Solving $8yy'^2 - 2xy' + y = 0$ I'm solving the differential equation $8yy'^2 - 2xy' + y = 0$
My attempt:
We divide both sides by $x$, obtaining:
$$8\frac{y}{x}y'^2 - 2y' + \frac{y}{x} = 0$$
Then, we introduce $t = y'$, hence the differential equation becomes:
$$8\frac{y}{x}t^2 - 2t + \frac{y}{x} = 0$$
from which follows:
$$\frac{y}{x} = \frac{2t}{8t^2 +1}$$
Hence, in parametric equation we obtain:
$$\begin{cases} y' = t \\ \frac{y}{x} = \frac{2t}{8t^2 +1}\end{cases}$$
From the second equation, after differentiating with respect to $t$, we obtain:
$$\frac{dy}{dt} = \frac{dx}{dt}\frac{2t}{8t^2 + 1} + \frac{-16t^2 + 2}{(8t^2 +1)^2}$$
Hence:
$$dy = dx\frac{2t}{8t^2 + 1} + \frac{-16t^2 + 2}{(8t^2 +1)^2}dt$$
From the first equation we have $dy = tdx$
Therefore:
$$tdx= dx\frac{2t}{8t^2 + 1} + \frac{-16t^2 + 2}{(8t^2 +1)^2}dt$$
Or after rearranging:
$$dx = \frac{-2dt}{(8t^2+1)dt}$$
And by integrating:
$$x(t) = -2\ln|t| - \frac{1}{2}\ln|8t^2 + 1| + c$$
And by the first equation:
$$y(t) = \frac{2t}{8t^2+1}(-2\ln|t| - \frac{1}{2}\ln|8t^2 + 1| + c)$$
Hence, the solution in parametric equation, is:
$$\begin{cases} x(t) = -2\ln|t| - \frac{1}{2}\ln|8t^2 + 1| + c \\ y(t) = \frac{2t}{8t^2+1}(-2\ln|t| - \frac{1}{2}\ln|8t^2 + 1| + c) \end{cases}$$
Can someone verify whether this is correct? The answer my book gives is $$y^2 - 4cx + 32c^2 = 0$$ with singular integral $$8y^2 - x^2 = 0$$ How would I derive this answer?
Thanks in advance.

You lost the factor $x$ in the last term while differentiating the parametric equation.

It is even easier to multiply with $y$ and then substitute $u=y^2$ to get
$$
2u'^2-xu'+u=0\iff u=xu'-2u'^2
$$
which is a Clairaut differential equation. This has the lines
$$
u=cx-2c^2
$$
as solutions and their envelope which is the non-linear solution to $0=(x-4u')u''$. Thus inserting $u'=x/4$ gives
$$
u=\frac{x^2}4-\frac{x^2}8=\frac{x^2}8
$$
