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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.PE.125

In Exercises 125–128 solve the differential equation.
125. dy/dx = √y cos(√y)

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Rewrite the differential equation \( \frac{dy}{dx} = \sqrt{y} \cos(\sqrt{y}) \) to separate variables. Express it as \( \frac{dy}{\sqrt{y} \cos(\sqrt{y})} = dx \).
Make the substitution \( u = \sqrt{y} \), which implies \( y = u^2 \) and \( dy = 2u \, du \).
Rewrite the left side integral in terms of \( u \): \( \frac{dy}{\sqrt{y} \cos(\sqrt{y})} = \frac{2u \, du}{u \cos(u)} = \frac{2 \, du}{\cos(u)} \).
Set up the integral \( \int \frac{2}{\cos(u)} \, du = \int dx \), which simplifies to \( 2 \int \sec(u) \, du = x + C \), where \( C \) is the constant of integration.
Integrate \( \sec(u) \) using the standard formula \( \int \sec(u) \, du = \ln | \sec(u) + \tan(u) | + C \), then substitute back \( u = \sqrt{y} \) to express the solution implicitly.

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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 y and a function of x, 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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Substitution Method

Substitution involves introducing a new variable to simplify the differential equation, especially when the equation contains composite functions like √y. For example, letting u = √y can transform the equation into a more manageable form for integration.
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Integration Techniques

Solving the separated equation requires integrating functions involving trigonometric and algebraic expressions. Familiarity with integrating functions like cos(u) and handling integrals after substitution is essential to find the explicit or implicit solution.
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