How does one solve a system of ODEs of the type $$ \ddot{x}=Ax $$ without reducing it to a system of four 1st-order equations?
2026-04-08 17:46:33.1775670393
Solving a 2nd-order system of ODEs
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As usual, bring $A$ into Jordan normal form $A=T^{-1}JT$. Then using $y=Tx$ the equation transforms to $\ddot y= Jy$ and you can treat each Jordan block separately. If the dimension is $1$, then $\ddot y=λy$ has a standard solution. For blocks of higher dimension you get coupled equations, for instance in dimension 2 \begin{align} \ddot y_1&=λy_1+y_2\\ \ddot y_2&=λy_2 \end{align} that can be solved in reverse order.