Answer :

Solution :

L [f(t)] = $\frac{s+3}{(s+1)(s+2)}$

The above equation through partial fractions can be written as :

$\frac{s+3}{(s+1)(s+2)}$ = $\frac{A}{(s+1)}$ + $\frac{B}{(s+2)}$

s + 3 = A(s + 2) + B(s + 1)

s + 3 = (A + B)s + 2A + B

Comparing co-efficients on both sides, we get

A + B = 1 and 2A + B = 3

We get, A = 2, B = -1

So, $\frac{s+3}{(s+1)(s+2)}$ = $\frac{2}{(s+1)}$ - $\frac{1}{(s+2)}$

f(t) = L^-1$\left(\frac{2}{s+1}\right)$ - L^-1$\left(\frac{1}{s+2}\right)$

f(t) = 2e^-t - e^-2t

So, f(0) = 2e^-0 - e^-0 = 2 - 1 = 1

∴ f(0) = 1