Solve $4y(y')^2-4x(y')^3-2(y')^2-1=0$ I recently came across the following differential equation: $$4y(y')^2-4x(y')^3-2(y')^2-1=0$$
I am not completely sure how to approach a general solution to such an equation, since I am only familiary with simple linear differential equations. I nevertheless attempted to fit the form $y=ax^n+b$ to the equation. This resulted in the following equation: $$4(ax^n+b)(anx^{n-1})^2-4x(anx^{n-1})^3-2(anx^{n-1})^2-1=0$$ $$4a^3n^2x^{3n-2}(1-n)+2a^2n^2x^{2n-2}(2b-1)=1$$
I then deduced that either $3n-2=0$ or $2n-2=0$, since we have a constant term on the right hand side. Splitting into cases, this gives the equation $2a^2(2b-1)=1$ for $n=1$ and the solution $y=(\frac{27}{16})^{\frac{1}{3}}x^{\frac{2}{3}}+\frac{1}{2}$ for $n=\frac{2}{3}$.
Is my work thus far correct and valid for the situation, especially assuming that one of the exponents of $x$ had to be $0$? Also, this does not necessarily find all possible solutions, since it assumes that $y$ is a power law type function. How could I find other solutions?
 A: $$4y(y')^2-4x(y')^3-2(y')^2-1=0$$
I suppose $y' \ne 0$ . Divide by $(y')^2$:
$$4y-4xy'=\dfrac 1 {(y')^2}+2$$
$$y=xy'+\dfrac 1 {4(y')^2}+\frac 12$$
It's of the form:
$$y=xp+f(p)$$
Where $p=y'$. This is Clairaut's differential equation. Look here Clairaut DE
A: Let $\frac{dy}{dx}=p$, the given ODE becomes
$$(4y-4xp-2)=p^{-2}~~~(1)$$
D. w.r.t. $y$, we get
$$ 4-4p\frac{dx}{dy}-4x\frac{dp}{dy}=-2p^{-3} \frac{dp}{dy}\implies -\frac{4x}{p}\frac{dp}{dy}=-2p^{-3}\frac{dp}{dy}\implies (-4x/p+2p^{-3})\frac{dp}{dy}=0$$
$$\frac{dp}{dy}=0 \implies p=C~~~(2), ~or~ p=\frac{1}{\sqrt{2x}}~~~(3)$$
So the generak solution of (1) is $$(4y-4cx-2)=C^{-2}\implies y=Cx+\frac{1-2C^2}{4}~~~(4)$$
The singular solution is given by (3) as 
$$\frac{dy}{dx}=-(2x)^{3/2} \implies y=-(2x)^{-3/2}~~~(5)$$
with no constant.
Finally (4) and (5) give the complete solution of (1).
A: If you were not finding the connection to Clairaut, you could still try to get an explicit DE of second order by taking the derivative of the given ODE. (One would have to find an initial value for the first derivative by selecting one solution of the original DE as polynomial in $y'$ in the initial point.)
$$
4y'^3+8yy'y''-4y'^3-6xy'^2y''-4y'y''=0
$$
All the terms that remain contain a factor $y''$, so the equation factors as
$$
4(2y-3xy'-1)y'y''=0
$$
Every solution now can be subdivided into segments where one of the factors is zero. At points connecting two segments, two of the factors have to be zero. This means either $y'=C=const.$, which inserted into the original equation gives directly a linear solution, or $y'=0$ which is a special case of that, or
$$
2y-3xy'-1=0\iff xy'-\frac23(y-\tfrac12)=0
\implies y=\tfrac12+Cx^{\frac23},~~y'=\tfrac23Cx^{-\frac13}
$$
On the other hand, reducing the original equation by this factor gives
$$
2xy'^3-1=0\implies y'=2^{-\frac13}x^{-\frac13}\implies y=D+\tfrac322^{-\frac13}x^{\frac23}
$$ 
which gives, comparing both approaches, $C=\tfrac322^{-\frac13}$ and $D=\frac12$.
