When \(k = 1\), \[ \dot {x} = -x - y, \dot {y} = x - y. \]
Hence, \begin {align*} \ddot {x} & = - \dot {x} - \dot {y} \\ & = - \dot {x} - (x - y) \\ & = -\dot {x} - x + y \\ & = -\dot {x} - x + (-x - \dot {x}) \\ & = -2\dot {x} - 2x, \end {align*}
and this gives \[ \ddot {x} + 2\dot {x} + 2x = 0. \]
The auxiliary equation to this differential equation is \[ \lambda ^2 + 2\lambda + 2 = 0, \] which solves to \[ \lambda = -1 \pm i. \]
The general solution for \(x\) is hence \[ x(t) = \exp (-t) \left (A \sin t + B \cos t\right ). \]
This means \begin {align*} \dot {x}(t) & = - \exp (-t) \left (A \sin t + B \cos t\right ) + \exp (-t) \left (A \cos t - B \sin t\right ) \\ & = -x(t) + \exp (-t) \left (A \cos t - B \sin t\right ), \end {align*}
and hence \[ y(t) = - \exp (-t) \left (A \cos t - B \sin t\right ) = \exp (-t) \left (B \sin t - A \cos t\right ). \]
When \(t = 0\), \(x = x(0) = B = 1\), \(y = y(0) = -A = 0\). Hence, \[ x(t) = \exp (-t) \cos t, y(t) = \exp (-t) \sin t. \]
The graph of \(y\) against \(t\) looks as follows:
\(y\) is greatest at the first stationary point of \(y\), as shown in the graph. Note that \[ \dot {y} = x - y = \exp (-t) \left (\cos t - \sin t\right ), \] and hence \[ \dot {y} = 0 \iff \cos t = \sin t \iff \tan t = 1, \] and the smallest positive solution to this is \(t = \frac {\pi }{4}\). The coordinate of the point is hence \[ (x, y) = \left (\exp \left (-\frac {\pi }{4}\right ) \cdot \frac {\sqrt {2}}{2}, \exp \left (-\frac {\pi }{4}\right ) \cdot \frac {\sqrt {2}}{2}\right ). \]
Similarly, the graph of \(x\) against \(t\) looks as follows:
\(x\) is smallest at the first stationary point of \(x\), as shown in the graph. Note that \[ \dot {x} = -x - y = -\exp (-t) \left (\cos t + \sin t\right ), \] and hence \[ \dot {x} = 0 \iff \cos t = - \sin t \iff \tan t = -1, \] and the smallest positive solution to this is \(t = \frac {3\pi }{4}\). The coordinate of the point is hence \[ (x, y) = \left (- \exp \left (-\frac {3\pi }{4}\right ) \cdot \frac {\sqrt {2}}{2}, \exp \left (-\frac {3\pi }{4}\right ) \cdot \frac {\sqrt {2}}{2}\right ). \]
Without the \(\exp (-t)\) factor, the \(x\)-\(y\) graph will simply be a circle, and with this factor, it will be a spiral with exponentially decreasing radius. This is the polar curve \(r = \exp (-\theta )\). Hence, the \(x\)-\(y\) graph looks as follows.
Since \(\dot {x} = -x\), we must have \(x(t) = A \exp (-t)\), and since \(x(0) = 1\), we have \(A = 1\) and \(x(t) = \exp (-t)\).
We have \[ \dot {y} = \exp (-t) - y, \] and hence \[ \dot {y} + y = \exp (-t). \]
Multiplying both sides by \(\exp (t)\), we have \[ e^t \dot {y} + e^t y = 1, \] and hence \[ \DiffFrac {ye^t}{t} = 1, \] which gives \[ ye^t = t + B, \] and hence \[ y = \exp (-t) (t + B). \]
Since \(y = 0\) when \(t = 0\), we must have \(B = 0\), and hence \[ y = t \exp (-t). \]
Note that \[ \DiffFrac {y}{t} = \exp (-t) - t \exp (-t), \] and hence \(\DiffFrac {y}{t} = 0\) when \(t = 1\), which is when \[ (x, y) = \left (e^{-1}, e^{-1}\right ). \]
Note that \[ \DiffFrac {x}{y} = - \exp (-t), \] and hence \(\DiffFrac {x}{t} = 0\) when \(t = 0\), which is when \[ (x, y) = (1, 0), \] and the tangent to the curve at this point will be vertical.
Hence, the graph will look as follows:
