Problem source

For $x \ge 0$ the curve $C$ is defined by 
$$
{\frac{\d y} {\d x}} = \frac{x^3y^2}{(1 + x^2)^{5/2}}
$$
with $y = 1$ when $x=0\,$. Show that  
\[
\frac 1 y = \frac {2+3x^2}{3(1+x^2)^{3/2}} +\frac13
\]
and hence that for large positive $x$ 
$$
y \approx 3 - \frac 9 x\;.
$$ 
Draw a sketch of $C$.
On a separate diagram, draw a sketch of the two curves defined for $x \ge 0$ by
$$
\frac {\d z} {\d x} = \frac{x^3z^3}{2(1 + x^2)^{5/2}}
$$
with $z = 1$ at $x=0$ on one curve, and $z = -1$ at $x=0$ on the other.

Solution source

\begin{align*}
&& {\frac{\d y} {\d x}} &= \frac{x^3y^2}{(1 + x^2)^{5/2}} \\
\Rightarrow &&\int \frac{1}{y^2} \d y &= \int \frac{x^3}{(1+x^2)^{5/2}} \d x \\
\Rightarrow && -\frac1y  &= \int \frac{x^3+x-x}{(1+x^2)^{5/2}} \d x \\
&&&= \int \left ( \frac{x}{(1+x^2)^{3/2}}-\frac{x}{(1+x^2)^{5/2}} \right) \d x \\
&&&= \frac{-1}{(1+x^2)^{1/2}} + \frac{1}{3(1+x^2)^{3/2}} + C \\
&&&= \frac{1-3(1+x^2)}{3(1+x^2)^{3/2}} + C \\
&&&= \frac{-3x^2-2}{3(1+x^2)^{3/2}} + C \\
(x,y) = (0,1): &&-1 &= -\frac23 + C \\
\Rightarrow && C &= -\frac13 \\
\Rightarrow && \frac1y &=  \frac{3x^2+2}{3(1+x^2)^{3/2}} + \frac13
\end{align*}

\begin{align*}
y &= \frac{1}{\frac13 +\frac{3x^2+2}{3(1+x^2)^{3/2}} } \\
&= \frac{3}{1+ \frac{3x^2+2}{3(1+x^2)^{3/2}}} \\
&= \frac{3}{1+ \frac{3}{x} + \cdots} \\
&\approx 3 - \frac{9}{x} 
\end{align*}


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/(1/3 + (3*(#1)^2+2)/(3*(1+(#1)^2)^(3/2))))};
    \def\xl{-5};
    \def\xu{25};
    \def\yl{-1};
    \def\yu{4};
    
    % 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
    }
    % 3 - 9/x + 27/x^2 - 147/(2 x^3) + 198/x^4 - 4275/(8 x^5) + 1443/x^6 - 62343/(16 x^7) + 10521/x^8 - 3636267/(128 x^9) + O((1/x)^10)
% (Laurent series)
    % 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=0:17.5, samples=100] 
            plot (\x, {\functionf(\x)});  

        \draw[thick, blue, smooth, domain=0.01:0.1, samples=100]
            plot(1/\x, {3 - 9*\x + 27*\x^2 - 147/2 *\x^3 + 198*\x^4-4275/8 * \x^5 });
        \draw[thick, red, dashed, smooth, domain=0.1:25, samples=100] 
            plot (\x, {3-9/\x});  
        % \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
        % \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        \draw[red] (25,3.25) node [below left] {$y = 3-\frac9x$};
    \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 z} {\d x} &= \frac{x^3z^3}{2(1 + x^2)^{5/2}} \\
\Rightarrow && \int \frac{1}{z^3} \d z &= \int \frac{x^3}{2(1+x^2)^{5/2}} \\
&& -\frac{1}{2z^2} &= -\frac{3x^2+2}{3(1+x^2)^{3/2}} - C \\
(x,z) = (0, \pm 1): && \frac{1}{2} &= \frac{2}{3} + C \\
\Rightarrow && C &= -\frac16 \\
\Rightarrow && \frac{1}{z^2} &= \frac{6x^2+4}{3(1+x^2)^{3/2}} - \frac13
\end{align*}

So as $x \to \infty$ $z \sim \pm (3 + \frac{2}{x} + \cdots)$ and so:

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sqrt(1/(((6*(#1)^2+4)/(3*(1+(#1)^2)^(3/2)))-1/3))};
    \def\xl{-1};
    \def\xu{6};
    \def\yl{-20};
    \def\yu{20};
    
    % 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
    }
    % 3 - 9/x + 27/x^2 - 147/(2 x^3) + 198/x^4 - 4275/(8 x^5) + 1443/x^6 - 62343/(16 x^7) + 10521/x^8 - 3636267/(128 x^9) + O((1/x)^10)
% (Laurent series)
    % 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=0:5.8, samples=100] 
            plot (\x, {\functionf(\x)});  
        \draw[thick, blue, smooth, domain=0:5.8, samples=100] 
            plot (\x, {-\functionf(\x)});  

        % \draw[thick, blue, smooth, domain=0.01:0.1, samples=100]
            % plot(1/\x, {3 - 9*\x + 27*\x^2 - 147/2 *\x^3 + 198*\x^4-4275/8 * \x^5 });
        % \draw[thick, red, dashed, smooth, domain=0.1:25, samples=100] 
        %     plot (\x, {3-9/\x});  
        % \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
        % \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        \draw[red] (25,3.25) node [below left] {$y = 3-\frac9x$};
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$z$};
    
    \end{tikzpicture}
\end{center}