Textbook Question
In Exercises 59–86, find the derivative of y with respect to the given independent variable.
65. y = (cos θ)^(√2)
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.4.17
Solve the differential equation in Exercises 9–22.
17. (dy/dx) = 2x(y - 1), y > 1
Verified step by step guidance1
Recognize that the given differential equation \( \frac{dy}{dx} = 2x(y - 1) \) 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 equation to separate variables: divide both sides by \( y - 1 \) and multiply both sides by \( dx \) to get \( \frac{1}{y - 1} dy = 2x \, dx \).
Integrate both sides: integrate \( \int \frac{1}{y - 1} dy \) on the left and \( \int 2x \, dx \) on the right.
After integration, express the result as \( \ln|y - 1| = x^2 + C \), where \( C \) is the constant of integration.
Since the problem states \( y > 1 \), you can drop the absolute value and solve for \( y \) by exponentiating both sides to get \( y - 1 = e^{x^2 + C} \), then rewrite as \( y = 1 + Ce^{x^2} \) where \( C = e^C \) is a new constant.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3mWas this helpful?
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 solution.
Recommended video:
06:06
Solving Separable Differential Equations
Integration Techniques
Solving separable equations requires integrating both sides after separation. Familiarity with basic integration rules and methods, such as substitution or recognizing standard integral forms, is essential to find the explicit solution.
Recommended video:
06:18Integration by Parts for Definite Integrals
Initial Conditions and Domain Restrictions
The problem specifies y > 1, which restricts the solution domain. Understanding how initial conditions or domain constraints affect the solution ensures the correct branch of the solution is chosen and the solution is valid within the given context.
Recommended video:
05:03Initial Value Problems
Related Practice
Textbook Question
44. Silver cooling in air The temperature of an ingot of silver is 60°C above room temperature right now. Twenty minutes ago, it was 70°C above room temperature. How far above room temperature will the silver be
b. 2 hours from now?
Textbook Question
In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = ∫(from 0 to lnx) sin(e^t) dt
Textbook Question
In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
35. y = sinh⁻¹(tan x)
Textbook Question
Evaluate the integrals in Exercises 87–96.
93. ∫₀^(π/2) 7^(cos t) sin t dt
Textbook Question
Evaluate the integrals in Exercises 53–76.
57. ∫dx/(x√(25x²-2))