Numerade

Calculus: Early Transcendentals

James Stewart

ISBN #9781285741550

8th Edition

6,422 Questions

2,819,387 Students Helped

Summary

This section explored second-order linear homogeneous differential equations, emphasizing constant coefficient cases. The key method involves assuming an exponential solution which leads to the auxiliary (characteristic) equation. Based on the discriminant, solutions are classified into three cases: real and distinct, real and repeated, and complex. The principle of superposition guarantees that any linear combination of two independent solutions is a solution, enabling the construction of the general solution. Applications include analysis of mechanical oscillations and electric circuits.

Learning Objectives

Define and classify second?order linear homogeneous differential equations with constant coefficients.

Explain the concept of linear independence of solutions and the principle of superposition in constructing the general solution.

Develop and solve the auxiliary (characteristic) equation to determine the structure of the solution based on the discriminant.

Apply methods for solving initial-value and boundary-value problems for second?order linear differential equations.

Key Concepts

CONCEPT

DEFINITION

Second-Order Linear Differential Equation

An equation of the form P(x)y'' + Q(x)y' + R(x)y = G(x). When G(x)=0, the equation is homogeneous.

Homogeneous Equation

A differential equation in which the function G(x) is zero for all x, leading to P(x)y'' + Q(x)y' + R(x)y = 0.

Linear Combination

The sum c1*y1(x) + c2*y2(x) of two solutions y1 and y2 of a homogeneous linear differential equation, which is also a solution.

Linearly Independent Solutions

Two functions y1 and y2 are linearly independent if neither is a constant multiple of the other, ensuring they form a basis for the solution space.

Auxiliary (Characteristic) Equation

An algebraic equation formed by substituting y = e^(rx) into the differential equation; for constant coefficients, it typically has the form ar^2 + br + c = 0.

Discriminant

The expression b^2 - 4ac from the auxiliary equation, which determines whether the roots are real and distinct, repeated, or complex.

Step-by-Step Explanations

QUESTION

How do you solve a second-order homogeneous differential equation with constant coefficients?

STEP-BY-STEP ANSWER:

Step 1: Write the differential equation in standard form: ay'' + by' + cy = 0.
Step 2: Assume a solution of the form y = e^(rx) and substitute into the equation to get the auxiliary equation: ar^2 + br + c = 0.
Step 3: Solve the auxiliary (quadratic) equation for r using factoring or the quadratic formula, r = (-b ± √(b^2 - 4ac)) / (2a).
Step 4: Classify the roots:
Step 5: (If needed) Use initial or boundary conditions to determine the constants c1 and c2.
Final Answer: The general solution is expressed in one of the forms listed above depending on the nature of the roots.

Solving a Constant-Coefficient Homogeneous Equation

QUESTION

Why is the linear combination of two solutions also a solution?

STEP-BY-STEP ANSWER:

Step 1: Notice that the differential operator L[y] = P(x)y'' + Q(x)y' + R(x)y is linear.
Step 2: If y1 and y2 are solutions, then L[y1] = 0 and L[y2] = 0.
Step 3: For any constants c1 and c2, L[c1*y1 + c2*y2] = c1L[y1] + c2L[y2] = 0 + 0 = 0.
Final Answer: The linear combination y = c1*y1 + c2*y2 is also a solution by the linearity of the differential operator.

Verifying Superposition Principle

Common Mistakes

Confusing homogeneous equations with nonhomogeneous ones by neglecting the G(x) term.

Failing to check the linear independence of the two solutions; for instance, using two similar exponential functions which are multiples of each other.

Incorrect computation of the discriminant, leading to misclassification of the roots (real vs. complex).

Omitting the x factor in the solution when dealing with repeated real roots.

Calculus: Early Transcendentals

Summary

Example 1

Example 2

Example 3

Example 4

Example 5

Common Mistakes