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[0.5(e-2t +e-4t) lo.5(e-2t-e-4t) 0.5(e-t-e-4t)1 0.5(e-2t + e-4t) Find x₂(t), t≥ 0, when x₁ (0) = 2 and x₂ (0) = 0. (a) 0.5(e-2t-e-4t), t≥ 0 (b) (e-2t-e-4t), t≥ 0 (c) 0.5(e-2t + e-*), t≥ 0 (d) (e-2t + e-4t), t≥ 0 Ф (t) = [0.5(e x₁ (0)= = 1 and x₂ (0) = 1, at what time will x₁ (t) equal 0.5? Answer is rounded to nearest hundredth. (a) t = 0.50 seconds (b) t= 0.35 seconds (c) t = 0.17 seconds (d) t = 0.00 seconds

[0.5(e-2t +e-4t) lo.5(e-2t-e-4t) 0.5(e-t-e-4t)1 0.5(e-2t + e-4t) Find x₂(t), t≥ 0, when x₁ (0) = 2 and x₂ (0) = 0. (a) 0.5(e-2t-e-4t), t≥ 0 (b) (e-2t-e-4t), t≥ 0 (c) 0.5(e-2t + e-*), t≥ 0 (d) (e-2t + e-4t), t≥ 0 Ф (t) = [0.5(e x₁ (0)= = 1 and x₂ (0) = 1, at what time will x₁ (t) equal 0.5? Answer is rounded to nearest hundredth. (a) t = 0.50 seconds (b) t= 0.35 seconds (c) t = 0.17 seconds (d) t = 0.00 seconds

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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I'm sorry, I can't transcribe the image directly. However, I can describe the structure of the content. The image contains a problem involving differential equations and initial conditions. 1. **Equations**: - \(\phi(t) = \frac{0.5(e^{-2t} + e^{-4t})}{0.5(e^{-2t} - e^{-4t})}\) - Find \(x_2(t)\) for \(t \geq 0\), when \(x_1(0) = 2\) and \(x_2(0) = 0\). 2. **Multiple Choices**: - (a) \(0.5(e^{-2t} - e^{-4t})\), \(t \geq 0\) - (b) \(0.5(e^{-2t} - e^{-t})\), \(t \geq 0\) - (c) \(0.5(e^{-2t} + e^{-t})\), \(t \geq 0\) - (d) \(e^{-2t} + e^{-4t}\), \(t \geq 0\) 3. **Question**: - Solve for \(t\) when \(x(0) = 1\) and \(x(0) = 1\) at what time will \(x(t)\) equal 0.5? Answer is rounded to the nearest hundredth. 4. **Options for Time (\(\tau\))**: - (a) \(\tau = 0.50\) seconds - (b) \(\tau = 0.35\) seconds - (c) \(\tau = 0.17\) seconds - (d) \(\tau = 0.00\) seconds This exercise involves solving differential equations with given initial conditions and identifying correct solutions from multiple choices.
Transcribed Image Text:I'm sorry, I can't transcribe the image directly. However, I can describe the structure of the content. The image contains a problem involving differential equations and initial conditions. 1. **Equations**: - \(\phi(t) = \frac{0.5(e^{-2t} + e^{-4t})}{0.5(e^{-2t} - e^{-4t})}\) - Find \(x_2(t)\) for \(t \geq 0\), when \(x_1(0) = 2\) and \(x_2(0) = 0\). 2. **Multiple Choices**: - (a) \(0.5(e^{-2t} - e^{-4t})\), \(t \geq 0\) - (b) \(0.5(e^{-2t} - e^{-t})\), \(t \geq 0\) - (c) \(0.5(e^{-2t} + e^{-t})\), \(t \geq 0\) - (d) \(e^{-2t} + e^{-4t}\), \(t \geq 0\) 3. **Question**: - Solve for \(t\) when \(x(0) = 1\) and \(x(0) = 1\) at what time will \(x(t)\) equal 0.5? Answer is rounded to the nearest hundredth. 4. **Options for Time (\(\tau\))**: - (a) \(\tau = 0.50\) seconds - (b) \(\tau = 0.35\) seconds - (c) \(\tau = 0.17\) seconds - (d) \(\tau = 0.00\) seconds This exercise involves solving differential equations with given initial conditions and identifying correct solutions from multiple choices.
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