For F(s) = L(f) = 1 /(s+a)(s+b)

Question

where a, b constants,

find f(t). Show the detail of your works.

please help

Answer 1

The inverse Laplace transform of F(s) is f(t). Let 1/((s+a)(s+b)) =A/(s+a) +B/(s+b) and now 1=A(s+b)+B(s+a) so 1 =As+Ab +Bs+Ba. Now match coefficients. Hence 1=Ab+Ba and 0= A+B since there is no s term on the left side of the equation. Therefore A=-B and uisng this in the other equation 1= Ab-Aa and A(b-a)=1 so A= 1/(b-a) and B=-1/(b-a). So now taking the inverse transform of both sides gives f(t) = £^-1((1/(b-a)(1/(s+a) -1/(s+b))) which gives f(t)= (1/(b-a))(e^-at-

e^-bt).