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 one year ago
I have a question about the unit step function in a differential equation. Here is a link to the "quiz" this comes from: http://ocw.mit.edu/courses/mathematics/1803scdifferentialequationsfall2011/unitiiifourierseriesandlaplacetransform/unitstepandunitimpulseresponse/MIT18_03SCF11_s25_3quizq.pdf
I will post the result I obtained in a followup post where I can actually use LaTeX.
 one year ago
I have a question about the unit step function in a differential equation. Here is a link to the "quiz" this comes from: http://ocw.mit.edu/courses/mathematics/1803scdifferentialequationsfall2011/unitiiifourierseriesandlaplacetransform/unitstepandunitimpulseresponse/MIT18_03SCF11_s25_3quizq.pdf I will post the result I obtained in a followup post where I can actually use LaTeX.

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Waynex
 one year ago
Best ResponseYou've already chosen the best response.0For v(t), I obtained: \[ce^{kt}, t < 0; ce^{kt}+\frac{1}{k}, t > 0.\] With the given initial condition, c = 0, and v(t) becomes: \[0, t < 0; \frac{1}{k}, t > 0.\] Then the derivative of that is delta(t), is it not? Then delta(0+) = 0. However, the soln given is delta(0+) = 1. http://ocw.mit.edu/courses/mathematics/1803scdifferentialequationsfall2011/unitiiifourierseriesandlaplacetransform/unitstepandunitimpulseresponse/MIT18_03SCF11_s25_3quiza.pdf

Waynex
 one year ago
Best ResponseYou've already chosen the best response.0It looks like where I went wrong was in assuming c to be the same for both parts of the piecewise solution. What I should have had for v(t) is:\[c_1e^{kt}, t < 0; c_2e^{kt}+\frac{1}{k}, t > 0.\]such that \[c_1 = 0; c_2 = \frac{1}{k}\]
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