Textbook Question
A second-order equation Consider the equation
t² y′′(t) + 2ty′(t) − 12y(t) = 0
b. Assuming the general solution of the equation is
y(t) = C₁ tᵖ¹ + C₂ tᵖ²,
find the solution that satisfies the conditions
y(1) = 0, y′(1) = 7
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Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 9.R.16
11–18. Solving initial value problems Use the method of your choice to find the solution of the following initial value problems.
y′(x) = 4x csc y, y(0) = π/2
Verified step by step guidance1
Identify the type of differential equation given: \( y'(x) = 4x \csc y \). Notice that it is separable because the right-hand side can be expressed as a product of a function of \( x \) and a function of \( y \).
Rewrite the differential equation in separable form by expressing \( y' = \frac{dy}{dx} \) and rearranging terms to isolate \( y \) and \( x \) on opposite sides: \( \sin y \, dy = 4x \, dx \).
Integrate both sides: compute \( \int \sin y \, dy \) on the left and \( \int 4x \, dx \) on the right. Remember to include the constant of integration after integrating.
Apply the initial condition \( y(0) = \frac{\pi}{2} \) to solve for the constant of integration. Substitute \( x = 0 \) and \( y = \frac{\pi}{2} \) into the integrated equation.
Rewrite the implicit solution to express \( y \) explicitly in terms of \( x \) if possible, or leave it in implicit form as the solution to the initial value problem.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Separable Differential Equations
A separable differential equation can be written as a product of a function of x and a function of y, allowing variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the general solution.
Recommended video:
06:06
Solving Separable Differential Equations
Initial Value Problems (IVP)
An initial value problem specifies a differential equation along with a condition that the solution must satisfy at a particular point. This condition helps determine the unique solution by solving for the constant of integration after finding the general solution.
Recommended video:
05:03Initial Value Problems
Integration of Trigonometric Functions
Solving the given differential equation involves integrating expressions containing trigonometric functions like cosecant (csc y). Understanding how to integrate these functions, often using identities or substitution, is essential to find the explicit solution.
Recommended video:
6:04Introduction to Trigonometric Functions
Related Practice
Textbook Question
Logistic growth in India The population of India was 435 million in 1960 (t=0) and 487 million in 1965 (t=5). The projected population for 2050 is 1.57 billion.
e. Discuss some possible shortcomings of this model. Why might the carrying capacity be either greater than or less than the value predicted by the model?
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Textbook Question
17–20. Verifying solutions of initial value problems Verify that the given function y is a solution of the initial value problem that follows it.
y(t) = 8t⁶ - 3; ty'(t) - 6y(t) = 18, y(1) = 5
Textbook Question
5–10. First-order linear equations Find the general solution of the following equations.
v'(y) − v/2 = 14
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Textbook Question
Orthogonal trajectories Use the method in Exercise 44 to find the orthogonal trajectories for the family of circles x² + y² = a²
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Textbook Question
Logistic growth The population of a rabbit community is governed by the initial value problem
P′(t) = 0.2 P (1 − P/1200), P(0) = 50
d. What is the population when the growth rate is a maximum?