The equation $(x-2xy-y^2)\frac{dy}{dx}+y^2=0$ Can one give a hint how to solve the following equation? 
$(x-2xy-y^2)\frac{dy}{dx}+y^2=0$
Thanks in advance.
 A: For this one we look for an integrating factor which will make the equation exact.
Let $M = x - 2xy - y^2$ and $N = y^2$.
Then $M_x = 1-2y$ and $N_y = 2y$.
Thus $$\frac{M_x-N_y}{N} = \frac{1-2y}{y^2} = \frac{1}{y^2} -2 \frac{1}{y}$$ which is purely a function of $y$. This tells us that there is an integrating factor $\mu (y)$ which makes the equation exact which satisfies
$$
\frac{d\mu}{dy} = \left(  \frac{1}{y^2} -2 \frac{1}{y} \right)\mu
$$
Clearly
$$\mu(y) = e^{\displaystyle\int\left(  \frac{1}{y^2} -2 \frac{1}{y} \right)dy} = e^{\frac{-1}{y}-2\log y} = y^{-2} e^{y^{-1}}.$$
Try (first checking that my algebra is correct) and then multiplying the differential equation by this integrating factor, it should become an exact differential equation which I'm guessing you know how to solve if you are asking this question.
A: Hint: Your equation can be turned into a first order linear differential equation in $\frac{dx}{dy}$ (i.e., just multiply the equation by $\frac{dx}{dy}$ and it is a standard form....)
A: As it looks some people misunderstood (I apologize, the fault may be my side) my first answer, here is the complete solution after reducing the problem into standard form:
As I wrote in the first answer, multiplying both sides by $\frac{dx}{dy}$ (the hint was that it will make the equation linear in $x$),
$\frac{dx}{dy}+\left(\frac{1-2y}{y^2}\right).x=1$. This is a linear equation (in $x$) of the standard form $ \frac{dx}{dy}+P(y).x=Q(x)$. So, $e^{\int P(y)dy}=e^{\int \left(\frac{1-2y}{y^2}\right)dy}=\frac{e^{\frac{-1}{y}}}{y^2}$ is an integrating factor (IF). So, multiplying by the IF and integrating the general solution is given by
$x.\frac{e^{\frac{-1}{y}}}{y^2}=\int\frac{e^{\frac{-1}{y}}}{y^2}dy+c=e^{\frac{-1}{y}}+c$
Or, $x=y^2(1+ce^{\frac{1}{y}})$.
