# Find the inverse laplace transform of the given problem

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

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.

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

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

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.