For the equation $′′+9=\delta(−4)$, with $(0)=0$ and $′(0)=0$; where the $g(t)$ is the step function.
Taking the Laplace of the left sides gives: $L\{\}\{ s^2 +9 \}$.
For the left side, from the table, I see that $\delta(−4)$ is transformed into $\ \frac{e^{-4s}}{s}\ $.
Moving everything to the right side to solve I have to take the inverse of : $\ \frac{e^{-4s}}{s} \{\frac{1}{s^2 +9}\}$.
However, looking at the solution notes I have I'm seeing that the inverse should not include that "S" under the $\ \ e^{-4s}\ $. As in the solution just took the inverse of $e^{-4s} \{\frac{1}{s^2 +9}\}$ and got
$\frac{1}{3} U_4(t) \sin(3(t-4))$ as an answer.
Does that "$S$" get factored out along with the $e^{-4s}$ while we are focused on transforming the other expression or does it get distributed through?
the laplace transform of $\delta(−4)$ is $e^{-4s} $.
$$ L^{-1}[ e^{-c s}L(f)]=u(t-c) f(t-c)$$
Hen ce $$\frac{1}{3} L^{-1}[ e^{-4 s}\frac{3}{s^2 +9}]=u(t-4) \sin(3(t-4))$$