Problem source

Show that if
\[
{\mathrm{d}y \over \mathrm{d} x}=\f(x)y + {\g(x) \over y}
\]
then the substitution $u = y^2$ gives a linear differential equation for $u(x)\,$.
Hence or otherwise solve the differential equation
\[
{\mathrm{d}y \over \mathrm{d} x}={y \over x} - {1 \over y}\;.
\]
Determine the solution curves of this equation which pass through $(1 \,,  1)\,$, $(2\, ,  2)$ and $(4 \, , 4)$ and sketch graphs of all three curves on the same axes.

Solution source

\begin{align*}
&& \frac{\d y}{\d x} &= f(x) y + \frac{g(x)}{y} \\
&& y \frac{\d y}{\d x} &= f(x) y^2 + g(x) \\
u = y^2: && \frac12 \frac{\d u}{\d x} &= f(x) u + g(x)
\end{align*}

Which is a linear differential equation for $u$.

\begin{align*}
&& \frac12 u' &= \frac1x u -1 \\
\Rightarrow && u' - \frac2xu &= -1 \\
\Rightarrow && \frac{1}{x^2} u' - \frac{2}{x^3} u &= -\frac{1}{x^2} \\
\Rightarrow && (\frac{u}{x^2})' &= - \frac{1}{x^2} \\
\Rightarrow && \frac{u}{x^2} &= \frac1x + C \\
\Rightarrow && u &= Cx^2 + x \\
\Rightarrow && y^2 &= Cx^2 + x
\end{align*}

If $(1,1)$ is on the curve then $1 = C + 1 \Rightarrow C = 0 \Rightarrow y^2 = x$.

If $(2,2)$ is on the curve then $4 = 4C + 2 \Rightarrow C = \frac12 \Rightarrow y^2 = x + \frac12 x^2$.

If $(3,3)$ is on the curve then $9 = 9C + 3 \Rightarrow C = \frac23 \Rightarrow y^2 = x + \frac23 x^2$

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){((1+(#1))^5-(1-(#1))^5)/((1+(#1))^5+(1-(#1))^5)};
    \def\functiong(#1){((1+(#1))^5+(1-(#1))^5)/((1+(#1))^5-(1-(#1))^5)};
    \def\xl{-2};
    \def\xu{10};
    \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
    }
    % 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=-5:5, samples=100] 
            plot ({(\x)^2}, {\x});
        \draw[thick, green, smooth, domain=0:10, samples=100] 
            plot ({\x}, {sqrt(\x + 0.5*(\x)^2)});
        \draw[thick, green, smooth, domain=0:10, samples=100] 
            plot ({\x}, {-sqrt(\x + 0.5*(\x)^2)});
        \draw[thick, red, smooth, domain=0:10, samples=100] 
            plot ({\x}, {sqrt(\x + 2/3*(\x)^2)});
        \draw[thick, red, smooth, domain=0:10, samples=100] 
            plot ({\x}, {-sqrt(\x + 2/3*(\x)^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}