Problem source

Find the general solution of the differential equation  $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{xy}{x^2+a^2}\;$,  where $a\ne0\,$, and show that it can be written in the form $\displaystyle y^2(x^2+a^2)= c^2\,$, where $c$ is an arbitrary constant.
Sketch this curve. 
 
Find an expression for $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (x^2+y^2)$ and  show that
\[
 \frac{\mathrm{d^2}}{\mathrm{d}x^2} (x^2+y^2) = 
2\left(1 -\frac {c^2}{(x^2+a^2)^2} \right) + \frac{8c^2x^2}{(x^2+a^2)^3}\;.
\]
\begin{questionparts}
\item Show that, if $0 < c < a^2$, the points on the curve whose distance from the origin is least are $\displaystyle \l 0\,,\;\pm \frac{c}{a}\r$. 
\item If $c > a^2$, determine the points on the curve whose distance from the origin is least. 
\end{questionparts}

Solution source

\begin{align*}
&& \frac{\d y}{\d x} &= - \frac{xy}{x^2+a^2} \\
\Rightarrow && \int \frac{1}{y} \d y &= \int -\frac{x}{x^2+a^2} \d x \\
&& \ln |y| &= -\frac12 \ln |x^2 + a^2| + C \\
\Rightarrow && C' &= \ln y^2 + \ln (x^2+a^2) \\
\Rightarrow && c^2 &= y^2(x^2+a^2)
\end{align*}

(where the final constant $c^2$ can be taken as a square since it is clearly positive).

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sqrt(1/((#1)^2+4))};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-.8};
    \def\yu{.8};
    
    % 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[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {-\functionf(\x)});
        \node[blue, above, rotate=-20] at (3, {\functionf(3)}) {\tiny $y^2(x^2+a^2) = c^2$}; 
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

\begin{align*}
&& \frac{\d }{\d x} \left (x^2 + y^2 \right) &= 2x - \frac{2xy^2}{x^2+a^2} \\
&&&=2x - \frac{2x c^2}{(x^2+a^2)^2} \\
&&&= 2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \\
\\
&& \frac{\d ^2}{\d x^2} \left (x^2 + y^2 \right) &= \frac{\d }{\d x} \left (2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right)  \right) \\
&&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + 2x \left (\frac{2c^2 \cdot 2x}{(x^2+a^2)^3} \right) \\
&&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + \frac{8x^2c^2 }{(x^2+a^2)^3} \\
\end{align*}


\begin{questionparts}
\item The shortest distance from the origin will have the first derivative as $0$, ie $x = 0$ or $x^2 + a^2 = c$, but if $c  < a^2$ this can only occur for $x = 0$, so the closest to the origin is $(0, \pm \frac{c}{a})$

\item If $c > a^2$ then we can have $x = 0$ or $x = \pm \sqrt{c-a^2}$. Looking at the second derivative, when $x = 0$ we have $2(1- \frac{c^2}{a^4}) < 0$ which is a local maximum. When $x = \pm\sqrt{c-a^2}$ we have $8(c-a^2)c^2/c^3 > 0$ which is the minimum, therefore the points are $(\pm \sqrt{c-a^2}, c)$
\end{questionparts}


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sqrt(1/((#1)^2+1))};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-5};
    \def\yu{5};
    
    % 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);


        \foreach \a in {1, ..., 20} {
            \draw[thin, green, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\a/3*\functionf(\x)});
        \draw[thin, green, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {-\a/3*\functionf(\x)});
        };
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {4*\functionf(\x)});
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {-4*\functionf(\x)});

        \foreach \a in {1, ..., 20} {
            \draw[red, dashed, thin] (0,0) circle ({\a/2});
        };

%         \foreach \x in {0,1,2,3,4,5} {
% % %     \draw (\x,0.1) -- (\x,-0.1) node[below] {$x_{\x}$};
% % %   }
        \node[blue, above, rotate=-20] at (3, {4*\functionf(3)}) {\tiny $y^2(x^2+a^2) = c^2$}; 
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}