Answer to Question #127080 in Differential Equations for jse

Question

Find the inverse Laplace transform L−1{F(s)} of the given function.

F(s) = 6s^2−10s+16 / s(s^2+4)

Your answer should be a function of t. Enclose arguments of functions in parentheses. For example, sin(2x).

L^−1 {F(s)} = _______________-

Answer 1

F(s) = (6s²-10s+16)/(s(s²+4))

by partial fraction

(6s²-10s+16)/(s(s²+4)) = A/s + (Bs+C)/(s²+4)

(6s²-10s+16) = A(s²+4)+(Bs+C)s

by comparing coefficients

A=4, B=2, C=-10

so F(s)= 4/s + (2s-10)/(s²+4)

= 4/s + 2s/(s²+4) - 10/(s²+4)

apply laplace inverse

f(t) = 4 + 2cos(2t) - 5sin(2t)

since laplace inverse(1/s)=1

laplace inverse(s/(s²+a²))=cos(at),

laplace inverse (a/(s²+a²))=sin(at).