Solving the differential equation $xy' + y+ y^2 = 0$ How do you tell if something is Homogeneous? Do all the powers need to be the same on the top, as well as on the bottom to make it into the form of $\cfrac yx$?
Anyways, I have this problem
So the first thing I did was try to see if it was homogenous:
$$y' = \frac{-y}{x}-\frac{y^2}{x}$$
I got stuck here, and believed this was not homogeneous since there's noway I can represent it as a function of $\cfrac yx$.
Anyways, I then tried to see if it was a separable equation:
$$x\frac{dy}{dx} = -y^2-y$$
$$\frac{dy}{-y^2-y} = \frac{dx}{x}$$
then I figured I could factor out a $y$ from the LHS and do partial fractions. Not too sure to be honest.
 A: Hint
$$y' + \frac{y}{x}=-\frac{y^2}{x}$$
Bernouilli's equation ..............
Or separable as you nicely did 
$$\int\frac{dy}{y(y+1)}=-\int\frac{dx}{x}=-\ln(x)+K$$
$$\int\frac{dy}{y}-\int\frac{dy}{y+1}=-\ln(x)+K$$
A: On an interval where $y$ is a $1$-to-$1$ function of $x$ and $y'\ne 0$ and where $0\ne y\ne 1,$ consider $x$ as a function of $y.$ Then $y'=\frac {dy}{dx}=\frac {1}{dx/dy}=1/x'.$ So $x/x'=-y-y^2, $ so $x'/x=-1/(y+y^2)=$ $=\frac {1}{1+y}-\frac {1}{y}, $ so $$\log |x|=C+\int\frac {1}{1+y}dy -\int \frac {1}{y}dy=C+\log |1+y| =\log |y|.$$
A: The equation is equivalent to
$$ (xy)' + y^2 = 0 $$
So you could try the substitution $u = xy$
$$ u' + \frac{u^2}{x^2} = 0 $$
This can be integrated easily
$$ \frac{u'}{u^2} = -\frac{1}{x^2} $$
$$ \frac{1}{u} = C - \frac{1}{x} = \frac{Cx-1}{x} $$
$$ u = \frac{x}{Cx-1} $$

Another approach is to observe that this is a classic Bernoulli equation. First divide through by $y^2$
$$ \frac{xy'}{y^2} + \frac{1}{y} + 1 = 0 $$
Then substitute $z = y^{-1}$ to get
$$ -xz' + z + 1 = 0 $$
which is once again separable and integrates to
$$ \frac{z'}{z+1} = \frac{1}{x} $$
$$ \ln(z+1) = \ln x + \ln C $$
$$ z+1 = Cx $$

Either way, the general solution can be obtained as
$$ y(x) = \frac{1}{Cx-1} $$
A: Given:
$ xy'+y+y^2 = 0$
$ xy'= -y - y^2 $
This is using Variable Separable Form
$  \frac{dy}{y + y^2}=-\frac{dx}{x}$
On Integration:
$ \int \frac{dy}{y+ y^2}= ln(\frac{1}{x}) + ln(C)$
You've done it upto here. You were going the right way.
Using Completing the Square Method on the denominator of LHS,
$ \int \frac{dy}{(y+\frac{1}{2})^2- (\frac{1}{2})^2}= ln(\frac{1}{x}) + ln(C)$
Integrating this, we get
$ \frac{1}{2*0.5}* ln(\frac{y}{y+1})= ln(\frac{1}{x}) + ln(C)$
Eliminating the ln's and by using ln properties,
$ \frac{y}{y+1}=\frac{C}{x}$
When you put x=1 and y = 2, C=2/3. Put it back up in the last equation and you're done.
