Textbook Question
In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = xe^x-e^x
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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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 7.4.10
Solve the differential equation in Exercises 9–22.
10. (dy/dx) = x²√y, y > 0
Verified step by step guidance1
Recognize that the given differential equation is separable. The equation is \(\frac{dy}{dx} = x^{2} \sqrt{y}\), where \(y > 0\).
Rewrite the equation to separate variables \(y\) and \(x\). Express it as \(\frac{dy}{\sqrt{y}} = x^{2} dx\).
Integrate both sides: integrate \(\frac{1}{\sqrt{y}} dy\) on the left and \(x^{2} dx\) on the right.
After integrating, you will get an expression involving \(y\) and \(x\) plus an integration constant \(C\). Write this implicit solution.
Finally, solve the implicit equation for \(y\) if possible, to express \(y\) explicitly as a function of \(x\).
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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 solution.
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Integration of Power Functions
Integrating power functions involves applying the power rule for integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C for n ≠ -1. This is essential when integrating terms like x² or y raised to a fractional power.
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Initial Conditions and Domain Restrictions
The condition y > 0 restricts the solution to positive values of y, which affects the domain and the form of the solution. Understanding initial or domain conditions ensures the solution is valid and meaningful within the given constraints.
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05:03Initial Value Problems
Related Practice
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Textbook Question
Which of the functions graphed in Exercises 1–6 are one-to-one, and which are not?
Textbook Question
135. Find the area of the “triangular” region in the first quadrant that is bounded above by the curve y = e^(2x), below by the curve y = e^x, and on the right by the line x = ln(3).