For a homogeneous 2nd-order ODE like


I have a system of ODE's from an (unknown) 2nd-order ODE:

$$ \begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} 2 & -5 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$

The eigenvalues are $\pm i$ and one of the eigenvectors is

$$k=\begin{pmatrix} 5 \\ 2-i \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} +i \begin{pmatrix} 0 \\ -1 \end{pmatrix} = a+ib$$

The general solution for complex roots is

$$x = c_1 e^{\lambda t} (a \cos(\mu t)-b \sin(\mu t)) + c_2 (a \sin(\mu t)+b \cos(\mu t))$$

where $\lambda$ is the real part of the root and $\mu$ is the complex part. Substituting the values in I get

$$ \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = c_1 \begin{pmatrix} 5 \cos(t) \\ 2\cos(t)+\sin(t) \end{pmatrix} + c_2 \begin{pmatrix} 5 \sin(t) \\ 2 \sin(t)-\cos(t) \end{pmatrix}$$

From the original 2nd-order ODE above, $x_1 = y$ and $x_2 = y'$ so I can write

$$y = c_1 5 \cos(t) + c_2 5 \sin(t)$$


$$y'=c_1 (2 \cos(t)+\sin(t)) + c_2 (2 \sin(t)-\cos(t))$$

However, differentiating the above expression for $y$ does not give the obtained expression for $y'$ except for the trivial case of $c_1 = c_2 = 0$. Why is this?

In contrast, when the roots are real, for example


can be written as

$$ \begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$

This system matrix has two real eigenvalues $-1$ and $3$ with eigenvectors $(1,3)$ and $(-1,1)$ giving a final solution of

$$ \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 3 \end{pmatrix} e^{3t} + c_2 \begin{pmatrix} -1 \\ 1 \end{pmatrix} e^{-t}$$

and here, $x_1'=x_2$ as expected.


As you have seen in your example, if the vectorization to the first order system is via $x=(y,y')^T$ then the top-left matrix entry has to be zero.

You can reverse-engineer a second order equation from the system, easiest starting with the second equation $$ x_1=x_2'+2x_2 $$ which can be inserted into the first equation to get $$ x_2''+2x_2'=x_1'=2(x_2'+2x_2)-5x_2 $$ and sorted $$ x_2''+x_2=0. $$

Of course you also get $x_1''+x_1=0$ which is then complemented by $x_2=\frac15(2x_1-x_1')$

  • $\begingroup$ Thank you for the answer, and also for explaining clearly how to reverse-engineer such an equation! $\endgroup$ – Gerhard Mar 5 '17 at 10:17

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