Question

Find the inverse laplace transform of e^(-3pi*s)/(s^2+2s+3).

Answer 1

Have you completed the square in the denominator?

Answer 2

do you have an online book i can check

Answer 3

Yes, it's 1/(s+1)^2+2), right?

Answer 4

Yes.

Answer 5

But what's next? What do I do with the numerator?

Answer 6

\(\mathcal{L}^{-1}\left\{e^{-ct}F(s)\right\}=u(t-c)f(t-c)\)

Here, \(F(s)=\dfrac{1}{(s+1)^2+2}\)

Answer 7

So the answer should be y=1/sqrt(2)*u3pi(t)e^(-(t-3pi))*sin(sqrt(2))*(t-3pi), but how do I get there?

Answer 8

Did you post the answer?

Answer 9

The \(e^{-3\pi s}\) contributes the \(u(t-3\pi)=u_{3\pi}(t)\). So you're left with
\[\mathcal{L}^{-1}\left\{\frac{e^{-3\pi s}}{(s+1)^2+2}\right\}=u(t-3\pi)\color{red}{\mathcal{L}^{-1}\left\{\frac{1}{(s+1)^2+2}\right\}}\]
Keep in mind that the red part includes a shift, as per the rule \(\mathcal{L}^{-1}\left\{e^{-ct}F(s)\right\}=u(t-c)f(t-c)\).

So the remaining step is to find
\[\mathcal{L}^{-1}\left\{\frac{1}{(s+1)^2+2}\right\}\]
For the next step, use \(\mathcal{L}^{-1}\left\{\dfrac{b}{(s-a)^2+b^2}\right\}=e^{at}\sin bt\). Here, \(b=\sqrt 2\), since \(b^2=2\).

However, you're missing a \(\sqrt 2\) in the numerator, so you do this:
\[\mathcal{L}^{-1}\left\{\frac{\sqrt2}{\sqrt2}\frac{1}{(s+1)^2+2}\right\}=\frac{1}{\sqrt2}\mathcal{L}^{-1}\left\{\frac{\sqrt2}{(s+1)^2+2}\right\}=\frac{1}{\sqrt 2}e^{-t}\sin\sqrt2 t\]

Answer 10

Now, putting it all together, you have
\[\frac{1}{\sqrt 2}u_{3\pi}(t)e^{-(t-3\pi)}\sin\left(\sqrt2(t-3\pi)\right)\]

Answer 11

Thank you so much.

Answer 12

You're welcome