Question

Find the inverse Laplace transform f(t) of the function F(s). Write uc for the Heaviside function that turns on at c, not uc(t).F(s) = (7e−7s)/(s2 − 49)f(t) =

Answer 1

Answer:

F(t)$=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$

Step-by-step explanation:

We have given $F(S)=\frac{7e^{-7s}}{s^2-49}$

Now  $F(S)=e^{-7s}G(s)$

Here $G(S)=\frac{7}{S^2-49}$

Now first find the Laplace inverse of G(S)

Using partial fraction

$\frac{7}{(s+7)(s-7)}=\frac{A}{(S+7)}+\frac{B}{S-7}$

$7=A(S-7)+B(S+7)$

On comparing the coefficient

$A=\frac{1}{2}$  and $B=\frac{-1}{2}$  

On putting the value of A and B  

$G(S)=\frac{-1}{2(S+7)}+\frac{1}{2(S+7)}$

Taking inverse Laplace

$G(t)=\frac{-1}{2}e^{7t}+\frac{1}{2}e^{-7t}$

Now in G(s) there is onether term $e^{-7s}$

So F(t)$=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$