Section 4.3 Separation of Variables Method (SOV)
We are now going to discuss the steps involved in the
separation of variables method. In order to use this method, the differential equation must be first-order and be
separable.
Identifying a differential equation as first-order is as simple as observing the presence of a first derivative and no higher derivatives. In comparison, determining that an equation is separable is not always immediately apparent. Sometimes, it requires some algebra to rearrange the equation into a form where the method can be applied.
Method 2. Separation of Variables (SOV).
If a differential equation can be written in the separable form
\begin{equation*}
\frac{dy}{dx} = f(x) \cdot g(y)\text{,}
\end{equation*}
then the general solution can be determined through the steps below.
- Step 1: Separate
\(\displaystyle \ds \frac{1}{g(y)}\ dy = f(x)\ dx\)
- Step 2: Integrate
\(\displaystyle \ds \int \frac{1}{g(y)}\ dy = \int f(x) \ dx\)
- Step 3: Isolate
Isolate \(y\) (if possible), and combine constants.
These problems serve as warm-up examples to familiarize you with the separation of variables method. You won’t encounter any tricky algebra or calculus steps, allowing you to focus solely on the general steps needed to apply this technique. However, pay close attention to the new concept of "combining constants", as it will appear frequently throughout this course.
Example 37.
\(\ \ \)Find the general solution to the following differential equations using the separation of variables method.
\(\ds \frac{dy}{dx} = x^2 \) Solution.
You actually don’t need separation of variables to solve for \(y\text{.}\) In fact, you might recognize that finding \(y\) is the same as finding the anti-derivative of \(x^2\text{,}\) which we know is \(y = \frac13x^3 + c\text{.}\) Nevertheless, let’s apply separation of variables to make sure we get the same answer.
The differential equation has order 1 and is separable since
\begin{equation*}
\ds \frac{dy}{dx} = \underset{f(x)}{\underset{\uparrow}{\big( x^2 \big)}} \cdot \underset{g(y)}{\underset{\uparrow}{\big( 1 \big)}}
\end{equation*}
Step 1: Separate. We use algebra to move all the
\(y\) terms to the left and all the
\(x\) terms to the right, giving
\begin{equation*}
\ds \frac{1}{\underset{g(y)}{\underset{\uparrow}{ 1 }}} \frac{dy}{dx} = \underset{f(x)}{\underset{\uparrow}{ x^2 }}
\end{equation*}
\begin{align*}
\int 1 \underset{=\ dy}{\underbrace{\frac{dy}{dx}dx}} \amp = \int x^2\ dx\\
\int 1\ dy \amp = \int x^2\ dx\\
y + c_1 \amp = \frac13 x^3 + c_2.
\end{align*}
\begin{align*}
y \amp = \frac13 x^3 + c_2 - c_1
\end{align*}
Since \(c_2 - c_1 \) is still a constant, we combine it into a single constant, \(c\text{,}\) and write our solution as
\begin{equation*}
y = \frac13 x^3 + c
\end{equation*}
\(\ds \frac{dy}{dx} = y^2 \) Solution.
The differential equation has order 1 and is separable since
\begin{equation*}
\frac{dy}{dx} = \underset{f(x)}{\underset{\uparrow}{\big( 1 \big)}} \cdot \underset{g(y)}{\underset{\uparrow}{\big( y^2 \big)}}
\end{equation*}
Step 1: Separate. We use algebra to move all the
\(y\) terms to the left and all the
\(x\) terms to the right, giving
\begin{equation*}
\frac{1}{\underset{g(y)}{\underset{\uparrow}{ y^2 }}} \frac{dy}{dx} = \underset{f(x)}{\underset{\uparrow}{ 1 }}
\end{equation*}
\begin{align*}
\int 1\frac{1}{y^2} \underset{=\ dy}{\underbrace{\frac{dy}{dx}dx}} \amp = \int 1\ dx\\
\int y^{-2}\ dy \amp = \int 1\ dx\\
\frac{y^{-1}}{-1} + c_1 \amp = x + c_2.
\end{align*}
\begin{align*}
-\frac{1}{y} \amp = x + c_2 - c_1 \\
-\frac{1}{y} \amp = x - (c_2 - c_1) \\
\frac{1}{y} \amp = -x + c \quad \leftarrow \text{combined constant} \\
y \amp = \frac{1}{-x + c}
\end{align*}
\(\ds \frac{dy}{dx} + \frac{x}{y^2} = 0 \) Solution.
\begin{equation*}
\frac{dy}{dx} = - \frac{x}{y} = -x\left(\frac{1}{y^2}\right) \quad \leftarrow \text{separable}
\end{equation*}
\begin{equation*}
y^2\frac{dy}{dx} = -x
\end{equation*}
\begin{align*}
\int y^2 \underset{=\ dy}{\underbrace{\frac{dy}{dx}dx}} \amp = \int -x\ dx\\
\int y^2 \ dy \amp = \int -x\ dx\\
\frac{y^{3}}{3} + c_1 \amp = -\frac{x^{2}}{2} + c_2.
\end{align*}
\begin{align*}
\frac{y^{3}}{3} \amp = -\frac{x^{2}}{2} + \underset{=\ c_3}{\underbrace{c_2 - c_1}} \\
y^{3} \amp = -3\cdot\frac{x^{2}}{2} + \underset{=\ c}{\underbrace{3\cdot c_3}} \\
y^{3} \amp = -\frac32 x^2 + c \\
y \amp = \left(-\frac32 x^2 + c \right)^{1/3}
\end{align*}
These examples illustrate the fundamental steps of the separation of variables method, giving you a solid foundation for solving simple separable differential equations. As you progress, the techniques you’ve learned here will be essential for tackling more challenging problems. In the next section, we’ll explore more complex examples that require additional algebraic manipulation and deeper understanding.
Reading Questions Check your Understanding
1. Why is it important to verify if a differential equation is separable?
2. What are the two key requirements for using the separation of variables method?
3. In the differential equation \(\ds\frac{dy}{dx} + \frac{x}{y^2} = 0\text{,}\) which step is crucial for verifying that it is separable?
In the differential equation \(\frac{dy}{dx} + \frac{x}{y^2} = 0\text{,}\) which step is crucial for verifying that it is separable?
- Integrating both sides immediately.
- Incorrect. Integration should come after the equation is verified to be separable.
- Substituting a value for \(x\text{.}\)
- Incorrect. Substitution is not part of the separation process; the goal is to separate the variables first.
- Expanding the terms on both sides.
- Incorrect. Expansion is not necessary here; the focus should be on separating the variables.
- Rearranging the equation to isolate \(\frac{dy}{dx}\) and separate the variables.
- Correct! Rearranging the equation is key to verifying that it can be separated into functions of \(x\) and \(y\text{.}\)
4. Separation of Variables Method Steps.
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