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First time question and never could format on these sites. Anyway, I have still a book from a course from years and years ago. Opening it up again to refresh some forgotten stuff, I try to solve a few initial value Laplace transformations but come up short each and every time. A small note in the margin indicates I didn't solve at least one of them back then either.

I can mention two, either one being shown how to solve would be amazing. I have the given answers from the book and will list them:

Problem 1:

y''-2y' +2y = cos t , where y(0) = 1 & y'(0) = 0.

The case of such initial value problems equalling 0 are fine, but once I have like e.g. cos t, it gets messed up, no matter what I try.

The solution is meant to be: y= 1/5 (cos t - 2sin t 4e^tcos t - 2e^tsin t)

I get as usual to the partial fractions, but something seems to go wrong.

I need some hand holding to get there. I tried backtracking from the solution on a similar problem and so close yet so far away from solving it, I am at a total loss.

To show where I am at perhaps? : ( let f = Y(s) )

(s^2)f - s - 2sf + 2f = s(s^2 + 1)^-1

f[(s - 1)^2 + 1] = s(s^2 + 1)^-1 + s

From here on, I guess I am starting to mess things up, so possibly here is where I could need some injection of hand holding. Have been attempting this whole day not to mention probably a stressful day or two back at uni days :)

EDIT: To JM. I have done the mentioned fractional method before and failed; as I said, I need(ed) some hand holding not to not knowing what to do, but that failing to achieve the answer, whilst doing so. :)

But I probably have failed to do something right and am going to cross-check yet again. In fact, it was the choice of using As+B/Q1 + Cs+D/Q2 I thought possibly be wrong since I was unable to reach the conclusion.

Ya thanks JM, somehow doing it again, I did get the coefficients trivially.. sigh! :)) Thanks for hand holding!!

EDIT 2: Hang on, I am not free yet to finish, but I don't have the N/5th in there so before I accept, just gonna make sure it all works out! It did. Thought of showing all steps to help others but ok, I kinda got busy :) I hope I accepted answer correctly; am not used to this forum.

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  • $\begingroup$ Heya.. hmm I do know how to solve an LT with fractional approach, it is this specific example though, I am having a nightmare with :) I did however come across 2 answers which might have jogged my memory on some hanky pankying of variables: Babaks: math.stackexchange.com/questions/492431/… and Rons: math.stackexchange.com/questions/379835/…. Thanks. But I will happily take any input on specific solving. $\endgroup$
    – Mal
    Feb 12 '17 at 22:23
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You have to decompose

$$\dfrac{s}{((s-1)^2+1)(s^2+1)}+\dfrac{s}{(s-1)^2+1}$$

into partial fractions. We get:

$$\tag{1}\dfrac{1}{5}\dfrac{4 - s}{(s-1)^2 +1} + \dfrac{1}{5}\dfrac{s - 2}{s^2 + 1}+\dfrac{s}{(s-1)^2+1}$$

$$\dfrac{4}{5}\dfrac{1}{(s-1)^2 +1} - \dfrac{1}{5}\dfrac{s}{(s-1)^2 +1}+ \dfrac{1}{5}\dfrac{s}{s^2 + 1}-\dfrac{2}{5}\dfrac{1}{s^2 + 1}+\dfrac{s}{(s-1)^2+1}$$

From here, you can retrieve simple originals, reading a Laplace table in the reverse order.


Edit: honestly, I obtained (1) by asking Wolfram Alpha. But it's good to know a "pedestrian" method for obtaining this partial fraction decomposition using paper and pencil (and eraser...). Let us see it on:

$$E=\dfrac{s}{(s^2-2s+2)(s^2+1)}$$

The more straighforward method is by using undeterminated coefficients. Here is how it works on this example. As you have a product of quadratics, say

$E=\dfrac{1}{Q_1}*\dfrac{1}{Q_2}$, you target a decomposition

$$E=\dfrac{A+Bs}{Q_1}+\dfrac{C+Ds}{Q_2}$$

(general rule: one degree less in the numerator than in the denominator).

Reducing the two fractions to a common denominator:

$$E=\dfrac{(A+Bs)Q_2+(C+Ds)Q_1}{Q_1Q_2}$$

There is nothing to do on the denominator.

The numerator in $E$ is

$$(A+Bs)(s^2+1)+(C+Ds)(s^2-2s+2)$$

$$=As^2 + A + Bs^3 + Bs + Cs^2 - 2Cs + 2C + Ds^3 - 2Ds^2 + 2Ds $$

$$=(A+2C)+(B-2C+2D)s+(A+C-2D)s^2+(B+D)s^3$$

a third degree polynomial which is equal to $$s$$ (the RHS is the initial numerator). Identity

$$(A+2C)+(B-2C+2D)s+(A+C-2D)s^2+(B+D)s^3=s$$ is equivalent to:

$$\begin{cases}A+2C&=&0\\B-2C+2D&=&1\\A+C-2D&=&0\\B+D&=&0\end{cases}$$

This is a linear system of four equations with four unknowns $A,B,C,D$, that you have to solve.

Remark: there are some more "witty" methods, but I don't advise to use them, because prone to errors that are more difficult to trace.

Bon courage!

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  • $\begingroup$ Merci beaucoup.. thanks JM, I appreciate your feedback and (1) might give me some ideas on how to get to (1), which is the question I am asking, I suppose. Since I gave the answer, and laplace transformations are unique for each f(t), then I could get the last line you wrote as well. I can add though, it is only the first fraction of the two fractions you start of with I need help with 'decomposing' to the form of (1). Thanks :) $\endgroup$
    – Mal
    Feb 13 '17 at 12:19
  • $\begingroup$ I thought at first that it is the idea of using partial fraction decomposition with which you had a problem. I am going to give a rapid idea of the "technique" for hand calculations as an edit to my answer. $\endgroup$
    – Jean Marie
    Feb 13 '17 at 12:36
  • $\begingroup$ I appreciate it. I admit, I feel like a blundering novice as usually Laplace transformations aren't always THAT hard, but this had me chuffed. Admitedly, I haven't done one in 5 years. No need to enclose technique with ""; it sure isn't as straight forward as one thinks. $\endgroup$
    – Mal
    Feb 13 '17 at 13:22
  • $\begingroup$ See the Edit I have added to my answer. $\endgroup$
    – Jean Marie
    Feb 13 '17 at 13:34
  • $\begingroup$ I have filled in the details... $\endgroup$
    – Jean Marie
    Feb 13 '17 at 18:07

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