3.2.2.1.7 Example 7 \(y^{\prime \prime \prime }+y^{\prime \prime }+y^{\prime }+y=0\)

This example is for higher order ode, showing how to easily handle IC if at zero or not or if some missing or not, all using same process. Given

\begin{equation} y^{\prime \prime \prime }+y^{\prime \prime }+y^{\prime }+y=0 \tag {1}\end{equation}

And lets say the IC’s given are

\begin{align*} y\left ( 1\right ) & =a\\ y^{\prime \prime }\left ( 0\right ) & =b \end{align*}

The idea is to always use \(c_{0},c_{1},c_{2}\) for \(y\left ( 0\right ) ,y^{\prime }\left ( 0\right ) ,y^{\prime \prime }\left ( 0\right ) \) and then at the very end solve for these from the given initial conditions. We will get two equations (since we only have 2 IC) and 3 unknowns. So some of the \(c_{0},c_{1},c_{2}\) will remain in the solution as unknowns which is OK. Applying Laplace transform on (1) gives

\[ s^{3}Y-y^{\prime \prime }\left ( 0\right ) -sy^{\prime }\left ( 0\right ) -s^{2}y\left ( 0\right ) +s^{2}Y-y^{\prime }\left ( 0\right ) -sy\left ( 0\right ) +sY-y\left ( 0\right ) +Y=0 \]

We now replace \(y^{\prime \prime }\left ( 0\right ) =c_{2},y^{\prime }\left ( 0\right ) =c_{1},y\left ( 0\right ) =c_{0}\) and simplify the above which becomes

\begin{align*} Y\left ( s^{3}+s^{2}+s+1\right ) -c_{2}-sc_{1}-s^{2}c_{0}-c_{1}-sc_{0}-c_{0} & =0\\ Y\left ( s^{3}+s^{2}+s+1\right ) -s^{2}c_{0}-s\left ( c_{1}+c_{0}\right ) -c_{2}-c_{1}-c_{0} & =0\\ Y & =\frac {c_{2}+c_{1}+c_{0}+s\left ( c_{1}+c_{0}\right ) +s^{2}c_{0}}{s^{3}+s^{2}+s+1}\end{align*}

Taking inverse Laplace gives the solution as

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

Now we only need to solve for the constants using the given initial conditions. This results in these two equations (since we have 2 IC only). Using \(y\left ( 1\right ) =a\) gives

\begin{equation} a=\frac {1}{2}\left ( c_{0}-c_{2}\right ) \cos \left ( 1\right ) +\frac {1}{2}e^{-1}\left ( c_{0}+c_{2}\right ) +\frac {1}{2}\sin \left ( 1\right ) \left ( c_{0}+2c_{1}+c_{2}\right ) \tag {3}\end{equation}

Taking derivatives twice of (2) and using \(y^{\prime \prime }\left ( 0\right ) =b\) gives the second equation

\[ y^{\prime \prime }\left ( t\right ) =\frac {1}{2}e^{-t}\left ( c_{0}+c_{2}\right ) +\frac {1}{2}\left ( -c_{0}-2c_{1}-c_{2}\right ) \sin \left ( t\right ) -\frac {1}{2}\left ( c_{0}-c_{2}\right ) \cos \left ( t\right ) \]

Using \(y^{\prime \prime }\left ( 0\right ) =b\) the above gives

\begin{align} b & =\frac {1}{2}\left ( c_{0}+c_{2}\right ) -\frac {1}{2}\left ( c_{0}-c_{2}\right ) \nonumber \\ & =c_{2} \tag {4}\end{align}

Now we need to solve (3,4) for \(c_{0},c_{1},c_{2}\). From (4) we see that \(c_{2}=b\). Substituting this into (3) gives

\begin{equation} a=\frac {1}{2}\left ( c_{0}-b\right ) \cos \left ( 1\right ) +\frac {1}{2}e^{-1}\left ( c_{0}+b\right ) +\frac {1}{2}\sin \left ( 1\right ) \left ( c_{0}+2c_{1}+b\right ) \tag {5}\end{equation}

We can now choose the free parameter as \(c_{0}\), hence

\[ c_{1}=-\frac {1}{2\sin \left ( 1\right ) }\left ( \cos \left ( 1\right ) +e^{-1}+\sin \left ( 1\right ) \right ) c_{0}+\frac {1}{2\sin \left ( 1\right ) }\left ( b\cos \left ( 1\right ) -be^{-1}-b\sin \left ( 1\right ) +2a\right ) \]

We are done. The solution (2) is now found by replacing \(c_{2},c_{1}\) into it. \(c_{0}\) remains are the only unknown.  This method works for any combination of IC given even if some at zero or not.