3.2.2.1.5 Example 5

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

This problem shows what to do when one IC is missing. Basically, if an IC is missing, it is just kept unknown. 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 given, then we let the missing IC be some unknown. In this case \(y\left ( 0\right ) =c_{1}\). And continue as before. The above becomes

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

Applying inverse Laplace transform gives

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

We can if we want, now replace \(c_{1}=y\left ( 0\right ) \) to make it more clear what the \(c_{1}\) represents.

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