Question

Laplace inverse (3s+4)/(s^2+4s+29)​

Answer 1

Complete the square in the denominator to reveal a sum of squares:

$s^2+4s+29=(s+2)^2+25=(s+2)^2+5^2$

Recall the Laplace transforms of sine and cosine,

$L(\sin(at))=\dfrac a{s^2+a^2}$

$L(\cos(at))=\dfrac s{s^2+a^2}$

as well as the frequency shift property,

$L(e^{at}f(t))=F(s-a)$

where $L(f(t))=F(s)$ is the Laplace transform of $f(t)$.

Rewrite the given transform as

$\dfrac{3s+4}{s^2+4s+29}=\dfrac{3(s+2)}{(s+2)^2+5^2}-\dfrac25\dfrac5{(s+2)^2+5^2}$

The inverse transforms then follows:

$F(s+2)=\dfrac{3(s+2)}{(s+2)^2+5^2}\implies f(t)=3e^{-2t}L^{-1}\left(\dfrac s{s^2+5^2}\right)$

$\implies f(t)=3e^{-2t}\cos(5t)$

$F(s+2)=-\dfrac25\dfrac5{(s+2)^2+5^2}\implies f(t)=-\dfrac25e^{-2t}L^{-1}\left(\dfrac5{(s+2)^2+5^2}\right)$

$\implies f(t)=-\dfrac25e^{-2t}\sin(5t)$

So we end up with (after some regrouping)

$L^{-1}\left(\dfrac{3s+4}{s^2+4s+29}\right)=\boxed{\dfrac{e^{-2t}}5(15\cos(5t)-2\sin(5t))}$