Answer:
y(t)= 11/3 e^(-t) - 5/2 e^(-2t) -1/6 e^(-4t)
Step-by-step explanation:
[tex] Y(s)=\frac{s+3}{(s^2+3s+2)(s+4)} + \frac{s+3}{s^2+3s+2} +\frac{1}{s^2+3s+2} [/tex]
We know that [tex] s^2+3s+2=(s+1)(s+2)[/tex], so we have
[tex] Y(s)=\frac{s+3+(s+3)(s+4)+s+4}{(s+1)(s+2)(s+4)} [/tex]
By using the method of partial fraction we have:
[tex] Y(s)=\frac{11}{3(s+1)} - \frac{5}{2(s+2)} -\frac{1}{6(s+4)} [/tex]
Now we have:
[tex] y(t)=L^{-1}[Y(s)](t) [/tex]
Using linearity of inverse transform we get:
[tex] y(t)=L^{-1}[\frac{11}{3(s+1)}](t) -L^{-1}[\frac{5}{2(s+2)}](t) -L^{-1}[\frac{1}{6(s+4)}](t) [/tex]
Using the inverse transforms
[tex] L^{-1}[c\frac{1}{s-a}]=ce^{at} [/tex]
we have:
[tex] y(t)=11/3 e^{-t} - 5/2 e^{-2t} -1/6 e^(-4t) [/tex]
