<s>[INST] Find the general solution to the ODE $x\frac{dy}{dx}=y-\frac{1}{y}$ I have been working through an ODE finding the general solution and following the modulus through the equation has left me with four general solutions, as shown below. Online ODE solvers, however, have only calculated the two (positive and negative), on the right hand side of the page. Am I incorrect with my use of the modulus operator after I integrated each side then?
$$\begin{align*}x\frac{dy}{dx}&=y-\frac{1}{y}\quad\quad(-1<y<1)\\
\int\frac{1}{y-\frac{1}{y}}\ dy&=\int\frac{1}{x}\ dx\\
\int\frac{y}{y^2-1}\ dy&=\int\frac{1}{x}\ dx\\
\frac{1}{2}\ln{|y^2-1|}&=\ln{|x|}+c\\
\sqrt{|y^2-1|}&=A|x|\ \text{, where $A=e^c$}\\
|y^2-1|&=A^2x^2\\
\implies y^2-1&=A^2x^2&1-y^2&=A^2x^2\\
y^2&=A^2x^2+1&y^2&=1-A^2x^2\\
\\
\therefore y&=\pm\sqrt{Kx^2+1}&y&=\pm\sqrt{1-Kx^2}\ \text{, where $K=A^2$}\\
\end{align*}$$
Thanks :)
 [/INST] Your solutions are $\pm\sqrt{1\pm Kx^2}$, where $K$ is positive. But this does not differ from $\pm\sqrt{1+Kx^2}$ where $K$ is unconstrained.

From
$$\frac12\log|y^2-1|=\log|x|+C$$ you draw
$$\log|y^2-1|=\log(e^Cx^2)$$ and
$$|y^2-1|=e^Cx.$$
Then
$$y^2=1\pm e^Cx^2$$ which you can very well rewrite as
$$y^2=1+Dx^2.$$
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