Solve $y'' - y' = yy'$ and find three Other Distinct Solutions Been stuck on this for a while. I need to solve the following differential equation by finding the constant solution y = c and three other distinct solutions.
$$y'' - y' = yy'$$
If someone could give me a complete step by step explanation, it would be greatly appreciated as I want to fully understand it.

Going from JJacquelin's answer, the equation can be written as
$$ \frac{2 \,dy}{y^2 + 2y + C} = dx$$
For $C = 1$ 
$$ \int \frac{2\,dy}{(y+1)^2} = \int dx $$
$$ -\frac{2}{y+1} = x + C_1 $$
$$ y = -1 - \frac{2}{x + c}$$
For $C < 1$ 
$$\int \frac{2\,dy}{(y-C_1)(y-C_2)} = \int dx $$
$$ \frac{2}{C_2-C_1}\ln \frac{y-C_2}{y-C_1} = x + C_3$$
$$ y = 2 \frac{c_2 - c_1 c_3 e^{(c_2-c_1)x}}{1 - c_3 e^{(c_2-c_1)x}}$$
For $C > 1$
$$ \int \frac{2\,dy}{(y+1)^2 + C_1^2} = \int dx $$
$$ \frac{2}{C_1} \arctan \frac{y+1}{C_1} = x + C_2 $$
$$ y = -1 + 2c_1\tan (c_1 x + c_2)$$
