Problem source

The coordinates of a particle at time $t$ are $x$ and $y$. For $t \geq 0$, they satisfy the pair of coupled differential equations
\[ \begin{cases} \dot{x} &= -x -ky \\
\dot{y} &= x - y \end{cases}\]
where $k$ is a constant. When $t = 0$, $x = 1$ and $y = 0$.
\begin{questionparts}
\item Let $k = 1$. Find $x$ and $y$ in terms of $t$ and sketch $y$ as a function of $t$.
Sketch the path of the particle in the $x$-$y$ plane, giving the coordinates of the point at which $y$ is greatest and the coordinates of the point at which $x$ is least.
\item Instead, let $k = 0$. Find $x$ and $y$ in terms of $t$ and sketch the path of the particle in the $x$-$y$ plane.
\end{questionparts}

Solution source

\begin{questionparts}
\item Let $k = 1$, then 

\begin{align*}
\dot{x} &= - x - y \\
\dot{y} &= x - y \\
\dot{x}-\dot{y} &= -2x \\
\ddot{x} &= -\dot{x}-\dot{y} \\
&= -\dot{x} - (\dot{x}+2x) \\
&= -2\dot{x}- 2x \\
\dot{x}+\dot{y} &= -2y \\
\ddot{y} &= \dot{x}-\dot{y} \\
&= -2y-2\dot{y}
\end{align*}

So we have an auxiliary equation for $x$ and $y$ which is $\lambda^2 + 2 \lambda+2 = 0 \Rightarrow \lambda = -1 \pm i$.

Therefore $x = Ae^{-t} \cos t + B e^{-t} \sin t, y = Ce^{-t}\cos t + De^{-t} \sin t$. We also must have that, $A = 1, C = 0$, so $x = e^{-t} \cos t + Be^{-t} \sin t$ and $y = De^{-t} \sin x$.

\begin{align*}
\dot{y} &= -De^{-t} \sin t +De^{-t} \cos t \\
&= e^{-t} \cos x + Be^{-t} \sin t- De^{-t} \sin t \\
\end{align*}

therefore $B = 0, D = 1$ and $x = e^{-t} \cos t, y = e^{-t} \sin t$

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-0.2};
    \def\xu{7};
    \def\yl{-1};
    \def\yu{1};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=0:5, samples=100] 
            plot (\x, {exp(-\x)*sin(5*deg(\x))});
        \draw[thick, red, dashed, smooth, domain=0:5, samples=100] 
            plot (\x, {exp(-\x)});
        \draw[thick, red, dashed, smooth, domain=0:5, samples=100] 
            plot (\x, {-exp(-\x)});

        \node[blue] at (1, {exp(-1)*sin(deg(1))}) {$y=e^{-t}\sin t$};
    \end{scope}


    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$t$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

\begin{align*}
y &= e^{-t} \sin t \\
\dot{y} &= -e^{-t} \sin t + e^{-t} \cos t \\
\dot{x} &= e^{-t} \cos t -e^{-t} \sin t
\end{align*}


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-1.2};
    \def\xu{1.2};
    \def\yl{-1.2};
    \def\yu{1.2};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=0:5, samples=100] 
            plot ({exp(-\x)*cos(deg(\x))}, {exp(-\x)*sin(deg(\x))});

        \filldraw ({exp(-pi/4)*cos(45)}, {exp(-pi/4)*sin(45)}) circle (1pt) node[above] {$\left (\frac1{\sqrt{2}} e^{-\pi/4},\frac1{\sqrt{2}} e^{-\pi/4}\right)$};
        \filldraw ({exp(-3*pi/4)*cos(135)}, {exp(-3*pi/4)*sin(135)}) circle (1pt) node[left] {$\left (-\frac1{\sqrt{2}} e^{-3\pi/4},\frac1{\sqrt{2}} e^{-3\pi/4}\right)$};
        
    \end{scope}


    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$t$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

\item \begin{align*}
\dot{x} = -x \\
\dot{y} = x-y
\end{align*}

So $x = e^{-t}$. $\dot{y} + y = e^{-t}$ so $y = (t+B)e^{-t}$ and so $y  =te^{-t}$.


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-.2};
    \def\xu{1.2};
    \def\yl{-.2};
    \def\yu{1};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=0:5, samples=100] 
            plot ({exp(-\x)}, {(\x)*exp(-\x)});
        
    \end{scope}


    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$t$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

\end{questionparts}