<s>[INST] 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?
 [/INST] Let $v = 4yx^2$ then we can write 
$$
\frac{d}{dx}\frac{v}{4x^2} = \frac{x\pm \sqrt{x^2+vx^2}}{2x^4} = \frac{1\pm\sqrt{1+v}}{2x^3} = \frac{1}{4x^2}v' -\frac{1}{2x^3}v =\frac{1\pm\sqrt{1+v}}{2x^3}
$$
we can re-arrange
$$
xv'-2v = 2\pm\sqrt{1+v} \implies xv' = 2(v+1)\pm 2\sqrt{1+v}
$$
we can clean up
$$
xu ' = 2u\pm 2\sqrt{u}
$$
where $u=v+1$.
then set $u=t^2$ to find
$$
x2tt' = 2t^2\pm 2t\implies xt' = t \pm 1
$$
can you take it from here and transform back to your original variables?
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