Inverse Laplace Transform

[ExtremeFusion]

Hello Guys, I have three Inverse Laplace problem that is troubling me literary..
I'm trying to solve it but i just can't figured it out..

Can you please provide a detailed solution on how you arrived on the answer that you going to give, so i can study it and how to do it..

Attached herein is my troublesome question.. BTW, if i don't answer this questions by monday (April 07, 2008) my professor is going to cancel my candidacy for graduation.. so please help me.. i could really use one.. thanks

 

[lippyka]

in advance guys..1. Find the inverse Laplace transform of F(s) = $\frac{2s^2+3s+5}{(s+2)(s+3)}$Solution:We use the partial fraction decomposition to find the inverse Laplace transform of F(s). Let F(s) = $\frac{A}{s+2} + \frac{B}{s+3}$. Then, $A(s+3) + B(s+2) = 2s^2 + 3s + 5$$A(s+3) = 2s^2 + 3s + 5 - B(s+2)$$A = \frac{2s^2 + 3s + 5 - B(s+2)}{s+3}$$A = \frac{2s^2 + 3s +5 - Bs - 2B}{s+3}$$A = \frac{2s^2 + (3-B)s + (5-2B)}{s+3}$Comparing coefficients of s on both sides, we get $3-B=3 \implies B=0$Comparing coefficients of constants on both sides, we get$5-2B = 5 \implies B = 0$Therefore, $A = \frac{2s^2 + 3s + 5}{s+3}$ Hence, F(s) = $\frac{2s^2 + 3s + 5}{(s+2)(s+3)} = \frac{2s^2 + 3s + 5}{s+3} - \frac{2s^2 + 3s + 5}{s+2}$Now, we use the formula $\mathcal{L^{-1}}\left(\frac{X(s)}{s+a}\right) = e^{at}x(t)$ to find the inverse Laplace transform of F(s)$\mathcal{L^{-1}}\left(\frac{2s^2 + 3s + 5}{s+3}\right) = e^{