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Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a delta function. y+y=6+8(t-1), 3(0) = 0. a. Find the Laplace transform of the solution. Y(s)=L(y(t)} = (6/s + e^(-s))/(s+1) b. Obtain the solution y(t). y(t)= 3+3e-t) + 3te^(-1) c. Express the solution as a piecewise-defined function and think about what happens to the graph of the solution at t = 1. -3.3e(-1) v(t)== 3e^(-1) if 0 < t < 1, if 1 ≤t < 0.

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a delta function. y+y=6+8(t-1), 3(0) = 0. a. Find the Laplace transform of the solution. Y(s)=L(y(t)} = (6/s + e^(-s))/(s+1) b. Obtain the solution y(t). y(t)= 3+3e-t) + 3te^(-1) c. Express the solution as a piecewise-defined function and think about what happens to the graph of the solution at t = 1. -3.3e(-1) v(t)== 3e^(-1) if 0 < t < 1, if 1 ≤t < 0.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a delta function. y+y=6+8(t-1), y(0) = 0. a. Find the Laplace transform of the solution. Y(s) C (y(t)) (6/s+ e^(-s))/(s+1) b. Obtain the solution y(t). y(t) 3+3e^(-1) + 3te^(-1) c. Express the solution as a piecewise-defined function and think about what happens to the graph of the solution at t = 1. -3+3e(-1) if 0 < t < 1, y(t) 3e^(-1) if 1 ≤ t < 0.
Transcribed Image Text:Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a delta function. y+y=6+8(t-1), y(0) = 0. a. Find the Laplace transform of the solution. Y(s) C (y(t)) (6/s+ e^(-s))/(s+1) b. Obtain the solution y(t). y(t) 3+3e^(-1) + 3te^(-1) c. Express the solution as a piecewise-defined function and think about what happens to the graph of the solution at t = 1. -3+3e(-1) if 0 < t < 1, y(t) 3e^(-1) if 1 ≤ t < 0.
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