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Loser66
 one year ago
Question in comment
Please, help
Loser66
 one year ago
Question in comment Please, help

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.01) Find the equation for the periodic function and its perioddw:1436381101289:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.02) find the Laplace transform of it

Loser66
 one year ago
Best ResponseYou've already chosen the best response.03) find the inverse laplace transform \(L^{1}(\dfrac{k}{as+b})\) where k, a, b in R/{0} and are constants

dan815
 one year ago
Best ResponseYou've already chosen the best response.0they are giving you the constants on your continous power series, u have to see what that gives you in t domain

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0dan, how to know whether a differential equation linear or not?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0why not? since the linear differential equation standard form is \(y^{(m)} +P_1(x) y^{(m1)}+..... = Q(x)\) Like y" +y' +xy =0. ??

dan815
 one year ago
Best ResponseYou've already chosen the best response.0that equation is linear

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i was confused with linear form there

dan815
 one year ago
Best ResponseYou've already chosen the best response.0by the way the question you asked earlier, what did u get for this 1) Find the equation for the periodic function and its period

dan815
 one year ago
Best ResponseYou've already chosen the best response.0wheres your work so farq

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0dan, remember the previous post ? http://math.stackexchange.com/questions/1354551/findingdifferentialequationfromthesolution

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Wow!!! they derive the problem into other way!! but it works also.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0dan, honestly, I don't know how to work on it.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Since the slopes of the lines are equal, we have y'= 1 , that gives us y= x +a. But it doesn't make sense since y = x +a are family of vertical lines, not horizontal like this.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0what I don't understand is heredw:1436404165210:dw

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1What about starting with $$ x(t) = t  \lfloor t \rfloor = t  \operatorname{floor}(t) $$ Then scaling? http://www.wolframalpha.com/input/?i=tfloor%28t%29&dataset=

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1Then Laplace  http://www.wolframalpha.com/input/?i=lapace+transform+%28tfloor%28t%29%29

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0how about initial value?dw:1436405123893:dw

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1What about multiplying everything by u(t), the step function: dw:1436405297232:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Wow!! I am dizzy!! this is the beginning level of DE. The students cannot know that far!!

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1If you agree that this might work, then the Laplace would be http://www.wolframalpha.com/input/?i=lapace+transform+theta%28x%29%28tfloor%28t%29%29

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thanks friend. I need time to digest it. :)

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1Ok  just some ideas, not sure if it will work

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1BTW, the period becomes clear if you use the function above, it's 1 second  which you can scale by \(a\).
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