Answer

Taking the Laplace transform on both sides

\[ \begin{aligned} s^{2}Y(s) + 6sY(s) + 8Y(s) &= \frac{1}{s+3} - \frac{1}{s+5}\newline Y(s) &= \roundbr{\frac{1}{s+3} - \frac{1}{s+5}}\frac{1}{s+2}\frac{1}{s+4}\newline &= \frac{1}{2}\roundbr{\frac{1}{s+3} - \frac{1}{s+5}}\roundbr{\frac{1}{s+2} - \frac{1}{s+4}}\newline &= \frac{1}{2}\roundbr{\frac{1}{(s+3)(s+2)} - \frac{1}{(s+5)(s+2)} - \frac{1}{(s+3)(s+4)} + \frac{1}{(s+5)(s+4)}}\newline &= \frac{1}{2}\roundbr{\frac{1}{s+2} - \frac{1}{s+3} - \frac{1}{3}\roundbr{\frac{1}{s+2} - \frac{1}{s+5}} - \frac{1}{s+3} + \frac{1}{s+4} + \frac{1}{s+4} - \frac{1}{s+5}}\newline \implies y &= \frac{1}{2}\roundbr{e^{-2t} - e^{-3t} - \frac{1}{3}\roundbr{e^{-2t} - e^{-5t}} - e^{-3t} + e^{-4t} + e^{-4t} - e^{-5t}}\newline &= \frac{1}{3}e^{-5t}\roundbr{e^{3t} - 3e^{2t} + 3e^{t} - 1}\newline &= \frac{1}{3}e^{-5t}\roundbr{e^{t} - 1}^{3} \end{aligned} \]