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Consider the following initial value problem: $y′′−4y′−45y=\sin(3t)y(0)=5$, $y′(0)=3$. How can we find the equation obtained by taking the Laplace transformation in terms of $Y(s)$?

Here is where i got to but i dont know how to finish the question to get the answer. please help give the correct answer:

$Y(s) = s/[(s² + 9)(s² - 4s - 45)] + (5s - 17)/(s² - 4s - 45)$

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You know that: $$\mathcal{L}(y'')=s^2\mathcal{L}(y)-sy(0)-y'(0),~~~\mathcal{L}(y')=s\mathcal{L}(y)-y(0)$$ so if we set $\mathcal{L}(y)=Y(s)$, in this OE, we have $$(s^2Y(s)-5s-3)-4(sY(s)-5)-45Y(s)=\frac{3}{s^2+9}$$ If you simplify to find $Y(s)$, you get: $$Y(s)=\frac{-150+5s^3+45s-17s^2}{(s^2+9)(s^2-4s-45)}=\frac{1425}{476(s+5)}+\frac{2s-27}{510(s^2+9)}+\frac{841}{420(s-9)}$$

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  • $\begingroup$ Very clear and concise! +1 $\endgroup$ – Amzoti Mar 30 '13 at 20:13
  • $\begingroup$ @Amzoti: Thanks my dear friend. $\endgroup$ – mrs Mar 30 '13 at 20:14
  • $\begingroup$ I agree with @Amzoti +1 $\endgroup$ – amWhy Mar 31 '13 at 0:22
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You need to take the inverse Laplace transform in order to find $y(t)$. See here for the techniques.

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  • $\begingroup$ This is not an answer. $\endgroup$ – Artem Mar 30 '13 at 19:06
  • $\begingroup$ @Artem: Read his question carefully. Here is what he asked "Here is where i got to but i dont know how to finish the question to get the answer". $\endgroup$ – Mhenni Benghorbal Mar 30 '13 at 19:07

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