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

Question

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 ?

Answer 1

Hint: Rewrite the DE as $$\frac{dx}{dy}-\frac{x}{y}=y$$ which is easy by integrating factor method

Answer 2

$${y}'\left ( x+ y^{2} \right )= y$$ $$xy'-y=-y'y^2$$ $$\left(\dfrac yx\right)'=-y'\dfrac {y^2}{x^2}$$ This DE is separable. $$\dfrac {x^2}{y^2}d\left(\dfrac yx\right)=-dy$$ Integrate.

Answer 3

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

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