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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 7.4.9

Solve the differential equation in Exercises 9–22.
9. 2√(xy) (dy/dx) = 1, x, y > 0

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Rewrite the given differential equation: \(2\sqrt{xy} \frac{dy}{dx} = 1\) with \(x, y > 0\).
Isolate \(\frac{dy}{dx}\) by dividing both sides by \(2\sqrt{xy}\): \(\frac{dy}{dx} = \frac{1}{2\sqrt{xy}}\).
Express \(\sqrt{xy}\) as \(\sqrt{x} \sqrt{y}\) to separate variables: \(\frac{dy}{dx} = \frac{1}{2 \sqrt{x} \sqrt{y}}\).
Rewrite the equation to separate variables \(y\) and \(x\): multiply both sides by \(2 \sqrt{y}\) and multiply both sides by \(dx\) to get \(2 \sqrt{y} dy = \frac{1}{\sqrt{x}} dx\).
Integrate both sides: \(\int 2 \sqrt{y} \, dy = \int \frac{1}{\sqrt{x}} \, dx\), then solve the resulting integrals to find the implicit or explicit solution for \(y\).

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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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Implicit Differentiation and Manipulation

Implicit differentiation involves differentiating both sides of an equation involving x and y without explicitly solving for y. In this problem, rewriting the equation to isolate dy/dx and expressing terms in a form suitable for separation is essential.
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Integration Techniques

Solving the separated equation requires integrating functions of x and y. Familiarity with integration methods, including substitution and handling radicals like √(xy), is necessary to find the explicit or implicit solution.
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