Separable differential equation

Definition

Form of the differential equation

The term separable is used for a first-order differential equation that, up to basic algebraic manipulation, is of the form:

$\frac{dy}{dx}=f(x)g(y)$

where $x$ is the independent variable and $y$ is the dependent variable.

Separable differential equations can be described as first-order first-degree differential equations where the expression for the derivative in terms of the variables is a multiplicatively separable function of the two variables.

Solution method and formula: general solution

It can be solved by rearranging and integrating:

$\int \frac{dy}{g(y)}=\int f(x)dx$

It suffices to have just one freely floating additive constant in the answer because the additive constants coming from the two integrals can be merged into one.

In general, the solution to this is in the form of an implicit function rather than an explicit description of $y$ as a function of $x$. However, it may be possible to manipulate the solution in order to write $y$ explicitly as a function of $x$.

In addition to the above formula of general solutions, it is also possible that there exist additional solutions that are stationary solutions. These are solutions of the form:

$y=k,\text{where }g(k)=0$

This family of solutions is usually a discrete, often finite, family of solutions. It may be possible to find an elegant way of writing the general solution that also manages to encompass the stationary solutions.

Particular cases

Where the derivative depends only on the dependent variable

This is the case of a first-order autonomous differential equation of degree one (usually, $x$ is replaced by the letter $t$ denoting time, and $y$ is replaced by the letter $x$):

$\frac{dy}{dx}=g(y)$

Here, we get:

$\int \frac{dy}{g(y)}=\int 1dx$

Note that performing the integration expresses $x$ in terms of $y$. We need to then do algebraic manipulation to express $y$ explicitly in terms of $x$.

Again, we need to take care of additional solutions of the form:

$y=k,\text{where }g(k)=0$

Where the derivative depends only on the independent variable

This is a situation where the function depends only on $x$:

$\frac{dy}{dx}=f(x)$

We get:

$y=\int f(x)dx$

This is a straightforward explicit functional description.

Note that in this case, there are no additional solutions that we need to worry about.

Examples

Example without stationary solutions

Consider the differential equation:

$\frac{dy}{dx}=(x^2+1)(y^2+4)$

Note first that $y^2+4=0$ has no solution, so the differential equation has no stationary functional solutions. We rearrange to get the general solution:

$\int \frac{dy}{y^2+4}=\int (x^2+1)dx$

Integrating, we get:

$\frac{1}{2}\arctan (y/2)=\frac{x^3}{3}+x+C$

Note that the constant $C$ is only put at one place rather than separate constants for each integration, because the multiple constants can be absorbed into a single one.

The above gives a family of relational solutions. If we wish, we could convert these to functional solutions, though we need to be somewhat careful when doing so in general. In this case, a naive manipulation would give:

$y=2\tan (2\frac{x^3}{3}+2x+2C)$

We do lose some information in the process (namely that $2\frac{x^3}{3}+2x+2C$ is in the range of $\arctan$) but that information was artificial anyway so this is not an issue.

Example with stationary solutions

Consider the differential equation:

$\frac{dy}{dx}=(x+1)(y+1)$

This has a stationary solution $y=-1$ (a constant function). Assuming $y\ne -1$, we can rearrange to get:

$\int \frac{dy}{y+1}=\int (x+1)dx$

This integrates to give:

$\ln |y+1|=\frac{x^2}{2}+x+C$

Exponentiating both sides, we get:

$|y+1|=e^C\exp (\frac{x^2}{2}+x)$

Let $k=e^C\mathrm{s}\mathrm{g}\mathrm{n}(y+1)$. Note that $k\ne 0$. We get:

$y+1=k\exp (\frac{x^2}{2}+x)$

Rearranging, we can get $y$ as an explicit function of $x$:

$y=k\exp (\frac{x^2}{2}+x)-1$

Note that currently we have the restriction $k\ne 0$. However, we see that allowing $k=0$ gives the stationary solution, so that we can combine the stationary solution into the general solution and get, with $k\in \mathbb{R}$ a parameter:

$y=k\exp (\frac{x^2}{2}+x)-1$