📙 Definition 40. Separable Form.
A first-order differential equation is separable if it can be written in the following separable form
\begin{equation}
\dfrac{dy}{dx} = f(x) \cdot g(y)\text{.}\tag{5.1}
\end{equation}
Section 5.1 Separable Form
Some differential equations can be rewritten so that all the \(x\) terms are on one side and all the \(y\) terms are on the other. Such equations are called separable, which is defined more precisely as follows.
Subsection Definition of Separable & Separable Form
The notation, \(f(x) \cdot g(y)\text{,}\) can seem strange at first glance, so let’s break it down.
Here, \(f(x)\) is just a stand-in for any expression that only involves the variable \(x\) and possibly a constant. For example, it might be:
\begin{equation*}
13\sin(3x), \quad 3x - 2e^{4x}, \quad \dfrac{2^x}{1+x}, \quad 1, \quad \text{etc.}
\end{equation*}
Likewise, \(g(y)\) stands for any expression involving only \(y\) and constants:
\begin{equation*}
\sin(y+\cos y), \quad 3y^2 + 1, \quad \dfrac{1}{y}, \quad 1, \quad \text{etc.}
\end{equation*}
So when we say \(f(x) \cdot g(y)\text{,}\) we just mean any product where one factor depends only on \(x\) and the other only on \(y\text{.}\)
Checkpoint 41.
Subsection Separable Differential Equations
Now that the notation makes sense, here are some examples of differential equations in separable form:
\begin{equation*}
\dfrac{dy}{dx} = \ub{(x^2 + 1) \cdot (y - 6)}_{\large f(x)\ \cdot\ g(y)}, \qquad
\dfrac{dy}{dx} = \ub{\sin(y) \cdot \cos(x)}_{\large g(y)\ \cdot\ f(x)}.
\end{equation*}
Differential equations can still be separable even if one of the variables does not explicitly appear. For example, all of these equations are separable:
-
\(\displaystyle \dfrac{dy}{dx} = 6x + 5 = (6x + 5) \cdot (1)\qquad \leftarrow\ f(x) \cdot g(y)\)
-
\(\displaystyle \dfrac{dP}{dt} = P\ (P-1) = (1) \cdot (\ P\ (P-1)\ )\qquad \leftarrow\ f(t) \cdot g(P)\)
-
\(\displaystyle \dfrac{dy}{dx} = 15 = (3) \cdot (5)\qquad \leftarrow\ f(x) \cdot g(y)\)
Subsection Non-Separable Differential Equations
As important as it is to show an equation is separable, it is equally important to identify when it is not. The next example demonstrates some non-separable equations.
🌌 Example 42.
Explain why the differential equations are NOT separable.
-
\(\displaystyle \dfrac{d^2y}{dx^2} = xy^2\)
-
\(\displaystyle \dfrac{dy}{dx} = -6x + 2y\)
-
\(\displaystyle \dfrac{dy}{dx} = \cos(xy)\)
Solution.
-
The first equation is a bit of a trick question to remind you of the first-order requirement of a separable equation. Since this equation is second-order, separability does not apply.
-
The second equation is first-order, but it is not separable since \(-6x + 2y\) cannot be expressed in the function-notation form \(f(x) \cdot g(y)\text{.}\)
-
The last equation is first-order, but there are no trigonometric identities that you could apply to \(\cos(xy)\) to separate the \(x\) and \(y\) parts by multiplication.
Checkpoint 43.
(a)
Which of the following differential equations is not separable?
- \(\quad\dfrac{dy}{dx} = \sin(x)\cos(y)\)
- Incorrect. This equation is separable because it can be expressed as a product of functions involving only \(x\) and \(y\text{.}\)
- \(\quad\dfrac{dy}{dx} = e^x \cdot y^2\)
- Incorrect. This equation is separable as the variables are already separated by multiplication.
- \(\quad\dfrac{dy}{dx} = x + y\)
- Correct! This equation is not separable because the terms involving \(x\) and \(y\) are added together, not multiplied.
- \(\quad\dfrac{dy}{dx} = \dfrac{x}{y}\)
- Incorrect. This equation is separable because \(\frac{x}{y}\) can be expressed as \(\frac{x}{1} \cdot \frac{1}{y}\text{.}\)
(b)
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