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Find the inverse laplace transform of $ \ Y(s)= \large \frac{\large \frac{82}{ \large s-6}-2s+2}{s^2+6s+10} \ $

Answer:

Let $ \ \mathcal{L}^{-1} \ $ be the inverse laplace operator.

Then,

$ y(t)=\mathcal{L}^{-1} [Y(s);t] \\ \Rightarrow y(t)= \mathcal{L}^{-1} \left[\frac{\large \frac{82}{ \large s-6}-2s+2}{s^2+6s+10} \right] \ = \mathcal{L}^{-1} \left[\frac{ \large -2s^2+12s+70}{ \large (s-6)(s^2+6s+10)} \right] $

Now,

$ \frac{ \large -2s^2+12s+70}{ \large (s-6)(s^2+6s+10)}= \frac{A}{s-6}+\frac{\large Bs+C}{s^2+6s+10} \ $ where $ \ A,B,C \ $ are unknown constants to be determined.

Is this the correct partial fraction?

Help me find the inverse laplace transform.

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You are on the right track, after you work out the constants $A$, $B$ and $C$ you should end up with

\begin{eqnarray} Y(s) &=& \frac{1}{s - 6} - \frac{3s + 10}{s^2 + 6s + 10} \\ &=& \frac{1}{s - 6} - \frac{3(s + 3) + 1}{(s + 3)^2 + 1} \\ &=& \frac{1}{s - 6} - 3\frac{(s + 3)}{(s + 3)^2 + 1} - \frac{1}{(s + 3)^2 + 1} \end{eqnarray}

Now use the fact that

$$ \mathcal{L}[e^{at}\sin bt] = \frac{b}{(s-a)^2 + b^2} $$

and

$$ \mathcal{L}[e^{at}\cos bt] = \frac{s-a}{(s-a)^2 + b^2} $$

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So far so good.

$$ \frac{ \large -2s^2+12s+70}{ \large (s-6)(s^2+6s+10)}= \frac {A}{s-6}+\frac {\large Bs+C}{s^2+6s+10} $$

Use Heavy-side method with $s=6$ to find $A= \frac {70}{82}$

Subtract $ \frac {70}{82(s-6)}$ from both sides to find $B$ and $C$.

Complete the square in $s^2 + 6s + 10 $ and use shifting formulas to finish the work.

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