Question

find the inverse laplace

F(s)=11s/ s^2-12s+52

Answer 1

Answer:

$11e^{6t}\cos 4t+\frac{33}{2}e^{6t}\sin 4t$

Step-by-step explanation:

We can write $\frac{11s}{s^2-12s+52}$ as follows:

$\frac{11s}{s^2-12s+52}\\=11\left [ \frac{s}{s^2-12s+52} \right ]\\=11\left [ \frac{s}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6+6}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16}$

To find:

$L^{-1}\left [ \frac{11s}{s^2-12s+52 \right ]}\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]$

We will use formulae:

$L^{-1}\left \{ \frac{s-a}{(s-a)^2+b^2} \right \}=e^{at}\cos bt\\L^{-1}\left \{ \frac{b}{(s-a)^2+b^2} \right \}=e^{at}\sin bt$

we get solution as :

$L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+4^2} \right ]+\frac{66}{4}\left [ \frac{4}{(s-6)^2+4^2} \right ] \right ]\\=11e^{6t}\cos 4t+\frac{33}{2}e^{6t}\sin 4t$