Diff Eq. Easy question on prob. Please help last step

[frozenguy]

Diff Eq. Easy question on prob. Please help...last step

Homework Statement

Solve

The Attempt at a Solution

So this is my attempt. I just need to figure out how to adjust Y(s) so I can take inverse laplace transform. I have made the denominator a single fraction, whose denominator is s²+1 and numerator is s³+s+s², then obviously inverse that but I can't really figure out what to do with that if its the right way. I even tried factoring it out a bit but leads to dead ends.
Thanks for your time..

 

[lippyka]

So the equation is$$\frac{s^3+s+s^2}{s^2+1}Y(s)=e^{-s}$$First, we need to rearrange the equation to put $Y(s)$ on one side:$$Y(s)=\frac{e^{-s}}{s^3+s+s^2}\frac{1}{s^2+1}$$Now, we can take the inverse Laplace transform of both sides:$$y(t)=\mathscr{L}^{-1}\left[\frac{e^{-s}}{s^3+s+s^2}\frac{1}{s^2+1}\right]$$This can be simplified as follows:$$y(t)=\mathscr{L}^{-1}\left[\frac{e^{-s}}{(s+1)(s^2+1)}\right]$$Using the properties of the inverse Laplace transform, we can write this as:$$y(t)=e^{-t}\mathscr{L}^{-1}\left[\frac{1}{s+1}\right]+\mathscr{L}^{-1}\left[\frac{1}{s^2+1}\right]$$Finally, using the inverse Laplace transform of $\frac{1}{s+1} = te^{-t}$ and $\frac{1}{s^2+1}=\frac{1}{2}\sin(t)+\frac{1}{2}\cos(t)$, we have:$$y(t)=e^{-t}te^{-t}+\frac{1}{2}\sin(t)+\frac{1}{2}\cos(t)$$$$y(t)=e^{-2t}t+\frac{1}{2}\sin(t)+\frac{1}{2}\cos(t)$$