I need to show that all the integral curves for the vector field $V(x,y) = (y, -3y-2x)$ meet at the origin in order to solve a problem on PDEs (that you can find here).
I did the following:
$ \frac{dx}{dt} = y \\ \frac{dy}{dt} = -3y-2x $
Assuming proper smoothness conditions, we have:
$\frac{d^2y}{dt^2} = -3\frac{dy}{dt} - 2\frac{dx}{dt} = -3\frac{dy}{dt} - 2y$.
I solved it with Wolfram Mathematica, that gave $y(t) = c_1e^{-2t} + c_2e^{-t}$. One then easily proceeds to get $x(t) = -\frac{c_1}{2}e^{-2t} - c_2e^{-t} + c_3$.
However, when the input given to the software is the original system, the solution it gives is $x(t) = -c_1 e^{-2 t}+2 c_1 e^{-t}-c_2 e^{-2 t}+c_2 e^{-t}$, $y(t) = 2 c_1 e^{-2 t}-2 c_1 e^{-t}+2 c_2 e^{-2 t}-c_2 e^{-t}$.
Using the second, it is easily seen that $lim_{t \rightarrow \infty}(x(t), y(t)=(0,0)$, what then solves the problem. However, how can I explain the discrepancies?
Thank you very much.
I figured this one out. We want to solve the following system of linear ODEs:
$\frac{dx}{dt} = y \\ \frac{dy}{dt} = -3y -2x $
The matrix of coefficients is $ A = \left[ \begin{array}{cc} 0 & 1 \\ -2 & -3 \end{array} \right] $. It has eigenvalues -1 and -2 associated with the eigenvectors $(-a, a); a \in \mathbb{R}$ and $(b, -2b); b \in \mathbb{R}$. The general solution is then
$ x(t) = c_1e^{-t} + c_2e^{-2t} \\ y(t) = -c_1e^{-t} -2 c_2e^{-2t} $.