3.2.2.1.1 Example 1 \(y^{\prime \prime }+2y^{\prime }+y=0\)

\begin{align*} y^{\prime \prime }+2y^{\prime }+y & =0\\ y\left ( 1\right ) & =2\\ y^{\prime }\left ( 0\right ) & =2 \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-sy\left ( 0\right ) -2\right ) +\left ( 2sY-2y\left ( 0\right ) \right ) +Y & =0 \end{align*}

Since not all initial conditions are at zero, and we need to have them at zero to use Laplace, then one way is to let \(y\left ( 0\right ) =y_{0}\) as unknown (we could also have used \(y\left ( 0\right ) =c_{1}\)). Find the solution, then solve for \(y_{0}\) using the initial condition \(y\left ( 1\right ) =2\). This shows how it is done. The above becomes

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

Applying inverse Laplace transform gives

\begin{equation} y\left ( t\right ) =\left ( y_{0}+2t+y_{0}t\right ) e^{-t} \tag {1}\end{equation}

But \(y\left ( 1\right ) =2\) hence

\begin{align*} 2 & =\left ( y_{0}+2+y_{0}\right ) e^{-1}\\ 2e & =2y_{0}+2\\ y_{0} & =e-1 \end{align*}

Therefore (1) becomes

\begin{align*} y\left ( t\right ) & =\left ( e-1+2t+\left ( e-1\right ) t\right ) e^{-t}\\ & =e^{-t}\left ( -1+e+t+et\right ) \end{align*}