Problem source

Find all the solution curves of the differential equation
\[
y^4 \l {\mathrm{d}y \over \mathrm{d}x }\r^{\! \! 4} = \l y^2 - 1 \r^2
\]
that  pass through either of the points
\begin{questionparts}
\item $\l 0, \, \frac{1}{2}\sqrt3 \r$,
\item $\l 0, \, \frac{1}{2}\sqrt5 \r$.
\end{questionparts}
Show also that $y = 1$ and $y = -1$ are solutions of the differential equation.
Sketch all these solution curves on a single set of axes.

Solution source

\begin{align*}
&& y^4 \left (\frac{\d y}{\d x} \right)^4 &= (y^2 - 1)^2 \\
\Rightarrow &&  y^2 \left (\frac{\d y}{\d x} \right)^2 &= |y^2 - 1| \\
&& y \left (\frac{\d y}{\d x} \right) &= \pm \sqrt{|y^2-1|} \\
\Rightarrow &&\int \frac{y}{\sqrt{|y^2-1|}} \d y &= \int \pm 1 \d x \\
\Rightarrow && \pm \sqrt{|y^2-1|} &= \pm x + C \\
\end{align*}



\begin{questionparts}
\item Since $y^2 < 1$, our solution curve should be of the from $-\sqrt{1-y^2} = \pm x + C$
Plugging in $(0, \tfrac12 \sqrt{3})$, we obtain $-\tfrac12 = C$, therefore our solution curves are

$\pm x = \frac12 - \sqrt{1-y^2}$

\item  Since $y^2 > 1$, our solution curve should be of the from $\sqrt{y^2-1} = \pm x + C$
Plugging in $(0, \tfrac12 \sqrt{5})$, we obtain $\tfrac12 = C$, therefore our solution curves are

$\pm x = \sqrt{y^2-1}-\frac12$
\end{questionparts}

Clearly if $y = \pm 1$, $y'=0$ and the equation is satisfied.


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1.5*sqrt(#1)-sqrt((#1)-1)};
    \def\xl{-2};
    \def\xu{2};
    \def\yl{-2};
    \def\yu{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] (\xl, 1) -- (\xu, 1);
        \draw[thick, blue, smooth] (\xl, -1) -- (\xu, -1);

        \draw[thick, blue, smooth, domain=-1:1, samples=100] 
            plot ({0.5-sqrt(1-\x*\x)}, {\x});
        \draw[thick, blue, smooth, domain=-1:1, samples=100] 
            plot ({-(0.5-sqrt(1-\x*\x))}, {\x});
        \draw[thick, blue, smooth, domain=1:\yu, samples=100] 
            plot ({sqrt(\x*\x-1)-0.5}, {\x});
        \draw[thick, blue, smooth, domain=1:\yu, samples=100] 
            plot ({-(sqrt(\x*\x-1)-0.5)}, {\x});
        % \draw[thick, blue, smooth, domain=\yl:-1, samples=100] 
        %     plot ({sqrt(\x*\x-1)-0.25}, {\x});
        % \draw[thick, blue, smooth, domain=\yl:-1, samples=100] 
        %     plot ({-(sqrt(\x*\x-1)-0.25)}, {\x});
        
    \end{scope}

    \filldraw (0, {sqrt(3/4)}) circle (1.5pt);
    \filldraw (0, {sqrt(5/4)}) circle (1.5pt);

    
    % 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}