<s>[INST] Solve the differential equation: $(y^\prime)^2 - \frac{2y}{x}y^\prime+1=0$ Solve the differential equation: $(y^\prime)^2 - \frac{2y}{x}y^\prime+1=0$
My first step was to solve for $y^\prime$ to get:
$$y^\prime=\frac{y}{x}\pm\sqrt{(\frac{y}{x})^2-1}$$
Solving each individually yields (using substitution $u = \frac{y}{x}$):
$$y^\prime = \frac{y}{x} + \sqrt{(\frac{y}{x})^2-1} \rightarrow u'x + u = u + \sqrt{u^2-1}$$
$$\int \frac{1}{\sqrt{u^2-1}}du = \int \frac{1}{x}dx \rightarrow \ln|u+\sqrt{u^2-1}|=\ln|x|+C$$
Raising to $e$ and subing back and multiplying each equation yields the following integral equation:
$$(y-Cx^2 + \sqrt{y^2-x^2})(y-Cx+\sqrt{y^2-x^2})=0$$
However, I am suppose to yield $(x^2C^2 + 1 - 2Cy)\times(x^2+C^2-2Cy)=0$ ... could anyone point out the issues in my approach?
 [/INST] $$(y^\prime)^2 - \frac{2y}{x}y^\prime+1=0$$
This is D'Alembert 's differential equation:
$$y=xf(y')$$
where $$f(y')=\dfrac {1+y'^2}{2y'}$$

From your attempt you should get
$$ u+\sqrt {u^2-1}=cx$$
$$u^2-1=(cx-u)^2$$
$$c^2x^2-2cxu+1=0$$
$$c^2x^2-2cy+1=0$$
$$ \implies y= \dfrac {c^2x^2+1}{2c}$$
There is also a singular solution:
$$y=\pm x$$
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