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Exploring Differential Equations An Interactive, Student-Centered Approach

Geoffrey W. Cox, Ph.D.

Section 5.3 Separation of Variables (SOV) Method

Once you’ve verified that an equation is separable, solving it becomes a systematic process. Here’s the step-by-step guide:
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Subsection Separation of Variables Steps

Separation of Variables.

1. Verify Separable, Separate Variables
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Check that the equation is first-order
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If possible, put into separable form:
\begin{equation*} \frac{dy}{dx} = f(x) \cdot g(y) \end{equation*}
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Separate \(x\) and \(y\) on opposite sides:
\begin{equation*} \frac{1}{g(y)}\ dy = f(x)\ dx \end{equation*}
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2. Integrate
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\begin{equation*} \int \frac{1}{g(y)}\, dy = \int f(x)\, dx. \end{equation*}
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3. Solve for \(y\)
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If possible, explicitly solve for \(y\text{.}\) If not, leave the solution in implicit form. Combine constants.
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(Optional) Justification of separating \(dy\) & \(dx\) in \(\dfrac{dy}{dx}\).

Although separating \(dy\) & \(dx\) in \(\sfrac{dy}{dx}\) is mathematically invalid, it is a widely accepted shortcut in the SOV method.
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To see why, start with the separable form,
\begin{equation*} \frac{dy}{dx} = f(x) \cdot g(y)\text{,} \end{equation*}
and divide both sides by \(g(y)\text{,}\) leaving the derivative intact:
\begin{equation*} \frac{1}{g(y)}\ \frac{dy}{dx} = f(x)\text{.} \end{equation*}
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From here, integrate both sides with respect to \(x\text{:}\)
\begin{equation*} \int \frac{1}{g(y)}\ \frac{dy}{dx}\ dx = \int f(x)\ dx\text{.} \end{equation*}
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Since the product, \(\sfrac{dy}{dx}\cdot dx\text{,}\) is defined as the differential, \(dy\text{,}\) we can substitute it into to get the form of the integral you see in Step 2:
\begin{equation*} \int \frac{1}{g(y)}\ dy = \int f(x)\ dx, \end{equation*}
which justifies the informal splitting of \(dy\) and \(dx\) in the derivative, \(\sfrac{dy}{dx}\text{.}\)
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Checkpoint 52. 📖❓ Reorder the Strategy.

Reorder the scrambled tasks used in the Separation of Variables method.
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Let’s begin with a few warm-up problems to build confidence in using separation of variables. The examples that follow are intentionally straightforward; each equation is either already separable or requires minimal algebraic manipulation. As you work through them, focus on the 3-step strategy described in the Separation of Variables method.
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Checkpoint 53. 📖❓ Review Separability.

Which of the following differential equations are separable?
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  • \(\quad\dfrac{dy}{dx} = x e^y\)
  • Is this the product of a function of \(x\) and \(y\text{?}\)
  • \(\quad\dfrac{dy}{dx} = \frac{x}{1 + y^2}\)
  • Note: \(\frac{x}{1 + y^2} = x \cdot \frac{1}{1 + y^2}\text{.}\)
  • \(\quad\dfrac{dy}{dx} - y = x\)
  • Moving \(y\) over you get
    \begin{equation*} \dfrac{dy}{dx} = x + y\text{.} \end{equation*}
    Can a sum like this be converted into the form \(f(x) \cdot g(y)\text{?}\)
  • \(\quad\ds y'' = \ln(x)\ln(y)\)
  • Note, this is a second-order differential equation...
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Subsection Warm-Up Examples

Find the general solution for each equation.
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🌌 Example 54.

\(\quad 1.\ \ \dfrac{dy}{dx} = x^4,\ \quad 2.\ \ \dfrac{dy}{dx} = y^2\)
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Solution.
We’ll solve these equations side by side, clearly demonstrating each step of the Separation of Variables method.
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Step 1 - Verify separability & separate variables:
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\begin{align*} \frac{dy}{dx} \amp = x^4 \cdot 1 \quad\checkmark\\ dy \amp = x^4\ dx \end{align*}
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\begin{align*} \frac{dy}{dx} \amp = 1 \cdot y^2 \quad\checkmark\\ y^{-2}\ dy \amp = dx \end{align*}
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Step 2 - Integrate:
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\begin{align*} \int dy \amp = \int x^4\ dx\\ y + c_1 \amp = \frac{1}{5}x^5 + c_2 \end{align*}
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\begin{align*} \int y^{-2}\ dy \amp = \int dx\\ -\frac{1}{y} + c_1 \amp = x + c_2 \end{align*}
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Step 3 - Solve for \(y\) and combine constants:
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\begin{equation*} \boxed{y = \frac{1}{5}x^5 + c} \end{equation*}
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\begin{equation*} \frac{1}{y} = -x + c \ \Rightarrow\ \boxed{y = \frac{1}{-x + c}} \end{equation*}
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where \(c = c_2 - c_1\) in both general solutions.
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🌌 Example 55.

\(\quad\dfrac{dy}{dx} + \dfrac{x}{y^2} = 0\)
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Solution.
Step 1 - Verify separability & separate variables:
\begin{align*} \frac{dy}{dx} \amp = -\frac{x}{y^2} = (-x)\cdot\frac{1}{y^2} \quad\checkmark\\ y^2\ dy \amp = -x\ dx. \end{align*}
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Step 2 - Integrate:
\begin{align*} \int y^2\ dy \amp = \int -x\ dx\\ \frac{y^3}{3} + c_1 \amp = -\frac{x^2}{2} + c_2 \end{align*}
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Step 3 - Combine constants and solve explicitly for \(y\text{:}\)
\begin{equation*} \frac{y^3}{3} = -\frac{x^2}{2} + c \quad\Rightarrow\quad \boxed{y = \left(-\frac{3}{2}x^2 + c\right)^{1/3}}. \end{equation*}
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Solve the initial-value problem.
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🌌 Example 56.

\(z' - 1 = z^2, \quad z(4) = 9\)
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Solution.
This differential equation is first order and separable since
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\begin{align*} z' - 1 \amp = z^2 \\ z' \amp = z^2 + 1 = \big( 1 \big) \big( z^2 + 1 \big) \quad\checkmark \end{align*}
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Separable
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Next, we separate and integrate:
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\begin{equation*} \frac{1}{z^2+1}dz = 1\ dt \end{equation*}
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Separate
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\begin{equation*} \int\frac{1}{z^2+1}\ dz = \int 1\ dt \end{equation*}
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Integrate
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\begin{equation*} \arctan z = t + c \end{equation*}
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So, the general solution is
\begin{equation*} z = \tan(t+c) \text{.} \end{equation*}
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Applying \(z(4) = 9\) to the general solution, we get an equation for \(c\text{:}\)
\begin{align*} 9 = \tan(4+c) \implies \amp\ \arctan(9) = 4 + c \\ \amp\ \arctan(9) - 4 = c \end{align*}
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Therefore,
\begin{equation*} z = \tan(t + \arctan(9) - 4) \end{equation*}
is the particular solution to the initial-value problem.
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Be cautious about rewriting constants in decimal form. For instance, if you approximate \(\arctan(9) - 4 \approx -2.54\text{,}\) the solution becomes:
\begin{equation*} z = \tan(t - 2.54), \end{equation*}
which may not exactly satisfy the original initial condition due to rounding errors. It’s usually better to leave constants in exact form.
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