Use Laplace transform to solve the following initial–value problem.
$y′′′′ + 2y′′ + y = 0, y(0) = 1, y′(0) = −1, y′′(0) = 0, y′′′(0) = 2$
Answer
$s^4 L(s) - s^3y(0) -s^2 y'(0) - s y''(0) - y'''(0) +2[s^2L(s)-sy(0)-y'(0)] +L(s) \\\\$ 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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$\begingroup$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.
$\endgroup$ 6 $\begingroup$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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