Recognize a separable equation
A slope made of two factors
The middle graph is the slope field of \(\frac{dy}{dx} = (x - 1)\,y\). The orange segment shows the slope at the orange point.
The slope is a product of two factors. Below the field is the graph of \(x - 1\), which uses only \(x\). To the left, turned on its side, is the graph of \(y\), which uses only \(y\).
The orange line on each graph is the value of that factor at the point. At \((2, 1.5)\) they are 1 and 1.5, so the slope is \(1 \times 1.5 = 1.5\).
Drag the point to the left of \(x = 1\). The x factor turns negative, and so does the slope. On the line \(x = 1\) the x factor is 0, so the segments are flat.
Slope equals g(x) times h(y)
The field now comes from \(\frac{dy}{dx} = g(x)\,h(y)\), with the two labelled graphs as its factors. Drag the point. Whatever shape the graphs have, the slope at the point is the product of the two orange heights.
Press the button. Dividing by \(h(y)\) turns the left graph into \(\frac{1}{h(y)}\), and the factor moves next to \(dy\), so each variable is on its own side.
Notation
An equation is separable if its right side is a product of two factors. One factor, \(g(x)\), uses only \(x\). The other factor, \(h(y)\), uses only \(y\).
\[\frac{dy}{dx} = g(x)\,h(y)\]To separate, you divide both sides by \(h(y)\). This step needs \(h(y) \ne 0\), because you cannot divide by 0. The orange segments are flat where \(h(y) = 0\).
\[\frac{1}{h(y)}\,\frac{dy}{dx} = g(x)\]Then you multiply both sides by \(dx\), so each side has one variable.
\[\frac{1}{h(y)}\,dy = g(x)\,dx\]Here \(\frac{dy}{dx}\) is treated as a ratio of two small changes. The next skill integrates both sides.
To test an equation, try to factor its right side into an \(x\) part times a \(y\) part. A sum such as \(x + y\) has no such factoring, so it is not separable.
When it fails
The rearrangement needs a product. Make the right side the sum \(g(x) + h(y)\) of the same two parts.
On the row of orange segments, \(h(y)\) is still 0, but the segments tilt, because their slope is now \(g(x)\). Dividing by \(h(y)\) gives \(\frac{g(x)}{h(y)} + 1\), which still mixes \(x\) and \(y\), so the equation does not separate.
Example 1: the product is already visible
Write \(\dfrac{dy}{dx} = x^3(y - 2)\) with each variable on its own side.
The right side is a product. The \(x\) factor is \(x^3\), and the \(y\) factor is \(y - 2\).
\[g(x) = x^3, \qquad h(y) = y - 2\]To separate, you divide both sides by \(y - 2\). You cannot divide by 0, so this step needs \(y \ne 2\). Every segment on the orange row \(y = 2\) is flat.
\[\frac{1}{y - 2}\,\frac{dy}{dx} = x^3\]Then you multiply both sides by \(dx\), and each variable is on its own side.
\[\frac{1}{y - 2}\,dy = x^3\,dx\]The case \(y = 2\) is a different question, which a later skill answers.
Example 2: the product is hidden
Write \(\dfrac{dy}{dx} = e^{x + y}\) with each variable on its own side.
The sum in the exponent splits into a product, because \(e^{a + b} = e^a e^b\). The \(x\) factor is \(e^x\), and the \(y\) factor is \(e^y\).
\[\frac{dy}{dx} = e^{x}\,e^{y}\]A tempting wrong move: Multiply both sides by \(dx\), and keep \(e^y\) with \(dy\).
\[e^{y}\,dy = e^{x}\,dx\]The factor \(e^y\) multiplies the right side, so to move it to the left side you must divide by it. The red curve, the wrong factor \(e^y\), grows as \(y\) increases, while the faint curve, the correct factor \(e^{-y}\), shrinks.
Divide both sides by \(e^y\). The factor moves to the left side as \(e^{-y}\), because \(\frac{1}{e^y} = e^{-y}\).
\[e^{-y}\,\frac{dy}{dx} = e^{x}\]Then multiply both sides by \(dx\).
\[e^{-y}\,dy = e^{x}\,dx\]The factor \(e^y\) is never 0, so the division needs no condition and there is no flat row.
Practice
Show that \(\dfrac{dy}{dx} = xy + 2x + 3y + 6\) is separable. Write it with each variable on its own side.
Factor \(x\) out of the first two terms and 3 out of the last two. On the figure, the segments on the column \(x = -3\) and the row \(y = -2\) are flat, so the right side has a factor for each.
Solution
Factor \(x\) out of the first two terms and 3 out of the last two. Both groups contain \(y + 2\), so you can factor it out.
\[\begin{aligned} &x(y + 2) + 3(y + 2) \\ &\quad = (x + 3)(y + 2) \end{aligned}\]Divide both sides by \(y + 2\), which needs \(y \ne -2\). Then multiply both sides by \(dx\).
\[\frac{1}{y + 2}\,dy = (x + 3)\,dx\]
Questions
The graph shows the slope field of \(\dfrac{dy}{dx} = (x - 2)(y + 1)\). Below it is the graph of the factor \(x - 2\). To the left, turned on its side, is the graph of the factor \(y + 1\).
The orange heights show the two factors at the point \((3, 1)\). What is the slope there?
- 2
- 3
- 1
Answer
2
Every segment along the vertical line \(x = 2\) is flat. Which factor is zero there?
- The x factor, \(x - 2\)
- The y factor, \(y + 1\)
- Both factors
Answer
The x factor, \(x - 2\)
Which equation is separable?
- \(\dfrac{dy}{dx} = \dfrac{3x\,e^{y}}{x^2 + 1}\)
- \(\dfrac{dy}{dx} = x^2 + y\)
- \(\dfrac{dy}{dx} = \dfrac{x + y}{y}\)
- \(\dfrac{dy}{dx} = x^2 y + 1\)
Answer
\(\dfrac{dy}{dx} = \dfrac{3x\,e^{y}}{x^2 + 1}\)
Summary
You write the right side as a function of \(x\) times a function of \(y\), divide by the y factor, and multiply by \(dx\).
This works because the slope at each point is the product of the two factors, so the factors can move to opposite sides.
You use it before you integrate, because each side then has one variable.
This is one of five skills in topic 7.6 of Which & Why: AP Calculus.