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Solve the given equation.. dy dx a. + 2xy = x, y(0) = -2 C. y(x)= xex-2e-x² ex-5e-x² 2 b. y(x) = 1/2 y(x)=3-5e-x² 1-5e-x² y(x) = d. e. y(x) = 1-5ex

Solve the given equation.. dy dx a. + 2xy = x, y(0) = -2 C. y(x)= xex-2e-x² ex-5e-x² 2 b. y(x) = 1/2 y(x)=3-5e-x² 1-5e-x² y(x) = d. e. y(x) = 1-5ex

Calculus: Early Transcendentals
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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**Problem Statement:** Solve the given differential equation: \[ \frac{dy}{dx} + 2xy = x, \quad y(0) = -2 \] **Options:** a. \( y(x) = xe^x - 2e^{-x^2} \) b. \( y(x) = \frac{e^x - 5e^{-x^2}}{2} \) c. \( y(x) = \frac{1 - 5e^x}{2} \) d. \( y(x) = 3 - 5e^{-x^2} \) e. \( y(x) = \frac{1 - 5e^{-x^2}}{2} \) **Note:** - Ensure to check the solution with the initial condition \( y(0) = -2 \). - Optionally, substitute the function back into the differential equation to verify if it satisfies both the equation and the initial condition.
Transcribed Image Text:**Problem Statement:** Solve the given differential equation: \[ \frac{dy}{dx} + 2xy = x, \quad y(0) = -2 \] **Options:** a. \( y(x) = xe^x - 2e^{-x^2} \) b. \( y(x) = \frac{e^x - 5e^{-x^2}}{2} \) c. \( y(x) = \frac{1 - 5e^x}{2} \) d. \( y(x) = 3 - 5e^{-x^2} \) e. \( y(x) = \frac{1 - 5e^{-x^2}}{2} \) **Note:** - Ensure to check the solution with the initial condition \( y(0) = -2 \). - Optionally, substitute the function back into the differential equation to verify if it satisfies both the equation and the initial condition.
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The given equation is;

dy/dx+2xy=x, y(0)=-2.

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