Implicit differential equation I'm having trouble understanding how to solve this equation to find the general solution, it's obviously not homogeneous, separable or linear or exact so it must be implicit:
$2(\frac{dy}{dx})^3 + \frac{dy}{dx} - y = 0$
So I know I let $p = \frac{dy}{dx}$ then the equation becomes $2p^3 + p - y = 0$
I then derive that and it becomes $6p^2 \frac{dp}{dx} + p + \frac{dp}{dx} - \frac{dy}{dx} = 0$. What do I do then? I don't see how that would simplify down xD
 A: Take the D.E.
$$2 y' ^3 + y' - y = 0$$
Differentiate wrt x.
$$ 6 y'^2 y'' + y'' - y' =0$$
Rearrange
$$ (6y'^2+1)y'' = y'$$
$$ (6 y'+ \frac 1 {y'})y'' = 1$$
Integrate wrt x
$$ \int (6 y'+ \frac 1 {y'})y'' \operatorname{d}x = x+ c_0$$
Apply the chain rule $\int f(u(x))\frac{\operatorname{d}u}{\operatorname{d}x} \operatorname{d}x = \int f(u)\operatorname{d}u$
$$ \int (6 y'+ \frac 1 {y'}) \operatorname{d}y' = x+ c_0$$
Integrate wrt (y')
$$  3 y'^2 + \ln(y') = x+ c_0$$
Raise to the exponential
$$  y' e^{3y'^2} = c_1 e^x$$
Square and multiply by 6
$$  6y'^2 e^{6y'^2} = c_2 e^{2x}$$
$\because  z e^z = a \iff z= \operatorname{W}_n(a)$, the product log. 
$$  6y'^2 = \operatorname{W}_n(c_2 e^{2x})$$
Divide by 6 then take the square root
$$  y' = \pm \sqrt{\frac16 \operatorname{W}_n(c_2 e^{2x})}$$
Integrate wrt x
$$ y = \pm \int \sqrt{\frac16 \operatorname{W}_n(c_2 e^{2x})} \operatorname{d} x$$
Substitute $e^x = u, \operatorname{d}x = u^{-1}\operatorname{d}u$
$$ y = \pm \sqrt{\frac{1}{6}} \int \frac1{u} \sqrt{\operatorname{W}_n(c_2 u^{2})} \operatorname{d} u$$
Substitute $c_2 u^2 = v, 2 c_2 u\operatorname{d}u = \operatorname{d}v$
$$ y = \pm \frac 1 2 \sqrt{\frac 1 6} \int \frac1{v} \sqrt{\operatorname{W}_n(v)} \operatorname{d} v$$

Given:
$$\frac{\operatorname{d}}{\operatorname{d}v} \operatorname{W}_n(v) = \frac{\operatorname{W}_n(v)}{v\operatorname{W}_n(v)+v}$$
Thus
$$\frac{\operatorname{d}}{\operatorname{d}v} \sqrt{ \operatorname{W}_n(v) } = \frac{1}{2 \sqrt{\operatorname{W}_n(v)}} \frac{\operatorname{W}_n(v)}{v\operatorname{W}_n(v)+v}$$
$$\frac{\operatorname{d}}{\operatorname{d}v} \sqrt{ \operatorname{W}_n(v) } = \frac{1}{2} \frac{\sqrt{\operatorname{W}_n(v)}}{v\operatorname{W}_n(v)+v}$$
Similarly
$$\frac{\operatorname{d}}{\operatorname{d}v} \operatorname{W}_n(v)^{\frac 32} = \frac 32 \frac{\operatorname{W}_n(v)^{\frac 32}}{v\operatorname{W}_n(v)+v}$$
So
$$\frac{\operatorname{d}}{\operatorname{d}v}\left(\frac{2}{3} \operatorname{W}_n(v)^{\frac 32} + 2 \operatorname{W}_n(v)^{\frac 12}\right) = \frac{\operatorname{W}_n(v)^{\frac 1 2}}{v}$$

$$ y = \pm \frac 1 2 \sqrt{\frac 1 6}  \int \frac1{v} \sqrt{\operatorname{W}_n(v)} \operatorname{d} v$$
$$ y = \pm \frac 1 2 \sqrt{\frac 1 6} \left(\frac{2}{3} \operatorname{W}_n(v)^{\frac 32} + 2 \operatorname{W}_n(v)^{\frac 12}\right) + c_3$$
Substitute $v=c_2 u^2, u=e^x \therefore v = c_2 e^{2x}$
$$ y = \pm \frac 1 3 \sqrt{\frac 1 6}\left(\operatorname{W}_n(c_2 e^{2x})+3\right) \operatorname{W}_n(c_2 e^{2x})^{\frac 12} + c_3$$
Done!
