Find the inverse Laplace of {2s+5/s^2+25}?
A) 2Sin5t+Cos5t
B) Cos5t-2Sin5t
C) 2Cos5t+Sin5t
D) 2Cos5t-Sin5t
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)
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