Solution of differential equation which is quadratic in $\frac{dy}{dx}$ Consider the following differential equation
$$ x\frac{dy}{dx} + y = x^4(\frac{dy}{dx})^2$$  
I used the quadratic formula and got $\frac{dy}{dx} = \frac{x \pm \sqrt {x^2 + 4yx^4}}{2x^4}$. Now how to proceed or is there any other method?  
Edit:
I proceeded further this way  
$\frac{dy}{dx} = \frac{1 \pm \sqrt {1 +4yx^2}}{2x^3}$  
Let $1 + 4yx^4 = t^2$  
then $8yx + 4x^2\frac{dy}{dx} = 2t\frac{dt}{dx}$  
multiplying throughout with $x$ I got $4yx^3 +2x^3\frac{dy}{dx} = tx\frac{dt}{dx}$  
now putting the value of $2x^3\frac{dy}{dx}$ and $t^2$ in 1st equation I got
$tx\frac{dt}{dx} - t^2 = \pm t$
which gave 2 solutions $$t = 0$$ $$and$$ $$x\frac{dt}{dx} = t \pm 1$$ 
Using $t  = 0$ I got one solution $$1 + 4x^2y = 0$$ and the other by solving the differential equation $x\frac{dt}{dx} = t \pm 1$ as $$c^2x^2 + 4y^2x^2 = 4cyx^2 + 2c$$  
Now the answer to this question which I have got is $$4x^2y + 1 = 0$$ $$and$$ $$xy - c^2x + c = 0$$
Why I couldn't get the second solution right?

The equation $\quad x\frac{dt}{dx}=t\pm 1\quad$ is correct.
The solutions of this linear ODE are :
$$t\pm 1=Cx$$
From $t^2=1+4yx^2 \quad\to\quad (Cx\pm 1)^2=1+4yx^2$
$$4yx^2=(Cx\pm 1)^2-1=C^2x^2\pm 2Cx$$
$$xy-\frac{C^2}{4}x\pm \frac{C}{2}=0$$
Let $c=\frac{C}{2}$
$$xy-c^2x\pm c=0$$
The solution is :
$$xy-c^2x+ c=0\quad\to\quad y=c^2-\frac{c}{x} $$
and $\quad xy-c^2x- c=0\quad\to\quad y=c^2+\frac{c}{x} \quad$ which is the same as the preceding one (just change $c$ into $-c$ ) since $c$ is any constant.
Of course, without forgetting the particular solution previously obtained for $t=0$ 
$$4yx^2+1=0\quad\to\quad y=-\frac{1}{4x^2}$$
NOTE : 
$y=-\frac{1}{4x^2}$ is the low envelop (drawn in red) of the family of curves $y=c^2-\frac{c}{x}$ in blue ($c<0$) and in black ($c\geq 0$). 

