to model the radioactivity of the material as time goes by.

The rate at which the amount of radioactive material changes is directly proportional to the amount present at that time:

$$ \frac{\mathrm{d} N}{\mathrm{d} t}=-\lambda N $$

where \(\lambda >0\) and is known as the decay constant, the reason there is a negative sign is that the amount of radioactive material is decreasing over time. Since the equation is of separable form, you get:

$$\begin{aligned} & \frac{\mathrm{d} N}{N}=-\lambda t \\ \implies \ & \int_{N_0}^N \frac{\mathrm{d} N}{N} = - \lambda \int_0^t \mathrm{d} t \\ \implies \ & \ln \left( \frac{N}{N_0} \right) = -\lambda t \\ \implies \ & N=N_0 e^{-\lambda t} \end{aligned}$$

Thus, you have a solution to the ordinary differential equation. One can deduce from the solution that the amount of radioactive material decreases exponentially over time. The graph of this equation can be plotted as below:

Figure 1.- Graph modelling the radioactive decay of a material

A very particular observation one can make is that the radioactive material nearly vanishes out over a very, very long period of time. You can let the physicists figure out the reasons behind that. In mathematical terms, one can observe that as \(t \rightarrow \infty\), \(N \rightarrow 0\). Let's look at an example regarding radioactive decay.

Newton's law of Cooling

A widely known application of several variable differential equations is in the physics of heat transfer. It models how an object loses its heat, which is known as Newton's law of cooling:

The rate of heat lost by an object to the surroundings is directly proportional to the difference in temperature between the object and the surroundings.

Let \(T\) be the temperature of the object at some arbitrary time \(t\) and \(T_s\) be the temperature of the surroundings. It should be noted that \(T>T_s\) at all times, since the body will not lose any heat if the surrounding is at a higher temperature. The flow of heat from the object will come to a stop when \(T=T_s\).

According to the law:

$$ \frac{\mathrm{d} T}{\mathrm{d} t}= -k(T-T_s) $$

Notice that you have the constant of proportionality as \(-k\) where \(k>0\) because the object is losing heat, and so the rate of change of it will be negative.

Separating the variables, you have:

$$ \begin{aligned} & \int \frac{\mathrm{d} T}{T-T_s}= \int -k \, \mathrm{d} t \\ \implies \ & \ln |c(T-T_s)|=-kt \\ \implies \ &c(T-T_s)=e^{-kt} \\ \implies \ &T=T_s+Ce^{-kt} \end{aligned}$$

where \(c=1/C\) for the sake of convenience.

You can determine \(C\) by substituting \(t=0\):

$$ \begin{aligned} &T_0=T_s+C \\ \implies \ &C=T_0-T_s \end{aligned} $$

where \(T_0\) is the temperature of the object at the time \(t=0\).

Thus, your solution is:

$$T=T_s+(T_0-T_s)e^{-kt}$$

where \(k\) is a constant that depends on the surface area and the material of the object.