Answer to Question #217597 in Differential Equations for Fhumu

Question

Inverse laplace (s-5/s^2+4s+20)

Answer 1

Solution;

Rewrite the denominator as follows;

s²+4s+20=s²+4s+4+16=(s+2)²+16

Rewrite the numerator as follows,

s-5=(s+2)-7

Therefore;

$L^{-1}[\frac{s-5}{s^2+4s+20}]$ =$L^{-1}[\frac{s+2}{(s+2)^2+16}]$ -$L^{-1}\frac{7}{(s+2)^2+16}$

=$L^{-1}[\frac{s+2}{(s+2)^2+4^2}]-\frac74L^{-1}[\frac{4}{(s+2)^2+4^2}]$

From the Laplace tables;

f(t)=e^-2tcos(4t)-$\frac 74$e^-2tsin(4t)