A continuous-time system is described by the differential equationdy(t)+2y(t)%3Du(t)dtwith y(0)-1. The output y(t) of the system is given byonA. y(t) = 0.25(1+3e-4t)u(t)B. y(t) = te-2t u(t)C. 24 + 1)u(t)-2ty(t)%3D 0.5(eD. y(t) = 0. 5(e-0.5t + e2 )u(t)

Find the output response of the given differential equation. Initial conditions are given.

A continuous-time system is described by the differential equation dy(t) + 2y(t) = u(t) with y(0)-1. The output y(t) of the system is given by A. y(t) = 0.25(1+3e-t)u(t) B. y(t) = te-2t u(t) C. y(t) = 0.5(e 2t +1)u(t) D. y(t) = 0. 5(e-0,5t + e-2t )u(t) %3D dt

A continuous-time system is described by the differential equation dy(t) + 2y(t) = u(t) with y(0)-1. The output y(t) of the system is given by A. y(t) = 0.25(1+3e-t)u(t) B. y(t) = te-2t u(t) C. y(t) = 0.5(e 2t +1)u(t) D. y(t) = 0. 5(e-0,5t + e-2t )u(t) %3D dt

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.16P

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Question

A continuous-time system is described by the differential equation
dy(t)
+2y(t)%3Du(t)
dt
with y(0)-1. The output y(t) of the system is given by
on
A. y(t) = 0.25(1+3e-4t)u(t)
B. y(t) = te-2t u(t)
C. 24 + 1)u(t)
-2t
y(t)%3D 0.5(e
D. y(t) = 0. 5(e-0.5t + e2 )u(t)

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