Question

Find the inverse Laplace of {2s+5/s^2+25}?

A) 2Sin5t+Cos5t

B) Cos5t-2Sin5t

C) 2Cos5t+Sin5t

D) 2Cos5t-Sin5t

Answer 1

Let us find the inverse Laplace of $\frac{2s+5}{s^2+25}=2\cdot\frac{s}{s^2+25}+\frac{5}{s^2+25}.$ Taking into account that \frac{s}{s^2+b^2}\to\cos (bt),\ and $\frac{b}{s^2+b^2}\to\sin (bt),$ we conclude that the inverse Laplace of $\frac{2s+5}{s^2+25}$ is

$2\cos(5t)+\sin(5t).$

Answer: C)