Hello. Trying to see where I went wrong on a quiz. How do you do the problem? **Problem Statement:**Solve the initial value problem.\[\frac{dy}{dx} = \frac{1}{x^3} + x, \quad x > 0; \quad y(2) = 0\]**Multiple Choice Answers:**1. $ y = -\frac{1}{2x^2} + \frac{x^2}{2} $2. $ y = \frac{4}{x^4} + \frac{x^2}{2} - \frac{9}{4} $3. $ y = -\frac{1}{2x^2} + \frac{x^2}{2} - \frac{15}{8} $ (Correct Answer)4. $ y = -\frac{1}{2x^2} + \frac{1}{8} $ (You Answered)**Explanation:**The task is to solve the differential equation subject to the initial condition provided. Among the options provided, the correct answer is indicated as the third choice:\[ y = -\frac{1}{2x^2} + \frac{x^2}{2} - \frac{15}{8} \]This solution satisfies both the differential equation and the initial condition. The fourth option was the user's answer but was incorrect.

Answered: Solve the initial value problem. dy +…

Math

Calculus

Solve the initial value problem. dy + x, x > 0; y(2) = 0 dx y = 2x2 x² 9 O y= Answer 15 y = - 2x2 swered y= 2x2 - 00

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Hello. Trying to see where I went wrong on a quiz. How do you do the problem?

**Problem Statement:**

Solve the initial value problem.

\[
\frac{dy}{dx} = \frac{1}{x^3} + x, \quad x > 0; \quad y(2) = 0
\]

**Multiple Choice Answers:**

1. $ y = -\frac{1}{2x^2} + \frac{x^2}{2} $

2. $ y = \frac{4}{x^4} + \frac{x^2}{2} - \frac{9}{4} $

3. $ y = -\frac{1}{2x^2} + \frac{x^2}{2} - \frac{15}{8} $ (Correct Answer)

4. $ y = -\frac{1}{2x^2} + \frac{1}{8} $ (You Answered)

**Explanation:**

The task is to solve the differential equation subject to the initial condition provided. Among the options provided, the correct answer is indicated as the third choice:

\[ y = -\frac{1}{2x^2} + \frac{x^2}{2} - \frac{15}{8} \]

This solution satisfies both the differential equation and the initial condition. The fourth option was the user's answer but was incorrect.

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