I am studying for an ODE exam, and working on this problem.

Let $\displaystyle x_1 = \begin{bmatrix} 1 \\ t \\ \end{bmatrix}$ and $\displaystyle x_2 = \begin{bmatrix} t \\ 2t^2 \\ \end{bmatrix}$, and call by (H) the homogeneous differential equation

$$x' = A(t) x$$

where $A$ is a continuous $2 \times 2$ matrix-valued function of $t$. I want to show that $x_1$ and $x_2$ cannot both be solutions to (H). Here is how I have gone about it:

I know that $x_1$ and $x_2$ are linearly independent, so if they are both solutions to (H) then $W(t) \neq 0$ on the domain for $t$. And I've calculated the $W(t) = t^2$, where $W$ is the Wronskian of $\displaystyle \begin{bmatrix} 1 & t \\ t & 2t^2 \\ \end{bmatrix}$. But since I am not given a domain, couldn't both $x_1$ and $x_2$ be solutions of (H) as long as the domain for $t$ does not include $0$, as that would be the only point where $W(t) = 0$?

The only thing I can think to apply is that the maximum interval of existence of each of these solutions is $\mathbb{R}$, which includes $0$. Should I generally assume that if I am not given a domain, I want the maximum interval of existence?

(If anyone is wondering, this is a problem from Q. Kong's A Short Course in Ordinary Differential Equations. There is a second part but I am wondering the same thing about that part.)

Edit: I've confirmed that it should be assumed that $A(t)$ is continuous on $\mathbb{R}$.

  • 1
    $\begingroup$ Examine whether, and where, the difference $x_1-x_2$ satisfies your (linear) ODE. I also recommend using V. Arnol'd's "Ordinary Differential Equations", which explains the essential physico-geometric intuition behind ODE. $\endgroup$
    – user8960
    Commented Sep 27, 2016 at 20:59
  • $\begingroup$ I understand why that would be helpful, but I'm not sure how I'd go about checking that, since I'm just given a general $A(t)$. (And Kong's hint is to use the Wronskian of the matrix solution to show that it is impossible.) $\endgroup$
    – mathiest
    Commented Sep 27, 2016 at 21:18

1 Answer 1


You are correct: if $A$ is not required to be continuous at $0$, they are both solutions of the homogeneous linear system with $$ A(t) = \pmatrix{-1/t & 1/t^2\cr -2 & 3/t\cr}$$ which is $\Phi'(t) \Phi(t)^{-1}$, where $\Phi(t)$ is the $2 \times 2$ matrix with columns $x_1$ and $x_2$.

I think the question meant to assume that $A(t)$ is continuous on all of $\mathbb R$.

  • $\begingroup$ Thanks for your response; I figured that was the case, since it would make the most sense with what we have been discussing in class. $\endgroup$
    – mathiest
    Commented Sep 27, 2016 at 21:21

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