Solve the differential equation ${y}'\left ( x+ y^{2} \right )= y$ 
Solve the differential equation
$${y}'\left ( x+ y^{2} \right )= y$$

by jp.Wolfram|Alpha _the result is $\left ( 2y+ c_{1} \right )^{2}= c_{1}^{2}+ 4x\Leftrightarrow x= y\left ( y+ c_{2} \right )\Leftrightarrow 1= {y}'\left ( 2y+ c_{1} \right ),$ how can I break this cycle ?
 A: Hint: Rewrite the DE as  $$\frac{dx}{dy}-\frac{x}{y}=y$$ which is easy by integrating factor method
A: $${y}'\left ( x+ y^{2} \right )= y$$
$$xy'-y=-y'y^2$$
$$\left(\dfrac yx\right)'=-y'\dfrac {y^2}{x^2}$$
This DE is separable.
$$\dfrac {x^2}{y^2}d\left(\dfrac yx\right)=-dy$$
Integrate.
A: As the answers here have already given a hint of solving the equation in particular, I would like to clear the other doubt of how to break the cycle. But before that here is a complete solution of Albus Dumbledore 's answer-

I.F. = $e^{-\int \left(\frac{1}{y}\right)dy}=e^{-\ln y}=\frac{1}{y}$. And so the equation becomes $\frac{x}{y}=\int \left(\frac{y}{y}\right)dy$. That is $\frac{x}{y}=y+c$

We have got the original equation ${y}'\left ( x+ y^{2} \right )= y$-----------------------(1)
We also have got Wolfram alpha's solution $y'\left(2y+c\right)=1$--------------(2)
Now dividing (1)/(2).
We get $\frac{\left(x+y^2\right)}{\left(2y+c\right)}=y$
On rearranging we get  $\frac{\left(x+y^2\right)}{y}=\left(2y+c\right)$
Which on simplfying we get our solution

$\left(\frac{x}{y}\right)=\left(y+c\right)$

