π Definition 47. Separable Form.
A first-order differential equation is separable if it can be written in following separable form
\begin{equation}
\frac{dy}{dx} = f(x) \cdot g(y)\text{.}\tag{9}
\end{equation}
Section Separable Differential Equations
The method of separation of variables applies only to a special class of first-order differential equations, those that can be written in a particular form. When an equation has this form, it can be rearranged so that integration leads directly to its general solution. So before we can use this powerful method, we first need to learn how to recognize when an equation is separable.
Subsection 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. These equations are called separable. Letβs make that idea precise.
Checkpoint 48. πβ.
Subsection What Does βSeparableβ Mean?
The notation \(f(x) \cdot g(y)\) can be intimidating at first. It compresses a lot of ideas into a small expression. But donβt worry, weβll break it down.
In the definition, \(f(x)\) is just a stand-in for any expression that only involves the variable \(x\text{.}\) For example, it might be:
\begin{equation*}
13\sin(3x), \quad (x^2 + 1)^{\cos x}, \quad 3x - 2e^{4x}, \quad \frac{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 \frac{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 \(y\text{.}\) Here are a few made-up examples:
\begin{equation*}
\ob{3(3x - 2e^x)}^{f(x)} \cdot \ob{(y^2-1.7)}^{g(y)}, \quad
\os{f(x)}{\os{\downarrow}{\left(\frac{1}{x+1}\right)}} \os{g(y)}{\os{\downarrow}{\left(\frac{y}{3}+7\right)}}, \quad
\os{f(x)}{\os{\downarrow}{(-1)}} \cdot \os{g(y)}{\os{\downarrow}{(\pi)}}, \quad \text{etc.}
\end{equation*}
Subsection Recognizing Separable Forms
Since the separation of variables method can only be applied to differential equations in separable form, it is crucial to be able to recognize this structure.
Examples of separable differential equations include:
\begin{equation*}
\frac{dy}{dx} = {\ub{(x^2 + 1) \cdot (y - 6)}_{\large f(x)\ \cdot\ g(y)}}^3 , \qquad
\frac{dy}{dx} = \ub{\sin(y) \cdot \cos(x)}_{\large g(y)\ \cdot\ f(x)}.
\end{equation*}
Non-separable equations include:
\begin{equation*}
\frac{dy}{dx} = \ub{-6x + 2y}_{\large f(x)\ +\ g(y)}, \quad
\frac{dy}{dx} = {\ub{(x^2 + y) \cdot (y - 6)}_{\large f(x ,\ y)\ \cdot\ g(y)}}^3, \quad
\frac{dy}{dx} = \ub{\cos(x - y)}_{\large f(x\ -\ y)}.
\end{equation*}
Checkpoint 49. πβ.
Which of the following differential equations is not separable?
- \(\quad\ds\frac{dy}{dx} = \sin(x)\cos(y)\)
- Incorrect. This equation is separable because it can be expressed as a product of functions involving only \(x\) and \(y\text{.}\)
- \(\quad\ds\frac{dy}{dx} = e^x \cdot y^2\)
- Incorrect. This equation is separable as the variables are already separated by multiplication.
- \(\quad\ds\frac{dy}{dx} = x + y\)
- Correct! This equation is not separable because the terms involving \(x\) and \(y\) are added together, not multiplied.
- \(\quad\ds\frac{dy}{dx} = \frac{x}{y}\)
- Incorrect. This equation is separable because the variables are divided, which can still be separated into a product of functions.
Subsection A Special Separable Form
As a special case, a differential equation can still be separable even if one or both variables do not explicitly appear. For example, the equations
\begin{equation*}
\frac{dy}{dx} = 6x + 5, \qquad
\frac{dy}{dx} = (y^2 + 7)^3, \qquad
\frac{dy}{dx} = 15\text{,}
\end{equation*}
are all separable since they can be written as
\begin{equation*}
\frac{dy}{dx} = (\underset{f(x)}{\underset{\uparrow}{ \vphantom{|} 6x }}) \cdot
(\underset{g(y)}{\underset{\uparrow}{ \vphantom{|} 1 }})
\qquad
\frac{dy}{dx} = (\underset{f(x)}{\underset{\uparrow}{ \vphantom{|} 1 }}) \cdot
\underset{g(y)}{\underset{\uparrow}{ \vphantom{|} (y^2 + 7)^3 }}
\quad
\frac{dy}{dx} = (\underset{f(x)}{\underset{\uparrow}{ \vphantom{|} 3 }}) \cdot
(\underset{g(y)}{\underset{\uparrow}{ \vphantom{|} 5 }})\text{.}
\end{equation*}
While all three of these are in separable form and can be solved with the separation of variables method, the first and third equations are better suited for direct integration.
Checkpoint 50. πβ.
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,
\begin{equation*}
\frac{dz}{dt} = \cos^2 z\text{,}
\end{equation*}
is separable
Subsection Showing An Equation is Separable
Separable equations are often not presented in a separable form. In such cases, youβll need to rewrite the equation to verify itβs separable. This process typically involves two main steps:
-
Isolate the derivative on one side of the equation \(\left(\frac{dy}{dx} = \ldots\right)\text{,}\)
-
Then form the product on the other side of the equation \(\left(\frac{dy}{dx} = f(x) \cdot g(y)\right)\text{.}\) Common algebra techniques to achieve this product include:
-
Factoring out common terms: \(\quad ab \pm ac = a \cdot (b \pm c)\)
-
Exponential sums to products : \(\quad e^{m\ \pm\ n} = e^m \cdot e^{\pm n}\)
-
Rearranging fractions: \(\quad \frac{a}{b} = a \cdot \frac{1}{b} \quad\) and \(\quad \frac{a\ \cdot\ b}{c\ \cdot\ d} = \frac{a}{c} \cdot \frac{b}{d}\)
-
To illustrate this, consider the equation
\begin{equation*}
\frac{dy}{dx} + 18xy = 6x.
\end{equation*}
Here, we isolate \(dy/dx\) and factor out \(x\text{:}\)
\begin{equation*}
\frac{dy}{dx} = 6x - 18xy = 6x \cdot \left(1 - 3y\right),
\end{equation*}
which shows the differential equation is separable.
π Example 52. Isolate \(dP/dt\) & Rearrange the Fraction.
Show that the differential equation is separable.
\begin{equation*}
P^2\frac{dP}{dt} + \frac{dP}{dt} = Pt.
\end{equation*}
Solution.
Since the variables in this equation are \(t\) and \(P\text{,}\) the separable form will look like
\begin{equation*}
\frac{dP}{dt} = f(P) \cdot g(t)\text{.}
\end{equation*}
To get there, we isolate \(dP/dt\) first:
\begin{align*}
P^2\frac{dP}{dt} + \frac{dP}{dt} \amp = Pt \\
(1+P^2)\frac{dP}{dt} \amp = Pt \\
\frac{dP}{dt} \amp = \frac{Pt}{1+P^2}.
\end{align*}
Now, we split the fraction so that \(P\) and \(t\) can be separated:
\begin{equation*}
\frac{dP}{dt}
= \frac{P\cdot t}{(P^2 + 1) \cdot (1)}
= \ub{
{\color{blue} \os{f(P)}{\os{\downarrow}{ \frac{P}{P^2 + 1}}} } \cdot
{\color{green} \os{g(t)}{\os{\downarrow}{ \frac{t}{1}}} }
}_{\text{separable form}}.
\end{equation*}
So, the equation is separable.
π Example 53. Splitting an Exponential.
Show that the differential equation is separable.
\begin{equation*}
\frac{dM}{d\omega} - 11e^{4\omega + 3M} = 0\text{.}
\end{equation*}
Solution.
Isolating the derivative and applying \(e^{A + B} = e^{A} \cdot e^{B}\) leads to the separable form:
\begin{equation*}
\frac{dM}{d\omega}
= 11e^{4\omega + 3M}
= 11(e^{4\omega} \cdot e^{3M})
= \ub{
{\color{blue} \os{f(\omega)}{ \os{\downarrow}{ 11e^{4\omega} }} } \cdot
{\color{green}\os{g(M)}{ \os{\downarrow}{ e^{3M} }} }
}_{\text{separable form}}.
\end{equation*}
Checkpoint 54. πβ Select the Separable Form.
How can the equation, below, be rewritten in separable form?
\begin{equation*}
\frac{dy}{dx} = \frac{y}{x+y}.
\end{equation*}
- None of the above.
- Correct! This equation is not separable, so it is impossible to write it in separable form.
- \(\quad\ds\frac{dy}{dx} = y\left(\frac{1}{x} + \frac{1}{y}\right)\)
- Incorrect. This rewriting does not separate the variables effectively.
- \(\quad\ds\frac{dy}{dx} = \frac{1}{x}\left(y + 1\right)\)
- Incorrect. This form still combines \(x\) and \(y\) terms in a way that is not separable.
- \(\quad\ds\frac{dy}{dx} = \frac{1}{x}\cdot y - \frac{1}{x}\)
- Correct! This form separates the variables, making it possible to identify whether the equation is separable.
π Example 55. Flexing Algebra Skills.
Show the differential equation is separable.
\begin{equation*}
y\frac{dy}{dx} + 8x^2e^{x + \cos y} = 6x^2e^x.
\end{equation*}
Solution.
This example uses a little of everything we discussed, so strap in.
Noticing that the equation is first-order, we start by isolating \(dy/dx\)
\begin{gather*}
y\frac{dy}{dx} + 8x^2e^{x + \cos y} = 6x^2e^x\\
\frac{dy}{dx} = \frac{6x^2e^x - 8x^2e^{x + \cos y}}{y}
\end{gather*}
Next, we can rewrite the exponent in the numerator as a product
\begin{equation*}
\frac{dy}{dx} = \frac{6x^2e^x - 8x^2e^{x}e^{\cos y}}{y}
\end{equation*}
and recognize that there is a common factor in the numerator
\begin{equation*}
\frac{dy}{dx} = \frac{2x^2e^x(3 - 4e^{\cos y})}{y}\text{.}
\end{equation*}
From here, we write the result as a product of fractions,
\begin{equation*}
\frac{dy}{dx} = \frac{2x^2e^x}{1} \cdot \frac{3 - 4e^{\cos y}}{y}\text{,}
\end{equation*}
and the equation is now clearly separable.
Checkpoint 56. πβ Showing an Equation is Separable.
As important as it is to show an equation is separable, it is equally important to identify when it is not.
π Example 57. Non-Separable Equations.
Show the differential equations are NOT separable.
\begin{equation*}
y^2\frac{d^2y}{dx^2} = x, \qquad
xy' + 3x = y.
\end{equation*}
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.
This equation is first-order, but it is not separable. To see this, we can rearrange the equation to isolate the derivative:
\begin{gather*}
xy' + 3x = y \\
xy' = y - 3x \\
y' = \frac{y - 3x}{x} \text{.}
\end{gather*}
Since there is no common factor of \(x\) or \(y\) in the numerator, the variables cannot be separated and the equation is not separable.
Learning to recognize when an equation is separable is the first and most important step in applying the method of separation of variables. In the next section, weβll see how this recognition leads to actual solutions.
Subsection π€ Wrap-Up
Check Your Understanding.
Checkpoint 58. π€π Separation of Variables Reading Questions.
(a) π€π Find the Separable Equations.
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