Take the Laplace transform of the following initial value problem and solve for $Y(s) = \mathcal{L}\lbrace y(t) \rbrace$: $$y'' + {6} y' + {25} y= \begin{cases} t, & 0 \leq t < 1 \cr 0, & 1 \leq t \end{cases} \hspace{0.5in} y(0)=0, \; y'(0)=0$$

$Y(s)=$ .
Now find the inverse transform to find $y(t) =$
$- u_1(t)$ times a similar function to the above and another one given by the second equation below. It's too long to enter in to a blank space. Note: $$\frac{1}{s^2(s^2+ {6}s + 25)} = \frac{-\frac{6}{625}}{s} + \frac{\frac{1}{25}}{s^2} + \frac{\frac{6}{625} (s + 3 ) +\frac{-7}{625} }{(s+3)^2 + 4^2}$$ and $$\frac{s}{s^2(s^2+ {6}s + 25)} = \frac{\frac{1}{25}}{s} + \frac{-\frac{1}{25} (s + 3 ) -\frac{3}{25} }{(s+3)^2 + 4^2}$$