Please help to solve with steps.
Determine the Laplace transform of: cos (3t b) f(t) J te 2tcos(t)
Solution
1) Since L {cos(3t)} = s/(s^2 + 9),
L{t*f(t))= -F'(s) ,
L {t cos(3t)} = -(d/ds) s/(s^2 + 9) = (s^2 -9)/(s^2+9)^2
2)since L(t) = 1/s^2
L(e^at*f(t))= F(s-a) ,
L(t*e^-t) = 1/(s+1)^2, - (1)
Since L {cos(t)} = s/(s^2 + 1),
L{t*f(t))= -F'(s) ,
L {2t cos(t)} = -(2)(d/ds) s/(s^2 + 1) = (2)(s^2 -1)/(s^2 + 1)^2. - (2)
adding both (1) + (2)
L( t*e^-t + 2t cos(t) ) = 1/(s+1)^2 + (2)(s^2 -1)/(s^2 + 1)^2.