Find the Laplace Transform of the following function e2tsin

Find the Laplace Transform of the following function e^2t(sin 2t + 5t - 2) Find the Inverse Laplace Transform of the following function s + 1/s^2 + 4

Solution

(2) Laplace transform of e^2t(sin2t+5t-2) is -

L {sin2t} +L{5t}-L{2} we know that,L{sinat}=(a/s²+a²) ,L{t}=(1/s²) and L{constant} =(1/s)

{2/s²+4}+{5/s²}-{2/s} -(A)

now, as L{e^at} = (1/s-a) substitute s-a in place of s in equation (A)

then we have, {2/(s-2)²+4}+{5/(s-2)²}-{2/(s-2)}

(3) L^-1 of { (s+1)/(s²+4) }

now splitting it as L^-1{ (s/s²+4) }+L^-1{ (1/s²+4) } -(B)

we know that, L^-1{ (s/s²+a²) } =cosat and L^-1 { (1/s²+a²) } =sinat/a

substituting in equation(B),we have

cos2t+(sin2t/2)