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Question in comment Please, help

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1) Find the equation for the periodic function and its period|dw:1436381101289:dw|
2) find the Laplace transform of it
3) find the inverse laplace transform \(L^{-1}(\dfrac{k}{as+b})\) where k, a, b in R/{0} and are constants

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they are giving you the constants on your continous power series, u have to see what that gives you in t domain
dan, how to know whether a differential equation linear or not?
no xy no y^2 and such
why not? since the linear differential equation standard form is \(y^{(m)} +P_1(x) y^{(m-1)}+..... = Q(x)\) Like y" +y' +xy =0. ??
no xy^2
that equation is linear
i was confused with linear form there
by the way the question you asked earlier, what did u get for this 1) Find the equation for the periodic function and its period
I didn't.
wheres your work so farq
dan, remember the previous post ? http://math.stackexchange.com/questions/1354551/finding-differential-equation-from-the-solution
Wow!!! they derive the problem into other way!! but it works also.
dan, honestly, I don't know how to work on it.
Since 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.
what I don't understand is here|dw:1436404165210:dw|
What about starting with $$ x(t) = t - \lfloor t \rfloor = t - \operatorname{floor}(t) $$ Then scaling? http://www.wolframalpha.com/input/?i=t-floor%28t%29&dataset=
Then Laplace - http://www.wolframalpha.com/input/?i=lapace+transform+%28t-floor%28t%29%29
how about initial value?|dw:1436405123893:dw|
What about multiplying everything by u(t), the step function: |dw:1436405297232:dw|
Wow!! I am dizzy!! this is the beginning level of DE. The students cannot know that far!!
If you agree that this might work, then the Laplace would be http://www.wolframalpha.com/input/?i=lapace+transform+theta%28x%29%28t-floor%28t%29%29
Thanks friend. I need time to digest it. :)
Ok - just some ideas, not sure if it will work
BTW, the period becomes clear if you use the function above, it's 1 second -- which you can scale by \(a\).

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