$s^4 L(s) - s^3y(0) -s^2 y"(0) - s y""(0) - y"""(0) +2~~ +L(s) \\$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

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

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

Mårten WMårten W

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