3.2.2.1.6 Example 6 \(y^{\prime \prime }+2y^{\prime }+y=0\)

\begin{align*} y^{\prime \prime }+2y^{\prime }+y & =0\\ y\left ( 0\right ) & =0 \end{align*}

Taking Laplace transform gives

\begin{align*} \left ( s^{2}Y-sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +2\left ( sY-y\left ( 0\right ) \right ) +Y & =0\\ \left ( s^{2}Y-y^{\prime }\left ( 0\right ) \right ) +2sY+Y & =0 \end{align*}

Since one IC is missing, then let \(y^{\prime }\left ( 0\right ) =c_{2}\). The above becomes

\begin{align*} \left ( s^{2}Y-c_{2}\right ) +2sY+Y & =0\\ Y\left ( s^{2}+2s+1\right ) -c_{2} & =0\\ Y & =\frac {c_{2}}{s^{2}+2s+1}\end{align*}

Applying inverse Laplace transform gives

\begin{equation} y\left ( t\right ) =c_{2}te^{-t} \tag {1}\end{equation}

We can if we want, now replace \(c_{2}=y^{\prime }\left ( 0\right ) \) to make it more clear what the \(c_{2}\) represents.

\begin{equation} y\left ( t\right ) =y^{\prime }\left ( 0\right ) te^{-t} \tag {2}\end{equation}