Conditions on initial value for system of ODEs

Question

Given $\dot{x} = x - y + e^{-t}$ and $\dot{y} = x+y + e^{-t}$. Find the set of initial conditions at $t = 0$ s.t $\lim_{t\rightarrow \infty} (x(t), y(t)) = (0, 0)$ whenever $(x(t), y(t))$ satisfies one of these initial conditions.

My progress: After using the "Variation Of Parameters" formula (aka, solutions to the ODE: $\dot{x} = A(t)x + g(t,x)$, have the form: $\phi(t) = \gamma(t)\gamma^{-1}(t_0)x_0 + \gamma(t)\int_{t_0}^{t_1} \gamma^{-1}(s)g(\phi(s), s)ds$, where $\gamma(t) =$ fundamental matrix solution to $\dot{x} = Ax$). Now, by letting $z_1 = x, z_2 = y$, and doing a bunch of algebraic manipulations, I ended up with the solution:

$z(t) = [z_{01}cos(t) - sin(t)(z_{02} + \frac{2(i-1)}{5}), sin(t)(z_{01}+\frac{1-i}{5})+ z_{02}cos(t)]$ where $z(0) = [z_{01} \ z_{02}]$.

Hopefully this is correct, as I checked over my computations several times. If that's the case, we see that for $\lim_{t\rightarrow \infty} (x(t), y(t)) = (0, 0)$, we need: $\lim_{t\rightarrow \infty} z(t) = (0, 0)$. But as $t\rightarrow \infty$, $\cot(t)$ oscillates between $(-\infty, \infty)$, and $z_{01}, z_{02}$ are some fixed numbers at $t = 0$, how can $\frac{z_{02} \ +\ \frac{2(i-1)}{5}}{z_{01}} = \cot{t}$ in that case? We cannot really let $z_{01} = 0, z_{02} = \frac{-2(i-1)}{5}$ as well, because then the 2nd component of $z(t)$ is not satisfies.

From these observations, I think there doesn't really exist any set of initial conditions that gives $\lim_{t\rightarrow \infty} (x(t), y(t)) = (0, 0)$, which is weird:P

My question: Can anyone please give this problem a try to see if my conclusion above is correct? Any help would really be appreciated.

Answer 1

Your calculations are obviously wrong, because you shouldn't have imaginary part in your solution. See that

$$e^{At} = e^{t} \begin{bmatrix} \cos t & -\sin t \\\sin t & \cos t \end{bmatrix}$$

Now, observe that

$$z(t) = e^{At} z(0) + \int_0^t e^{A(t-s)} \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{-s} ds$$

The integral part will contain the exponential term $e^{-t}$ after integrating, so it will go to $0$ at infinity. Since, $\lim e^{At} = \infty$, the only choice is $z(0) = 0$.

Answer 2

I am not totally sure about your solution, however I will give you a few hints.

Notice that this is nonhomogeneous linear differential equation, so the general solution would contain a linear combination of the solutions to the homogeneous differential equation and the particular solution, which satisfies the nonhomogeneity.

Do you have to find the complete solutions to homogeneous differential equation? Because only the solution corresponding to eigenvalues with negative real parts will disappear at $t\to\infty$.