Section 1.3 Variables
In this section, we’ll explore the two types of variables found in differential equations: dependent and independent variables. Understanding the roles of these variables is crucial when solving and interpreting differential equations.
Consider the differential equation:
\begin{equation}
\frac{dy}{dx} + 2y = 3x^2.\tag{1}
\end{equation}
This equation involves two variables: \(x\) and \(y\text{.}\) One of these variables is the unknown that we are trying to solve for, but which one?
Typically, the variable you want to solve for is the one with a derivative applied to it. Here, the term
\(\frac{dy}{dx}\) in
(1) indicates:
The derivative is being taken with respect to \(x\text{,}\) and
\(y\) is a function of \(x\) (since it wouldn’t make sense to take the derivative of \(y\) with respect to \(x\) unless \(y\) depends on \(x\)).
Therefore, \(y\) is called the dependent variable because it depends on \(x\text{.}\) On the other hand, \(x\) is the independent variable.
Definition 4. Dependent & Independent Variables.
 Dependent Variable:
The variable representing the unknown function we are trying to find. It is always the variable with a derivative applied to it.
 Independent Variable:
The variable that the dependent variable is a function of. It does not have a derivative applied to it.
Example 5.
\(\ \ \)Identify the dependent and independent variables in each equation and state which variable represents the unknown.
\(\ds \frac{dP}{ds} + \frac{P}{s^2} = 17s\) 
\(\hspace{27em}\) Solution.Since the derivative \(\ds\frac{dP}{ds}\) appears, it implies that:
\(P\) is a function of \(s\) and is the dependent variable.
\(s\) is the independent variable.
\(P(s)\) represents the unknown to this differential equation.

\(\ds u'' + t^2 \ u = 0\) 
\(\hspace{27em}\) Solution.
The second derivative \(u''\) suggests that:
\(u\) is a function of \(t\) and is the dependent variable.
\(t\) is the independent variable.
\(u(t)\) represents the unknown to this differential equation.

\(\ds Q'' = 11Q\) 
\(\hspace{27em}\) Solution.Here, \(Q''\) implies that \(Q\) is the dependent variable. Since no independent variable is explicitly shown, typically the context of the problem would provide it. If not, we can assume it ourselves—perhaps \(t\text{,}\) making \(Q(t)\) the unknown.

Reading Questions Check your Understanding
1. Which variable in \(\ds (1  x)y''  4xy' + 5y = \cos x \) is the independent variable?
Which variable in \(\ds (1  x)y''  4xy' + 5y = \cos x \) is the independent variable?
 \(x\)
Yes! \(x\) is the independent variable.
 \(y\)
Incorrect. Review the examples.
 \(y'\)
Incorrect. Review the examples.
2. Which variable in \(\ds \frac{dy}{dx} + 2y = 3x^2 \) is the dependent variable?
Which variable in \(\ds \frac{dy}{dx} + 2y = 3x^2 \) is the dependent variable?
 \(\ds\frac{dy}{dx}\)
Incorrect. \(\frac{dy}{dx}\) represents the derivative of the dependent variable with respect to the independent variable.
 \(x\)
 Incorrect. The dependent variable is the one being differentiated.
 \(y\)
 Correct! \(y\) is the dependent variable in this equation.
3. In a differential equation, the dependent variable has a derivative applied to it.
In a differential equation, the dependent variable has a derivative applied to it
 True
Correct! The dependent variable in a differential equation always has a derivative applied to it.
 False
Incorrect. By definition, a differential equation involves derivatives of the dependent variable.
4. Which variable in \(\ds \frac{dP}{ds} + \frac{P}{s^2} = 17s \) are we solving for?
Which variable in \(\ds \frac{dP}{ds} + \frac{P}{s^2} = 17s \) are we solving for?
 dependent variable, \(s\)
Incorrect. \(s\) is neither the dependent variable, nor what we are solving for.
 independent variable, \(s\)
Incorrect! \(s\) is the independent variable, but it is not what we are solving for.
 dependent variable, \(P\)
Yes! We are solving for the unknown, \(P\text{.}\)
 independent variable, \(P\)
Incorrect. We are solving for \(P\text{,}\) but it is not the independent variable.
5. In the differential equation \(\ds \frac{dx}{dy} = 2xy + 1\text{,}\) the dependent variable is .
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