Question: Consider the differential equation $\displaystyle 5y''' + 4y'' + 1y' - 3y = 0 $ with $\displaystyle y(0) = -3, y'(0) = 4, y''(0) = -3 $. Taking the Laplace transform and solving for $\displaystyle L(y) $ yields $\displaystyle L(y) = F(s) $ where $\displaystyle F(s) = . $

$\displaystyle 5\mathfrak{L}\{y'''\} +4\mathfrak{L}\{y''\} +\mathfrak{L}\{y'\} -3\mathfrak{L}\{y\} = \mathfrak{L}\{0\} $

$\displaystyle 5s^{3}Y(s) -5s^{2}y(0) -5sy'(0) -5y''(0)) +4s^{2}Y(s) - 4sy(0) -4y'(0) + sY(s) -y(0) -3Y(s) = 0 $

$\displaystyle 5s^{3}Y(s) + 15s^{2} -20s +15 +4s^{2}Y(s) +12s -16 + sY(s) + 3 - 3Y(s) = 0 $

$\displaystyle (5s^{3} + 4s^{2} + s - 3)Y(s) + 15s^{2} -8s + 2 = 0 $

$\displaystyle Y(s) = \frac{\-15s^{2}}{5s^{3} + 4s^{2} + s - 3} + \frac{\ 8s}{5s^{3} + 4s^{2} + s - 3} - \frac{\ 2}{5s^{3} + 4s^{2} + s - 3} $

It seems that after this step, the denominator doesn't factor with integers, so I am stuck at this point. I know to use partial fraction decomposition to find the $\displaystyle F(s) $. Any help is appreciated!

Thank you!