#Modeling with Differential Equations

#Table of Contents

What is a Differential Equation?
What is a First Order Differential Equation?
Why are Differential Equations Useful for Modeling?
General and Particular Solutions to Differential Equations
Verifying Solutions to Differential Equations
Practice Questions
Glossary
Summary and Key Takeaways

#What is a Differential Equation?

A differential equation is an equation that involves derivatives of a function. It expresses the relationship between a function and its derivatives.

For example,

\frac{dy}{dx}=12x{y}^{2}

is a differential equation.

Another example is $\frac{d^2x}{dt^2} - 5\frac{dx}{dt} + 7x = 5\sin(t).$

A differential equation includes both variables and the rates of change of those variables.

#What is a First Order Differential Equation?

A first order differential equation is a differential equation that contains only first derivatives and no higher-order derivatives.

For example,

\frac{dy}{dx}=12x{y}^{2}

is a first order differential equation.

However, $\frac{d^2x}{dt^2} - 5\frac{dx}{dt} + 7x = 5\sin(t)$ is not a first order differential equation because it contains the second derivative $\frac{d^2x}{dt^2}$ .

#Why are Differential Equations Useful for Modeling?

Many real-world problems involve rates of change. Differential equations help us model these problems mathematically.

Some examples include: - The rate of change of a population of animals in a geographic area - The rate of change of the number of people infected by a particular disease - The rate of change of the amount of medication in a person's bloodstream at different times after ingestion - The rate of change of velocity for a falling object - The rate of change of voltage across a component in an electrical circuit

If we can express the relationship between quantities and their rates of change as a differential equation, we can solve the equation to predict the behavior of these quantities in the real world.

#General and Particular Solutions to Differential Equations

#What is the Difference Between General and Particular Solutions for a Differential Equation?

The general solution to a differential equation represents all possible solutions.

For example,

y^2 = \frac{C - x^2}{4}

is the general solution to

\frac{dy}{dx} = -\frac{x}{4y}.

Here, $C$ is an arbitrary constant. The general solution is an infinite family of solutions, each corresponding to a different value of $C$ . The graph of the solution will change depending on the value of $C$ .

The particular solution to a differential equation is a specific solution that satisfies the equation under a particular set of conditions.

For instance, if we know that

y=1

when

x=0

, then

y^2 = \frac{4 - x^2}{4}

is the *only* solution that satisfies the differential equation with that set of conditions.

A condition like "y=1 when x=0" is known as an initial condition (or boundary condition). Finding a particular solution requires knowing an initial condition.

#Verifying Solutions to Differential Equations

#How Can I Use Differentiation to Verify Solutions for a Differential Equation?

To verify that a function is a solution to a differential equation, we differentiate the proposed solution and see if it matches the original differential equation.

Verify that

y = 1 - x - \cos(x)

is a solution to the differential equation

\frac{dy}{dx} = \sin(x) - 1.

Differentiate the proposed solution with respect to $x$ : $\frac{dy}{dx} = \frac{d}{dx}\left(1 - x - \cos(x)\right) = -1 + \sin(x) = \sin(x) - 1.$

This matches the original differential equation, so the solution has been verified.

For more complicated answers, additional techniques such as implicit differentiation and/or substitution may be required.

#Worked Example

Verify that $y^2 = \frac{1}{x^2 + C}$ is a solution to the differential equation $\frac{dy}{dx} = -xy^3.$

Answer:

Start by differentiating both sides of the proposed solution with respect to $x$ using implicit differentiation: $\begin{array}{rcl} y^2 &=& \left(x^2 + C\right)^{-1} \\ \frac{d}{dx}\left(y^2\right) &=& \frac{d}{dx}\left(\left(x^2 + C\right)^{-1}\right) \\ 2y\frac{dy}{dx} &=& -\left(x^2 + C\right)^{-2} \cdot 2x \\ 2y\frac{dy}{dx} &=& -\frac{2x}{\left(x^2 + C\right)^2} \end{array}$

Rearrange to make $\frac{dy}{dx}$ the subject: $\frac{dy}{dx} = -\frac{x}{y \left(x^2 + C\right)^2}.$

This may not look like $\frac{dy}{dx} = -xy^3$ , but remember that $y^2 = \frac{1}{x^2 + C}$ . Therefore: $\begin{array}{rcl} \frac{dy}{dx} &=& -\frac{x}{y} \cdot \frac{1}{\left(x^2 + C\right)^2} \\ &=& -\frac{x}{y} \cdot \left(\frac{1}{x^2 + C}\right)^2 \\ &=& -\frac{x}{y} \cdot \left(y^2\right)^2 \\ &=& -\frac{x}{y} \cdot y^4 \\ &=& -xy^3. \end{array}$

This confirms that $y^2 = \frac{1}{x^2 + C}$ is a solution to the differential equation $\frac{dy}{dx} = -xy^3.$

#Practice Questions

Practice Question

Question 1: Determine if the following function is a solution to the given differential equation. Function: $y = e^{2x}$ Differential Equation: $\frac{dy}{dx} = 2y$

Solution: Differentiate $y = e^{2x}$ : $\frac{dy}{dx} = 2e^{2x}.$ Since $y = e^{2x},$ we have: $\frac{dy}{dx} = 2y.$ Thus, the function $y = e^{2x}$ is a solution to the differential equation $\frac{dy}{dx} = 2y.$

Practice Question

Question 2: Solve the first order differential equation: $\frac{dy}{dx} = 3y.$

Solution: The general solution is obtained by separating variables: $\frac{1}{y} \frac{dy}{dx} = 3$ $\int \frac{1}{y} dy = \int 3 dx$ $\ln|y| = 3x + C$ $y = e^{3x + C} = Ae^{3x},$ where $A = e^C$ is an arbitrary constant.

Practice Question

Question 3: Verify that $y = \sin(x) + C$ is a solution to the differential equation $\frac{dy}{dx} = \cos(x).$

Solution: Differentiate $y = \sin(x) + C$ : $\frac{dy}{dx} = \cos(x).$ This matches the given differential equation, so $y = \sin(x) + C$ is a solution.

#Glossary

Differential Equation: An equation involving derivatives of a function.
First Order Differential Equation: A differential equation involving only first derivatives.
General Solution: The set of all possible solutions to a differential equation.
Particular Solution: A specific solution that satisfies the differential equation under given conditions.
Initial Condition: A condition that specifies the value of the function (or its derivatives) at a particular point.

#Summary and Key Takeaways

Summary:

Differential equations involve derivatives and are used to model real-world problems involving rates of change.
First order differential equations contain only first derivatives.
General solutions represent all possible solutions, while particular solutions satisfy specific initial conditions.
Verifying solutions involves differentiating the proposed solution and checking if it matches the original differential equation.

Key Takeaways:

Understanding the basic concepts of differential equations is crucial for solving real-world problems.
The ability to distinguish between general and particular solutions is important for finding specific solutions.
Verifying solutions through differentiation ensures the accuracy of your work.

Exam Tip

Remember to always check your solutions by differentiating and substituting back into the original differential equation.

Differential equations are a powerful tool in various fields like physics, biology, and engineering.