# Solving a 2nd-order system of ODEs

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?

• How about good old ansatz $v e^{\lambda t}$? Should work if $A$ is constant. – Evgeny Nov 16 '16 at 9:41

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.