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?
Best Answer
$$y''=y'y \implies y''=\frac 12 (y^2)' $$ 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