Find the $L^{-1} (\frac{1}{(s^2+4)(s^2+9)})$.

Question

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$\frac{1}{5} \left (\frac{sin⁡(2t)}{2}-\frac{sin⁡(t)}{3}\right )$

TuteeHUB · Answer

$\frac{1}{5} \left (\frac{sin⁡(2t)}{2}+\frac{sin⁡(3t)}{3}\right )$

TuteeHUB · Answer

$\frac{1}{5} \left (\frac{sin⁡(t)}{2}-\frac{sin⁡(3t)}{3}\right )$

TuteeHUB · Answer

$\frac{1}{5} \left (\frac{sin⁡(2t)}{2}-\frac{sin⁡(3t)}{3}\right )$

1.	Find the \(L^{-1} (\frac{1}{(s^2+4)(s^2+9)})\).
A.	\(\frac{1}{5} \left (\frac{sin⁡(2t)}{2}-\frac{sin⁡(t)}{3}\right )\)
B.	\(\frac{1}{5} \left (\frac{sin⁡(2t)}{2}+\frac{sin⁡(3t)}{3}\right )\)
C.	\(\frac{1}{5} \left (\frac{sin⁡(t)}{2}-\frac{sin⁡(3t)}{3}\right )\)
D.	\(\frac{1}{5} \left (\frac{sin⁡(2t)}{2}-\frac{sin⁡(3t)}{3}\right )\)
Answer» E.

Find the \(L^{-1} (\frac{1}{(s^2+4)(s^2+9)})\).