<s>[INST] reduce a differential equation $y'+\dfrac{x}{y}=0$ I want to reduce a differential equation.
$y'+\dfrac{x}{y}=0$
I reduced this. But my answer don't much with Wolfram alpha. Please tell me what is wrong.
$y'=-\dfrac{x}{y}=-\dfrac{1}{\left( \dfrac{y}{x}\right) }$
When I put, $u=\dfrac{y}{x}$,
$\dfrac{dy}{dx}=-\dfrac{1}{u}$
differentiate both sides with respect to x
$\dfrac{du}{dx}=-\dfrac{1+u^{2}}{ux}$
$\dfrac{u}{1+u^{2}}\dfrac{du}{dx}=-\dfrac{1}{x}$
Integrate both sides with respect to x
$\dfrac{1}{2}\log \left| u^{2}+1\right| =-\log \left| x\right| +C'$
Organize the formula
$u^{2}+1=\dfrac{C}{x}$
$\begin{aligned}\dfrac{y^{2}}{x}+1=\dfrac{C}{x}\\ y=\pm \sqrt{Cx-x^{2}}\end{aligned}$
 [/INST] Not sure exactly what our OP langhtorn means by "reduction", or how the calculations presented in the text of the question itself work, but here is my solution:
$y' + \dfrac{x}{y} = 0; \tag 1$
multiply through by $y$:
$yy' + x = 0; \tag 2$
$\dfrac{1}{2}(y^2)' + x = 0; \tag 3$
$(y^2)' = -2x; \tag 4$
integrate:
$y^2 = -x^2 + C = C - x^2; \tag 5$
then
$y = \pm \sqrt{C - x^2} = \pm (C - x^2)^{1/2}.  \tag 6$
Check:
$y' = \pm \dfrac{1}{2}(C - x^2)^{-1/2}(-2x) = \pm (-x(C - x^2)^{-1/2})$
$= \pm (-\dfrac{x}{(C - x^2)^{1/2}}) = -\dfrac{x}{\pm (C - x^2)^{1/2}} = -\dfrac{x}{y}. \checkmark \tag 7$
In light of this verified solution, I am skeptical of our OPs calculations, but as of this writing have not scrutinized them carefully enough to find an error.
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