$s^4 L(s) - s^3y(0) -s^2 y"(0) - s y""(0) - y"""(0) +2 +L(s) \\$I acquire the partial fraction part and obtained stuck, need help!

$L(s) =fracs^3 - s^2 +2ss^4 +2s^2 +1= fracs-1s^2 +1+fracs+1(s^2+1)^2 ::$Factoincreasing the denominator I get: $(s^2+1)^2$

Please some let me understand if Im heading in the wrong direction below.

You are watching: Use the laplace transform to solve the following initial value problem


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Great project acquiring to this suggest, now we simply have to get the answer in creates we have the right to work-related through or use the formal definitions for inverse Laarea transdevelops. I will use recognized develops.

We have (break-up up the numerators):

$$dfracs-1s^2 +1+dfracs+1(s^2+1)^2 = dfracss^2+1 - dfrac1s^2+1 + dfracs(s^2+1)^2 + dfrac1(s^2+1)^2$$

The inverse LT of this (using Laarea tables) is offered by:

$$y(x) = cos x - sin x + dfrac12 x sin x + dfrac12 (sin x -x cos x)$$

So, some easy algebra yields:

$$y(x) = cos x - dfrac12sin x + dfrac12 x sin x - dfrac12 x cos x$$

You have to verify this outcome satisfies the original ODE.


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edited Sep 13 "13 at 13:12
answered Sep 13 "13 at 12:52
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AmzotiAmzoti
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I guess it is the last fractivity, $frac1+s(s^2+1)^2$, that reasons you troubles.

For the (one sided) Lalocation transcreate we have actually the transcreate pairs$$sintstackrelrianowens.tvcalLlongmapstofrac1s^2+1,$$$$tf(t)stackrelrianowens.tvcalLlongmapsto-F"(s),$$and$$f"(t)stackrelrianowens.tvcalLlongmapstosF(s).$$The initially two suppose that$$tsintstackrelrianowens.tvcalLlongmapstofrac2s(s^2+1)^2,$$and also the third dominance offers a hint on how to get rid of the $s$ in the numerator.


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answered Sep 13 "13 at 12:54
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Mårten WMårten W
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