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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 7.4.15

Solve the differential equation in Exercises 9–22.
15. √x (dy/dx) = e^(y+√x), x > 0

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Rewrite the given differential equation to isolate \( \frac{dy}{dx} \): \[ \sqrt{x} \frac{dy}{dx} = e^{y + \sqrt{x}} \implies \frac{dy}{dx} = \frac{e^{y + \sqrt{x}}}{\sqrt{x}} \].
Express the right-hand side as a product of exponentials to separate variables more easily: \[ \frac{dy}{dx} = \frac{e^y \cdot e^{\sqrt{x}}}{\sqrt{x}}. \]
Rewrite the equation in differential form to separate variables \( y \) and \( x \): \[ \frac{dy}{e^y} = \frac{e^{\sqrt{x}}}{\sqrt{x}} dx. \]
Integrate both sides: \[ \int \frac{1}{e^y} dy = \int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx. \] The left integral simplifies to \( \int e^{-y} dy \).
For the right integral, use substitution: let \( t = \sqrt{x} \), so \( x = t^2 \) and \( dx = 2t dt \). Substitute and simplify the integral accordingly before integrating.

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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 so that all terms involving y are on one side and all terms involving x are on the other. This allows integration of each side independently. Recognizing separability is key to solving equations like √x (dy/dx) = e^(y+√x).
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Integration Techniques

Solving separable equations requires integrating functions of x and y separately. Familiarity with integrating exponential functions and substitutions, such as u = √x, helps simplify the integrals and find the general solution.
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Implicit and Explicit Solutions

After integration, solutions may be implicit or explicit. Understanding how to manipulate and interpret these forms is important, especially when the solution involves expressions like e^(y+√x), which may require logarithmic manipulation to isolate y.
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