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.
 A: Another method, through direct integration  of $y''-y'=yy'$
$$y'-y=\frac{1}{2}y^2+C$$
This is a separable ODE:  $\frac{dy}{y+\frac{1}{2}y^2+C}=dx$
$$2C_1\tan^{-1}(C_1(y+1))=x+C_2$$
$$y=2c_1\tan\big(c_1x+c_2\big)-1$$
Particular solutions are obtained with various values of $c_1$ and $c_2$. For example :
$c_1=0$ gives $y=$constant=$c$
$c_1=1$ and $c_2=0$ gives $y=2\tan(x)-1$
etc. You can obtain as many patricular solutions as you want.
Also one can observe that changing $c_2$ to $c_2+\pi/2$ changes the tan to -cot. So the solutions can also be expressed on the form (with different $c_2$) :
$$y=-2c_1 \cot\big(c_1x+c_2\big)-1$$
Moreover, the solutions are not limited to real $c_1 , c_2$, but can be imaginary numbers, for example changing $c_1$ to $ic_1$ changes $c_1\tan(c_1x)$ to $-c_1\tanh(c_1x)$. So, the solutions can also be expressed on the form :
$$y=-2c_1\tanh\big(c_1x+c_2\big)-1$$
or
$$y=2c_1 \coth\big(c_1x+c_2\big)-1$$
A: I'll assume the independent variable is $x$, you can make the necessary changes if it's $t$ or anything else.
Let $z=y'$.  Then
$$y''=\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}=z\frac{dz}{dy}$$
and substituting into the DE gives
$$z\frac{dz}{dy}-z=yz\ .$$
See if you can take it from here.
A: 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 $C > 1$
$$ \int \frac{2\,dy}{(y+1)^2 + C_1^2} = \int dx $$
$$ \frac{2}{C_1} \arctan \frac{y+1}{C_1} = x + C_2 $$
$$ y = -1 + 2c_1\tan (c_1 x + c_2)$$
