How to obtain the singular solution for this ODE $$y=2xy'+\frac{1}{y'}$$ After getting the general solution (by calling y' as p and differentiating the equation by x) , I tried to get the envelope but I failed because I got two parametric equations representing the general solution. here are the equations: $$x=\frac{ln(p)}{p^2}+\frac{c}{p^2}$$
$$y=2\frac{ln(p)}{p}+\frac{2c+1}{p}$$
So which equation do I have to differentiate wrt c ? If I differentiate any of them wrt c and equating by zero , I got 1=0 !
I agree with your parametric solution, except that absolute values should be in the logarithm : $$\begin{cases}x=\frac{\ln|p|}{p^2}+\frac{c}{p^2}\\ y=2\frac{\ln|p|}{p}+\frac{2c+1}{p}\end{cases}$$
The trouble comes from the singular point which exists on each curve.
For several values of $c$ the curve corresponding to the function $y(x)$ is drawn on the next graph:
The envelope of the family of curves is drawn in red. The equation is : $$y^2=8x$$ But the envelope is not tangential to the curves because the singular points are on the envelope. In this special case, the equation of the envelope is not solution of the ODE. That is why you don't obtain a singular solution.