Differential Equation $2x\frac{dy}{dx} = 2\tan y + (\frac{dy}{dx})^{3}\cos^{2} y$ Do I miss something when solve the o.d.e above with the following steps?
$2x\cos y \frac{dy}{dx} = 2 \sin y+(\cos y \frac{dy}{dx})^3$
Let $u=\cos y \frac{dy}{dx}$.
$2xu=2\sin y+u^3$
Taking the derivative w.r.t $x$, we have
$2u+2x\frac{du}{dx}=2\cos y\frac{dy}{dx}+3u^2\frac{du}{dx}=2u+3u^2\frac{du}{dx}$
$(2x-3u^2)\frac{du}{dx}=0$
Thus, $\frac{du}{dx}=0$ or $2x-3u^2=0$.
If $\frac{du}{dx}=0$, we have $\cos y\frac{dy}{dx}=C$. Here, if we solve this equation, we will get another constant besides $C$. But the general solution of the original first order o.d.e can only have one const. I wonder whether I miss something.
 A: You would get farther in a more direct way by setting $u=\sin y$, $u'=\cos(y)\,y'$ so that then from your first transformation
$$
2xu'=2u+u'^3\iff u=xu'-\frac12u'^3
$$
which is a Clairaut differential equation with solution family
$$
u=cx-\frac12c^3
$$
and singular solutions 
$$
x=\frac32u'^2\implies x\ge0\land u=\frac23xu'=\pm\left(\frac23x\right)^{3/2}
$$
A: Let $u = \sin y$, then $u' \frac{du}{dx} = \cos y \frac{dy}{dx}$
$$ 2xu' = 2u + {u'}^3 $$
Differentiate both sides
$$ 2u' + 2xu'' = 2u' + 3 {u'}^2 u'' $$
$$ u''(3{u'}^2 - 2x) = 0 $$
There are two possibilities: Either $u''=0$ or $3{u'}^2 - 2x = 0$

If $u'' = 0$ then $u(x) = c_0 + c_1 x$. Plugging back into the original equation
$$ 2xc_1 = 2(c_0+c_1x) + {c_1}^3 $$
$$ 2c_0 + {c_1}^3 = 0 $$
or $c_0 = -\dfrac{{c_1}^3}{2}$, so the solution is 
$$u(x) = c_1\left(x-\frac{{c_1}^2}{2}\right)$$

Otherwise we have
$$ {u'}^2 = \frac{2x}{3} $$
$$ u'= \pm \sqrt{\frac{2x}{3}} $$
$$ u = \pm \sqrt{\frac23}\cdot \frac23 x^{3/2} + K $$
Note that $u = {u'}^3 + K$. Plugging this in
$$ 2xu' = 3{u'}^3 + 2K $$
$$ u'(3{u'}^2-2x) + 2K = 0 $$
This implies $K=0$, therefore
$$ u(x) = \left( \frac{2x}{3} \right)^{3/2} $$
