Section 9.8 Solving higher-order IVPs
So far we only addressed the problem of solving first-order IVPs. Bad news: some of the most important IVPs in applications are second-order, namely involve second derivatives. Good news: IVPs of any order can be reduced to first-order IVPs. Of course there is a price to pay: while reformulating an ODE of order 2 or more as a first-order ODE, we need to introduce a new ODE each time we lower the order by one. We show below in two examples how to proceed. A second order ODE. The ODE describing the motion of a mass attached to a spring is
\begin{equation*}
\ddot x = -kx,\;x(0)=x_0,\;\dot x(0) = v_0.
\end{equation*}
In order to re-write this ODE as a first-order one, we set
\begin{equation*}
\dot x = v.
\end{equation*}
Then, the original ODE becomes now the system of 2 first-order ODEs
\begin{equation*}
\begin{cases}
\dot x = v,\; x(0)=x_0\\
\dot v = kx,\;v(0)=v_0
\end{cases}.
\end{equation*}