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!
Best Answer
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.