DE-BOOK Showing Separability

Section 5.2 Showing Separability

Separable equations are often not presented in a separable form. In such cases, you’ll need to rewrite the equation to verify it’s separable. This process typically involves two main steps:

Isolate the derivative on one side of the equation \(\left(\dfrac{dy}{dx} = \ldots\right)\text{,}\)
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Then form the product on the other side \(\left(\dfrac{dy}{dx} = f(x) \cdot g(y)\right)\text{.}\)
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Common algebra techniques to achieve this product include:

Sums ➜ Products 🔗: \(\quad ab + ac = a \cdot (b + c)\)
Exponents ➜ Products 🔗: \(\quad a^{m + n} = a^m \cdot a^{n}\text{,}\) \(\quad a^{m - n} = \sfrac{a^m}{a^n}\)
Fractions ➜ Products 🔗: \(\quad \dfrac{a}{b} = \dfrac{a}{1} \cdot \dfrac{1}{b}\text{,}\) \(\quad \dfrac{ab}{cd} = \dfrac{a}{c} \cdot \dfrac{b}{d}\)

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To illustrate this, consider the following warm-up examples.

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Subsection Warm-Up Examples

Show that the differential equations are separable.

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🌌 Example 44.

\(y' = y x^2 - y\)

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Solution.

To show this equation is separable, we convert Sums ➜ Products by factoring out the common \(y\text{:}\)

\begin{equation*} y' = \ub{y \cdot (x^2 - 1)}_{f(x)\ \cdot\ g(y)} \quad\leftarrow\text{separable}. \end{equation*}

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🌌 Example 45.

\(y' + 18xy = 6x\)

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Solution.

This time we isolate \(y'\) first, then factor out \(x\text{:}\)

\begin{equation*} y' = 6x - 18xy = 6x \cdot \left(1 - 3y\right) \quad\leftarrow\textit{separable}. \end{equation*}

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🌌 Example 46.

\(y y' - 1 = x\)

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Solution.

First, isolate \(y'\text{:}\)

\begin{equation*} yy' = x + 1 \quad \Rightarrow \quad y' = \frac{x + 1}{y}, \end{equation*}

then convert Fractions ➜ Products:

\begin{equation*} y' = \frac{x + 1}{1} \cdot \frac{1}{y} = \ub{(x+1) \cdot \frac{1}{y}}_{f(x)\ \cdot\ g(y)} \quad \leftarrow \text{separable}. \end{equation*}

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🌌 Example 47.

\(y' = {\large e^{x+y}}\)

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Solution.

At first glance, this doesn’t look separable because of the \(x+y\text{,}\) but applying Exponents ➜ Products we get:

\begin{equation*} {\large y' = e^{x+y} = \ub{e^x \cdot e^{y}}_{f(x)\ \cdot\ g(y)}} \quad\leftarrow\text{separable}. \end{equation*}

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The next set of examples requires more work to demonstrate separability.

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Subsection Challenging Examples

Show that the differential equations are separable.

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🌌 Example 48.

\(P^2\dfrac{dP}{dt} + \dfrac{dP}{dt} = Pt\)

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Solution.

Since the variables in this equation are \(t\) and \(P\text{,}\) the separable form will look like

\begin{equation*} \dfrac{dP}{dt} = f(P) \cdot g(t)\text{.} \end{equation*}

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To get there, we isolate \(dP/dt\) first:

\begin{align*} P^2\dfrac{dP}{dt} + \dfrac{dP}{dt} \amp = Pt \\ (1+P^2)\dfrac{dP}{dt} \amp = Pt \\ \dfrac{dP}{dt} \amp = \dfrac{Pt}{1+P^2}. \end{align*}

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Now, we split the fraction so that \(P\) and \(t\) can be separated:

\begin{equation*} \dfrac{dP}{dt} = \dfrac{P\cdot t}{(P^2 + 1) \cdot (1)} = \ub{ {\color{RoyalBlue} \os{f(P)}{\os{\downarrow}{ \dfrac{P}{P^2 + 1}}} } \cdot {\color{green} \os{g(t)}{\os{\downarrow}{ \dfrac{t}{1}}} } }_{\text{separable form}}. \end{equation*}

So, the equation is separable.

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🌌 Example 49.

\(y\dfrac{dy}{dx} + 8x^2e^{x + \cos y} = 6x^2e^x\)

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Solution.

This example uses a little of everything we discussed, so strap in.

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Noticing that the equation is first-order, we start by isolating \(dy/dx\)

\begin{gather*} y\dfrac{dy}{dx} + 8x^2e^{x + \cos y} = 6x^2e^x\\ \dfrac{dy}{dx} = \dfrac{6x^2e^x - 8x^2e^{x + \cos y}}{y} \end{gather*}

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Next, we can rewrite the exponent in the numerator as a product

\begin{equation*} \dfrac{dy}{dx} = \dfrac{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*} \dfrac{dy}{dx} = \dfrac{2x^2e^x(3 - 4e^{\cos y})}{y}\text{.} \end{equation*}

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From here, we write the result as a product of fractions,

\begin{equation*} \dfrac{dy}{dx} = \dfrac{2x^2e^x}{1} \cdot \dfrac{3 - 4e^{\cos y}}{y}\text{,} \end{equation*}

and the equation is now clearly separable.

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Learning to recognize when an equation is separable is the first step in applying the method of separation of variables, which we will discuss in the next section.

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Checkpoint 50.

(a) Steps to Show Separable.

Suppose you want to show that the differential equation

\begin{equation*} \dfrac{dq}{dt} - 3qt^2 = 2q \end{equation*}

is separable.

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Match the scrambled tasks (on the left) to the order in which they should be performed to show separability.

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check the equation is 1st-order
The first task is to ...
move \(3qt^2\) to the left side via subtraction
Then, the second task is to ...
factor out a common factor of \(q\) on the left side
The final step is to ...
check the equation is linear
integrate both sides
divide both sides by \(3q\)

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(b) 📖❓ Match the Needed Technique.

Match the technique required to demonstrate that each equation is separable.

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Drag the technique from the left to the equation on the right.

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Review it

\(e^{A+B} = e^A \cdot e^B\)
\(\quad w' = w e^{t+w} \)
\(\quad\dfrac{A\cdot B}{C\cdot D} = \dfrac{A}{C} \cdot \dfrac{B}{D}\)
\(\quad\dfrac{dy}{d\theta} = \dfrac{3y}{(y+9)\theta} \)
\(\text{Isolate the derivative}\)
\(\quad xr' + 2r = 0 \)
\(\text{Take out a common factor}\)
\(\quad\dfrac{dP}{dt} = e^t P^2 - e^t \)
\(\text{No technique is needed}\)
\(\quad w' = e^Q e^{x}\)

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(c)

How can the equation, below, be rewritten to show it is separable?

\begin{equation*} \dfrac{dy}{dx} = \dfrac{y}{x+y}. \end{equation*}

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\(\quad\dfrac{dy}{dx} = y\left(\dfrac{1}{x} + \dfrac{1}{y}\right)\)
Incorrect. This rewriting does not effectively separate the variables.
\(\quad\dfrac{dy}{dx} = \dfrac{1}{x}\left(y + 1\right)\)
Incorrect. This form still combines \(x\) and \(y\) terms in a way that is not separable.
\(\quad\dfrac{dy}{dx} = \dfrac{1}{x}\cdot y - \dfrac{1}{x}\)
Incorrect. This rewriting is not equivalent and does not separate the variables.
This equation is not separable.
Correct! This equation is not separable, so it is impossible to write it in separable form.

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(d) 🤔💭 Find the Separable Equations.

Click on each of the separable differential equations below.

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A differential equation is separable if you can rearrange it to express it as a product of functions involving only \(x\) and \(y\) separately. Look for equations that can be manipulated into this form, even if they do not initially appear separable.

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\(\quad\dfrac{dy}{dx} = \dfrac{x + y}{xy}\)		\(\quad\dfrac{dy}{dx} = \dfrac{1}{x^2} + \dfrac{y}{x^2}\)
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\(\quad\dfrac{dy}{dx} = \dfrac{\sin(x+y)}{x}\)		\(\quad\dfrac{dy}{dx} = y \cdot e^{-x}\)
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\(\quad\dfrac{dy}{dx} = \dfrac{3}{x+y}\)		\(\quad\dfrac{dy}{dx} = \cos(xy)\)
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\(\quad\dfrac{dy}{dx} = x - 3y\)		\(\quad\dfrac{dy}{dx} = x^2 + y\)
\(\)

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🌌 Example 51.

Explain why the differential equations are NOT separable.

\(\displaystyle \normalsize\dfrac{d^2y}{dx^2} = xy^2\)

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\(\displaystyle \normalsize\dfrac{dy}{dx} = -6x + 2y\)

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\(\displaystyle xy' + 3x = y\)

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\(\displaystyle y' = x^2y^2 + y\)

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\(\displaystyle y' = \cos(xy)\)

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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.

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This equation is first-order but not separable. To see this, we can rearrange the equation to isolate the derivative:

\begin{gather*} xy' + 3x = y \\ xy' = y - 3x \\ y' = \dfrac{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.

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