Book cover for Calculus

Calculus

Deborah Hughes-Hallett, Andrew M. Gleason, William G. McCallum

ISBN #9781119379331

7th Edition

5,101 Questions

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33,688 Students Helped

Homework Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This section introduces the fundamental concepts of differential equations, including what constitutes a solution and how general solutions containing arbitrary constants are obtained through antiderivatives. It also covers the method of verifying a solution by substitution and demonstrates the use of initial conditions to arrive at a particular solution. Additionally, the distinctions between first- and second-order differential equations are clarified, laying the groundwork for more complex applications in various fields.

Learning Objectives

1

Define what a differential equation is and understand its components such as derivatives and unknown functions.

2

Explain the concept of a general solution and how arbitrary constants appear after integration.

3

Demonstrate how to verify a proposed solution by substituting it into the differential equation.

4

Describe the process of using initial conditions to determine a particular solution from the general solution.

5

Differentiate between first- and second-order differential equations and understand their implications in solution methods.

Key Concepts

CONCEPT

DEFINITION

Differential Equation

An equation that relates a function with one or more of its derivatives.

Solution

A function that, when substituted into the differential equation, satisfies the equation.

General Solution

A family of solutions that incorporates one or more arbitrary constants resulting from the integration process.

Particular Solution

A specific solution obtained by assigning particular values to the arbitrary constants, usually derived from given initial conditions.

Initial Value Problem (IVP)

A differential equation combined with one or more conditions specifying the value of the solution at a given point.

Equilibrium Solution

A constant solution where the derivative is zero, indicating no change.

First-Order Differential Equation

A differential equation that involves the first derivative of the unknown function and no higher derivatives.

Second-Order Differential Equation

A differential equation that involves the second derivative of the unknown function and no higher derivatives.

Example Problems

Example 1

Is $y=x^{3}$ a solution to the differential equation. $$ x y^{\prime}-3 y=0 ? $$

Example 2

In Exercises $2-7$, which differential equation, ( 1 )-(VI), has the function as a solution? I. $y^{\prime}=-2 x y$ II. $y^{\prime}=-x y$ III. $y^{\prime}=x y$ IV. $y^{\prime}=-x^{2} y$ V. $y^{\prime}=x^{-3} y$ VI. $y^{\prime}=2 x y$ $$y=e^{-x^{2}}$$

Example 3

In Exercises $2-7$, which differential equation, ( 1 )-(VI), has the function as a solution? I. $y^{\prime}=-2 x y$ II. $y^{\prime}=-x y$ III. $y^{\prime}=x y$ IV. $y^{\prime}=-x^{2} y$ V. $y^{\prime}=x^{-3} y$ VI. $y^{\prime}=2 x y$ $$y=e^{-0.5 x^{2}}$$

Example 4

In Exercises $2-7$, which differential equation, ( 1 )-(VI), has the function as a solution? I. $y^{\prime}=-2 x y$ II. $y^{\prime}=-x y$ III. $y^{\prime}=x y$ IV. $y^{\prime}=-x^{2} y$ V. $y^{\prime}=x^{-3} y$ VI. $y^{\prime}=2 x y$ $$y=0.5 e^{-x^{2}}$$

Example 5

In Exercises $2-7$, which differential equation, ( 1 )-(VI), has the function as a solution? I. $y^{\prime}=-2 x y$ II. $y^{\prime}=-x y$ III. $y^{\prime}=x y$ IV. $y^{\prime}=-x^{2} y$ V. $y^{\prime}=x^{-3} y$ VI. $y^{\prime}=2 x y$ $$y=0.5 e^{x^{2}}$$
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Step-by-Step Explanations

QUESTION

Show that y = 100 + C·e^(–t) is a solution to the differential equation dy/dt = 100 – y.

STEP-BY-STEP ANSWER:

Step 1: Differentiate y = 100 + C·e^(–t) with respect to t. The derivative is dy/dt = –C·e^(–t).
Step 2: Substitute y into the right-hand side of the differential equation: 100 – (100 + C·e^(–t)) = –C·e^(–t).
Step 3: Compare the left-hand side (–C·e^(–t)) with the right-hand side (–C·e^(–t)). Since they match, the proposed function is indeed a solution.
Final Answer: y = 100 + C·e^(–t) satisfies the differential equation.

Verification of a Candidate Solution

QUESTION

Determine the particular solution given the initial condition y(0) = 0 for the general solution y = 100 + C·e^(–t).

STEP-BY-STEP ANSWER:

Step 1: Substitute t = 0 into the general solution: y(0) = 100 + C·e^(0) = 100 + C.
Step 2: Set the equation equal to the initial condition: 100 + C = 0.
Step 3: Solve for C: C = –100.
Final Answer: The particular solution is y = 100 – 100·e^(–t).

Finding the Particular Solution

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Common Mistakes

  • Failing to include or correctly determine the arbitrary constant(s) when integrating to obtain the general solution.
  • Misinterpreting the process of verifying a solution by not properly substituting the proposed solution back into the differential equation.
  • Confusing the roles of the general solution and the particular solution, especially in the context of initial conditions.
  • Overlooking the significance of the order of the differential equation and the corresponding number of integration constants needed.