*different*solutions to be generated.

Consider my example below. Forget about ODE's for now. For the first case I'll use partial fraction decomposition.

\( \frac{2s+3}{s^2 + 6s + 8} = \frac{2s + 3}{(s + 2)(s + 4)} \)

\( \frac{2s + 3}{(s + 2)(s + 4)} \equiv \frac{A}{(s+2)} + \frac{B}{(s+4)} \Rightarrow 2s + 3 \equiv A(s+4) + B(s+2) \)

\( A=-\frac{1}{2}\ and\ B=\frac{5}{2} \)

\( \frac{2s + 3}{(s + 2)(s + 4)} \equiv \frac{5}{2} \times \frac{1}{s+4} - \frac{1}{2} \times \frac{1}{s+2}\)

\( L^{-1}[ \frac{5}{2} \times \frac{1}{s+4} - \frac{1}{2} \times \frac{1}{s+2}] = -\frac{1}{2}e^{-4t}(e^{2t} - 5)\)

Intuition, however, says otherwise. This time I'll take the denominator and complete the square.

\( \frac{2s+3}{s^2 + 6s + 8} = \frac{2s+3}{(s+3)^2 - 1} \)

Now I'll add zero onto the end. There is nothing to stop me from doing this.

\( \frac{2s+3}{(s+3)^2 - 1} = \frac{2s+3}{(s+3)^2 - 1} + \frac{3}{(s+3)^2 - 1} - \frac{3}{(s+3)^2 - 1}\)

\( \frac{2s+3}{(s+3)^2 - 1} = \frac{(2)(s+3)}{(s+3)^2 - 1} - \frac{3}{(s+3)^2 - 1} \)

Taking the inverse Laplace transform,

\( L^{-1}[\frac{2s+3}{(s+3)^2 - 1}] = e^{-3t}\times[2ch(3t) - 3sh(t)] \)

Wolfram Alpha agrees with the former. But I cannot see what the problem is with the latter expression as a solution?