Question

find the inverse transform of {s+1/s²+6s+25}

Answer 1

Answer:

(s+1)/[(s+3)^2+4^2]

Explanation:

(s+1)/(s^2+6s+25)

can be rewritten by completing the square:

(s+1)/(s^2+6s+9+16)

= (s+1)/[(s+3)^2+4^2]

And then split into the form A*(s-a)/[(s-a)^2+b^2] +B*b/[(s-a)^2+b^2]:

(s+3-2)/[(s+3)^2+4^2]

= (s+3)/[(s+3)^2+4^2] - (2/4)*4/[(s+3)^2+4^2]

= L^-1{e^(-3t)cos(4t)} - (1/2)L^-1{e^(-3t)cos(4t)}

Since the Inverse Laplace Transform is a linear operation, the Inverse Laplace of the original expression is

e^(-3t)cos(4t) - (1/2)e^(-3t)cos(4t)