Definition 34.
A first-order differential equation is separable if it can be written in the separable form:
\begin{equation}
\frac{dy}{dx} = f(x) \cdot g(y)\tag{5}
\end{equation}
Section 4.1 Separable Form
The separation of variables method works when we can separate the dependent and independent variables by multiplication. Let’s define this property formally.
The notation \(f(x)\) and \(g(y)\) are place-holders for expressions that only contain the variable \(x\) and the variable \(y\text{,}\) respectively. Equivalently, the definition could be written as
\begin{equation*}
\frac{dy}{dx} = \left(
\begin{array}{c}
\text{ONLY } x \text{ terms} \\
\text{ and/or constants}\\
\end{array} \right) \cdot \left(
\begin{array}{c}
\text{ONLY } y \text{ terms} \\
\text{and/or constants}\\
\end{array} \right),
\end{equation*}
but it is much easier to just write \(f(x) \cdot g(y)\text{.}\)
For example, the following differential equations are separable:
\begin{equation*}
\frac{dy}{dx} = 6xy, \quad \frac{dy}{dx} = (x^2 + 1)(y - 6)^3, \quad \frac{dy}{dx} = \sin(y)\cos(x)\text{.}
\end{equation*}
Whereas the following differential equations are not separable:
\begin{equation*}
\frac{dy}{dx} = 6x + y, \quad \frac{dy}{dx} = (x + y)(y - 6)^3, \quad \frac{dy}{dx} = \cos(x - y)\text{.}
\end{equation*}
Special Separable Forms.
A differential equation can still be separable even if one or both of the variables, \(x\) or \(y\text{,}\) do not appear in the equation. For example, the differential equation,
\begin{equation*}
\frac{dy}{dx} = y^2
\end{equation*}
is still separable since it can be written as
\begin{equation*}
\frac{dy}{dx} = (\underset{f(x)}{\underset{\uparrow}{ \vphantom{|} 1 }}) \cdot (\underset{g(y)}{\underset{\uparrow}{ \vphantom{|} y^2 }}) \text{.}
\end{equation*}
Similarly, the following equations are also separable:
\begin{equation*}
\frac{dy}{dx} = 6x + 5, \quad \frac{dy}{dx} = (y - 6)^3, \quad \frac{dy}{dx} = 15\text{.}
\end{equation*}
Understanding the separable form of differential equations is crucial for effectively applying the separation of variables method. By recognizing when an equation can be expressed as a product of functions involving only the independent and dependent variables, you unlock a straightforward path to finding solutions.
Reading Questions Check your Understanding
1. The term separable refers to the idea that the dependent and independent variables can be separated by .
- multiplication
- Correct!
- addition
- Read the first sentence of this page.
- integration
- Read the first sentence of this page.
- gender
- Really?
The term separable refers to the idea that the dependent and independent variables can be separated by .
2. Which of the following differential equations is not separable?
- \(\ds\frac{dy}{dx} = \sin(x)\cos(y)\)
- Incorrect. This equation is separable because it can be expressed as a product of functions involving only \(x\) and \(y\text{.}\)
- \(\ds\frac{dy}{dx} = e^x \cdot y^2\)
- Incorrect. This equation is separable as the variables are already separated by multiplication.
- \(\ds \frac{dy}{dx} = x + y\)
- Correct! This equation is not separable because the terms involving \(x\) and \(y\) are added together, not multiplied.
- \(\ds\frac{dy}{dx} = \frac{x}{y}\)
- Incorrect. This equation is separable because the variables are divided, which can still be separated into a product of functions.
Which of the following differential equations is not separable?
3. The differential equation, \(\ds\frac{dz}{dt} = \cos^2 z \text{,}\) is separable.
True.
- We can show it is separable by rewriting it as\begin{equation*} \frac{dz}{dt} = \underset{f(t)}{\underbrace{(\ \pi + 5 \ )}} \cdot \underset{g(t)}{\underbrace{(\ 1 \ )}} \qquad \text{ or } \qquad \frac{dP}{dt} = \underset{f(z)}{\underbrace{(\ \cos^2 z \ )}} \cdot \underset{g(t)}{\underbrace{(\ \pi + 5 \ )}} \end{equation*}
False.
- We can show it is separable by rewriting it as\begin{equation*} \frac{dz}{dt} = \underset{f(t)}{\underbrace{(\ \pi + 5 \ )}} \cdot \underset{g(t)}{\underbrace{(\ 1 \ )}} \qquad \text{ or } \qquad \frac{dP}{dt} = \underset{f(z)}{\underbrace{(\ \cos^2 z \ )}} \cdot \underset{g(t)}{\underbrace{(\ \pi + 5 \ )}} \end{equation*}
The differential equation, \(\displaystyle \frac{dz}{dt} = \cos^2 z \text{,}\) is separable
4. How can the equation \(\ds\frac{dy}{dx} = \frac{y}{x+y}\) be rewritten in separable form?
- \(\ds\frac{dy}{dx} = y\left(\frac{1}{x} + \frac{1}{y}\right)\)
- Incorrect. This rewriting does not separate the variables effectively.
- \(\ds\frac{dy}{dx} = \frac{1}{x}\left(y + 1\right)\)
- Incorrect. This form still combines \(x\) and \(y\) terms in a way that is not separable.
- \(\ds\frac{dy}{dx} = \frac{1}{x}\cdot y - \frac{1}{x}\)
- Correct! This form separates the variables, making it possible to identify whether the equation is separable.
- None of the above.
- Correct! This equation is not separable, so it is impossible to write it in separable form.
How can the equation \(\ds\frac{dy}{dx} = \frac{y}{x+y}\) be rewritten in separable form?
5. Select the Separable Differential Equations.
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