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Use Laplace transform to solve the following initial–value problem.

$y′′′′ + 2y′′ + y = 0, y(0) = 1, y′(0) = −1, y′′(0) = 0, y′′′(0) = 2$


$s^4 L(s) - s^3y(0) -s^2 y'(0) - s y''(0) - y'''(0) +2[s^2L(s)-sy(0)-y'(0)] +L(s) \\\\$ I get the partial fraction part and got stuck, need help!

$L(s) =\frac{s^3 - s^2 +2s}{s^4 +2s^2 +1}= \frac{s-1}{s^2 +1}+\frac{s+1}{(s^2+1)^2} \:\:$Factorising the denominator I get: $(s^2+1)^2$

Please some let me know if Im heading in the wrong direction here.

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why $+s^2y'(0)$? – Bennett Gardiner Sep 13 '13 at 11:31
@Bennett Gardiner: Just edited, it was error. – Avinesh Sep 13 '13 at 12:06
up vote 3 down vote accepted

Great job getting to this point, now we just have to get the answer in forms we can work with or use the formal definitions for inverse Laplace transforms. I will use known forms.

We have (split up the numerators):

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

The inverse LT of this (using Laplace tables) is given by:

$$y(x) = \cos x - \sin x + \dfrac{1}{2} x \sin x + \dfrac{1}{2} (\sin x -x \cos x)$$

So, some simple algebra yields:

$$y(x) = \cos x - \dfrac{1}{2}\sin x + \dfrac{1}{2} x \sin x - \dfrac{1}{2} x \cos x$$

You should verify this result satisfies the original ODE.

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Yes, I'm done with hospital detail. From here I'll just have some weekly appointments at with a couple doctors. Big sigh (of relief!) 8^) – amWhy Sep 13 '13 at 18:24
Thanks. Just wanted to know how you got $\frac{1}{2}xsinx$? – Avinesh Sep 13 '13 at 21:43
@Avinesh: Look at item 11 in this LT Table. See that extra product by 2 in the form? Since we do not have $2 a^3$, we only have $a^3$ where $a=1$, we have to divide out that product of $2$, hence we divide the "entire" quantity by $2$. Clear? Regards – Amzoti Sep 13 '13 at 23:02
Thanks. Unfortunately my instructor did not provide this detailed table. I didnt know about this extra transforms till you gave me this link(LT Table). Good place to learn new and amazing things. – Avinesh Sep 14 '13 at 2:51
Just wanted to have my final answer verified. If Im welcome to do so – Avinesh Sep 15 '13 at 3:07

I guess it is the last fraction, $\frac{1+s}{(s^2+1)^2}$, that causes you problems.

For the (one sided) Laplace transform we have the transform pairs $$\sin{t}\stackrel{\mathcal{L}}{\longmapsto}\frac{1}{s^2+1},$$ $$tf(t)\stackrel{\mathcal{L}}{\longmapsto}-F'(s),$$ and $$f'(t)\stackrel{\mathcal{L}}{\longmapsto}sF(s).$$ The first two mean that $$t\sin{t}\stackrel{\mathcal{L}}{\longmapsto}\frac{2s}{(s^2+1)^2},$$ and the third rule gives a hint on how to get rid of the $s$ in the numerator.

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