Date due and handed in April 13,2010
\[ y^{\prime \prime }\left ( t\right ) +3y^{\prime }\left ( t\right ) +2y\left ( t\right ) =u\left ( t\right ) \] Using the Laplace approach. First we note that the input is a delayed step input, hence \(u\left ( t\right ) =\xi \left ( t-1\right ) \) where \(\xi \left ( t\right ) \) is the unit step function. Laplace transform of a delayed unit step is \({\displaystyle \int \limits _{0}^{\infty }} \xi \left ( t-1\right ) e^{-st}dt={\displaystyle \int \limits _{1}^{\infty }} e^{-st}dt=\frac{\left [ e^{-st}\right ] _{1}^{\infty }}{-s}=\frac{e^{-s}}{s}\)
Applying the Laplace transformation on the ODE gives
\begin{align} s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) +3sY\left ( s\right ) -y\left ( 0\right ) +2Y\left ( s\right ) & =\frac{e^{-s}}{s}\nonumber \\ s^{2}Y\left ( s\right ) -1+3sY\left ( s\right ) +2Y\left ( s\right ) & =\frac{e^{-s}}{s}\nonumber \\ Y\left ( s\right ) \left ( s^{2}+3s+2\right ) -1 & =\frac{e^{-s}}{s}\nonumber \\ Y\left ( s\right ) & =\frac{1}{s^{2}+3s+2}+\frac{e^{-s}}{s\left ( s^{2}+3s+2\right ) }\tag{1} \end{align}
Considering the first term on the RHS of (1), calling it \(Y_{1}\left ( s\right ) =\frac{1}{s^{2}+3s+2}\), and using partial fractions gives\begin{align*} Y_{1}\left ( s\right ) & =\frac{1}{\left ( s+1\right ) \left ( s+2\right ) }=\frac{A}{s+1}+\frac{B}{s+2}\\ A & =\lim _{s->-1}\frac{1}{\left ( s+2\right ) }=1\\ B & =\lim _{s->-2}\frac{1}{\left ( s+1\right ) }=-1 \end{align*}
Hence \[ Y_{1}\left ( s\right ) =\frac{1}{s+1}-\frac{1}{s+2}\] Considering the second term on the RHS, calling it \(Y_{2}\left ( s\right ) =\frac{e^{-s}}{s\left ( s^{2}+3s+2\right ) }\), and using partial fractions gives
\[ \frac{Y_{2}\left ( s\right ) }{e^{-s}}=\frac{1}{s\left ( s+1\right ) \left ( s+2\right ) }=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+2}\]\begin{align*} A & =\lim _{s->0}\frac{1}{\left ( s+1\right ) \left ( s+2\right ) }=\frac{1}{2}\\ B & =\lim _{s->-1}\frac{1}{s\left ( s+2\right ) }=-1\\ C & =\lim _{s->-2}\frac{1}{s\left ( s+1\right ) }=\frac{1}{2} \end{align*}
Hence\[ \frac{Y_{2}\left ( s\right ) }{e^{-s}}=\frac{1}{2}\frac{1}{s}-\frac{1}{s+1}+\frac{1}{2}\frac{1}{s+2}\] Therefore \begin{align} Y\left ( s\right ) & =Y_{1}\left ( s\right ) +Y_{2}\left ( s\right ) \nonumber \\ & =\left ( \frac{1}{s+1}-\frac{1}{s+2}\right ) +\left ( \frac{1}{2}\frac{e^{-s}}{s}-\frac{e^{-s}}{s+1}+\frac{1}{2}\frac{e^{-s}}{s+2}\right ) \tag{2} \end{align}
The effect of \(e^{-as}\) is to cause a time delay when finding the inverse Laplace transform.
\[ e^{-as}F\left ( s\right ) \rightarrow f\left ( t-a\right ) \xi \left ( t-a\right ) \] Now, taking the inverse Laplace transform of (2) gives the solution\begin{align*} y\left ( t\right ) & =e^{-t}\xi \left ( t\right ) -e^{-2t}\xi \left ( t\right ) +\frac{1}{2}\xi \left ( t-1\right ) -e^{-\left ( t-1\right ) }\xi \left ( t-1\right ) +\frac{1}{2}e^{-2\left ( t-1\right ) }\xi \left ( t-1\right ) \\ & =\left ( e^{-t}-e^{-2t}\right ) \xi \left ( t\right ) +\frac{1}{2}\left ( 1-2e^{-\left ( t-1\right ) }+e^{-2\left ( t-1\right ) }\right ) \xi \left ( t-1\right ) \end{align*}
