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1) $w'' + w = t^2 + 2$; $w(0) = 1$, $w'(0) = -1$

2) $s^2W(s) - sw(0) - w'(0) = \frac{2 + 2s^2}{s^3}$

3) $s^5W(s) - s^4w(0) - s^3w'(0) = 2 + 2s^2$

4) $ W(s) = \frac{2 + 2s^2 - s^3 + s^4}{s^5}$

5) $W(s) = 2\left(\frac{1}{s^5}\right) + 2\left(\frac{1}{s^3}\right) - \frac{1}{s^2} + \frac{1}{s}$

6) $w(t) = \frac{2}{4!}t^4 + t^2 - t + 1$

Anyone?

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It seems that there is a defect in your solution. In fact, we have: $$s^2W(s)+1-s+W(s) = 2/s^3+2/s$$ instead which leads us to have: $$W(s)=\frac{2+2s^2-s^3+s^4}{s^3(s^2+1)}$$

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  • $\begingroup$ Incorrect. This is because the RHS is equivalent to $\frac{2+2s^2}{s^3}$. $\endgroup$ – Don Larynx Nov 21 '13 at 2:43
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    $\begingroup$ @DonLarynx: B.S. has the correct form for $W(s)$ and your answer is incorrect. Regards $\endgroup$ – Amzoti Nov 21 '13 at 3:02
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    $\begingroup$ Question: Did I do Laplace correctly? Answer: No. I am confused. What B.S. shows is what you should have for $W(s)$ and your $w(t)$ is wrong. Am I not making this clear enough for you? $\endgroup$ – Amzoti Nov 21 '13 at 3:05
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    $\begingroup$ Keep in mind that you are wrong and it is not even close! Perhaps before you throw a tirade, you should actually understand what you are doing. Look at your step 2 for your problem! Wow! $\endgroup$ – Amzoti Nov 21 '13 at 3:12
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    $\begingroup$ Never mind. I am fool. $\endgroup$ – Don Larynx Nov 21 '13 at 3:18

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