Solution to this Differential Equation $f''(x)=f(x)f'(x)$ needed I came up with this differential equation and I don't know how to solve it.
$$f''(x)=f(x)f'(x)$$
I attempted to solve it several times, but they were all fruitless. Wolfram Alpha says that the solution is 
$$f(x)=\sqrt{2a} \tan\left({\frac{\sqrt{2a}}{2} \cdot (x+b)}\right),$$ where $a$ and $b$ are constants.
How does one get this solution?
 A: $$y''=y'y \implies y''=\frac 12 (y^2)' $$
After integration 
$$\implies y'=\frac 12 y^2+K$$
$$y'=\frac 12(y^2+2K)$$
For $ K=0 \implies -\frac 1y=\frac x2 +a$
$$\implies y(x)=-\frac 2 {x+c}$$
For K negative
$$ \int \frac {dy}{y^2-c^2}=\frac x 2+b$$
$$\ln (\frac {y-c}{y+c})=cx+a$$
$$y(x)=-c \frac {ae^{cx}+1}{ae^{cx}-1}$$
For K positive, substitute $2K=c^2$
$$ \int \frac {dy}{y^2+c^2}=\frac 12\int dx=\frac x 2+b$$
Substitute $z=y/c \implies dz=dy/c$
$$ \int \frac {dz}{z^2+1}=c(\frac x 2+b)$$
$$ \arctan (z)=c(\frac x 2+b)$$
$$ y=c\tan (c(\frac x 2+b))$$
Substitute $c/2=k$ and $bc=a$
$$ y=2k\tan (kx+a)$$

$$y''=y'y$$
Substitute 
$$y'= p \implies y''=pp'$$
$$p'=y \implies p=\frac 12 y^2 +K$$
$$y'=\frac 12 y^2+K$$
This equation is separable
A: as an example, in Isham's answer, if we choose $K$ so that
$$ y' = \frac{1}{2} (y^2-1),  $$
the first thing we notice is two stationary solutions, $y = 1$ and $y = -1$ 
This is autonomous...
For $y > 1 $ or $y < -1$ we get
$$ y = \frac{-1}{\tanh \frac{x}{2}} $$
which blow up in finite time, one in each direction. Also translates.
For $-1 < y < 1$
$$ y = - \tanh \frac{x}{2} $$

A: Let us consider your differential equation:
$$f''(x)=f(x)\cdot f'(x)$$
Integrate with respect to $x$ on both sides. Recognize that $df'(x)=f''(x)\ dx$ and $df(x)=f'(x)\ dx$:
$$\int f''(x)\ dx=\int f(x)\cdot f'(x)\ dx\rightarrow \int df'(x)=\int f(x)\ df(x).$$
It follows that
$$f'(x)=\frac{(f(x))^2}{2}+a=\frac{(f(x))^2+2\cdot a}{2}.$$
Divide by $(f(x))^2+2\cdot a$ on both sides:
$$\frac{f'(x)}{(f(x))^2+2\cdot a}=\frac{1}{2}.$$
Integrate with respect to $x$ on both sides. Recognize that $df(x)=f'(x)\ dx$:
$$\int \frac{f'(x)\ dx}{(f(x))^2+2\cdot a}=\int \frac{dx}{2}\rightarrow \int \frac{df(x)}{(f(x))^2+2\cdot a}=\frac{x}{2}+b=\frac{x+2\cdot b}{2}.$$
Redefine $2\cdot b$ as $b$, since it is a constant:
$$\int \frac{df(x)}{(f(x))^2+2\cdot a}=\frac{x+b}{2}.$$
Let $f(x)=\sqrt{2\cdot a}\cdot \tan(s)$ such that $df(x)=\sqrt{2\cdot a}\cdot (\tan^2(s)+1)\ ds$:
$$\int \frac{ds}{\sqrt{2\cdot a}}=\frac{x+b}{2}\rightarrow \frac{s}{\sqrt{2\cdot a}}=\frac{x+b}{2}.$$
Isolate $s$ and let $s=\arctan \left (\frac{f(x)}{\sqrt{2\cdot a}}\right)$:
$$s=\frac{\sqrt{2\cdot a}}{2}\cdot (x+b)\rightarrow \arctan \left (\frac{f(x)}{\sqrt{2\cdot a}}\right)=\frac{\sqrt{2\cdot a}}{2}\cdot (x+b).$$
Therefore, an expression for your function $f(x)$ can be written as
$$f(x)=\sqrt{2\cdot a}\cdot \tan \left (\frac{\sqrt{2\cdot a}}{2}\cdot (x+b)\right).$$
The derived expression is equivalent to what you found with Wolfram Alpha.    
