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Solve the following differential equation with initial conditions: y"e-2t+10e4t; y(0) = 1, y'(0) = 0. The solution is: y=e4t+e-21-3t+ A. B. y=e-2t + e4f-1 C. y=-2t+4-2t+ - -He-2t + het-2 D. E. =4e-2t +16e4t-6t+19

Solve the following differential equation with initial conditions: y"e-2t+10e4t; y(0) = 1, y'(0) = 0. The solution is: y=e4t+e-21-3t+ A. B. y=e-2t + e4f-1 C. y=-2t+4-2t+ - -He-2t + het-2 D. E. =4e-2t +16e4t-6t+19

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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### Differential Equations Problem **Problem Statement:** Solve the following differential equation with initial conditions: \[ y'' = e^{-2t} + 10e^{4t}, \; y(0) = 1, \; y'(0) = 0. \] The solution is: **Options:** - **A.** \[ y = \frac{5}{8}e^{4t} + \frac{1}{4}e^{-2t} - 3t + \frac{1}{8} \] - **B.** \[ y = e^{-2t} + e^{4t} - 1 \] - **C.** \[ y = \frac{1}{4}e^{-2t} + \frac{5}{8} e^{4t} - 2t + \frac{1}{8} \] - **D.** \[ y = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2 \] - **E.** \[ y = 4e^{-2t} + 16e^{4t} - 6t + 19 \] **Explanation:** To solve this differential equation, one must use techniques related to solving second-order differential equations with constant coefficients and particular solutions for non-homogeneous differential equations. The provided options include potential solutions that satisfy both the differential equation and the given initial conditions. Students are encouraged to: 1. Verify the general solution of the associated homogeneous equation. 2. Find the particular solution that fits the non-homogeneous part. 3. Apply the initial conditions to determine the constants in the solution.
Transcribed Image Text:### Differential Equations Problem **Problem Statement:** Solve the following differential equation with initial conditions: \[ y'' = e^{-2t} + 10e^{4t}, \; y(0) = 1, \; y'(0) = 0. \] The solution is: **Options:** - **A.** \[ y = \frac{5}{8}e^{4t} + \frac{1}{4}e^{-2t} - 3t + \frac{1}{8} \] - **B.** \[ y = e^{-2t} + e^{4t} - 1 \] - **C.** \[ y = \frac{1}{4}e^{-2t} + \frac{5}{8} e^{4t} - 2t + \frac{1}{8} \] - **D.** \[ y = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2 \] - **E.** \[ y = 4e^{-2t} + 16e^{4t} - 6t + 19 \] **Explanation:** To solve this differential equation, one must use techniques related to solving second-order differential equations with constant coefficients and particular solutions for non-homogeneous differential equations. The provided options include potential solutions that satisfy both the differential equation and the given initial conditions. Students are encouraged to: 1. Verify the general solution of the associated homogeneous equation. 2. Find the particular solution that fits the non-homogeneous part. 3. Apply the initial conditions to determine the constants in the solution.
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