# Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

## Section1.3Variables

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:
$$\frac{dy}{dx} + 2y = 3x^2.\tag{1}$$
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:
1. The derivative is being taken with respect to $$x\text{,}$$ and
2. $$y$$ 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.

### Definition4.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.

### Example5.

$$\ \$$Identify the dependent and independent variables in each equation and state which variable represents the unknown.
 $$\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.

#### 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.