# Obtaining a second linearly independent solution in an ODE using a known one.

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.
• 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. – rannoudanames Mar 13 at 16:20
• 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 – rannoudanames Mar 13 at 19:15
• $$\begin{bmatrix} te^{-t} \\ (1-t)e^{-t} \end{bmatrix} = \ \begin{bmatrix} t e^{t} \\ (1 - t) e^{t} \end{bmatrix}$$ is that what you are saying? – rannoudanames Mar 13 at 19:35