Given the following problem: $\dot{x}(t) = A(t)x(t)$, say A is two by two. We are given a solution $\xi_1$. How can I use this in order to find a second linearly independent solution? Is there a general method such as the one one would use when we have $y''(t) + p(t)y'(t) + q(t)y(t) = 0$ along with one solution $x_1(t)$, where our second solution comes from solving for $v(t)$ in $x_2(t) = v(t)x_1(t)$

I have an example I have been working on: A = \begin{bmatrix} 2t & 1+2t \\ 1-2t & -2t \end{bmatrix} and $\xi_1 = \begin{bmatrix}e^{-t} \\ -e^{-t}\end{bmatrix}$

I started by taking some function $v(t)$ and try to satisfy $\xi_2(t) = v(t)\xi_1(t)$, but then I find that $v(t)$ is some constant, whereby $\xi_2$ not linearly independent from the original solution. Any insight on how to approach this? Thank you!


Notice that a second order linear ODE $$ \tag{1} y''(t) + p(t) y'(t) + q(t) y(t) = 0 $$ corresponds to the system $$ \tag{2} x'(t) = \begin{bmatrix} 0 & 1 \\ -q(t) & -p(t) \end{bmatrix}, $$ and a solution $x_1(t)$ of $(1)$ corresponds to the solution $$ \begin{bmatrix} x_1(t) \\ x_1'(t) \end{bmatrix} $$ of $(2)$. So, multiplication of $x_1(t)$ by $v(t)$ corresponds to replacing the above vector function by $$ \begin{bmatrix} v(t) x_1(t) \\ v'(t) x_1(t) + v(t) x_1'(t) \end{bmatrix}. $$ Incidentally, your system can be easily solved: observe that the derivative of the sum of both coordinates equals the sum of the coordinates, and, after some (not very pleasant) calculation you obtain that the vector functions $$ \begin{bmatrix} e^{-t} \\ -e^{-t} \end{bmatrix}, \ \begin{bmatrix} t e^{t} \\ (1 - t) e^{t} \end{bmatrix} $$ form a fundamental system of solutions.

  • $\begingroup$ I have been trying to set up this problem, I am still confused as to what system is being solved. How do I set up my equation for $v(t)$ in the $\dot{x} = A(t)x$ case. $\endgroup$ – rannoudanames Mar 13 at 16:20
  • $\begingroup$ I did not use that method to solve the system. $\endgroup$ – user539887 Mar 13 at 19:06
  • $\begingroup$ Then do you perhaps mind elaborating your answer? I understand the logic you follow to detail the matrix representation of the ODE of the second degree, I do not see how that translates to the case I am trying to understand. I would be expecting for $\frac{d te^t}{dt} = (1-t)e^t$, where $v(t) = te^{2t}$ which is not the case $\endgroup$ – rannoudanames Mar 13 at 19:15
  • $\begingroup$ How come you have a negative in your exponential? Because the solution you provide above fits. $\endgroup$ – rannoudanames Mar 13 at 19:25
  • $\begingroup$ $$ \begin{bmatrix} te^{-t} \\ (1-t)e^{-t} \end{bmatrix} = \ \begin{bmatrix} t e^{t} \\ (1 - t) e^{t} \end{bmatrix} $$ is that what you are saying? $\endgroup$ – rannoudanames Mar 13 at 19:35

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