Is there any function such that $x^\prime(t) = x(t)^t$? This seems like a simple ODE, yet I can't find any solution to it. It can be solved numerically, but I would be interested in an analytic solution, or proof that none exists. Thanks in advance.
 A: Hint:
Similar to Possible to solve this differential equation?:
$x'(t)=x(t)^t$
$\dfrac{dx}{dt}=(e^{\ln x})^t$
$\dfrac{dx}{dt}=e^{t\ln x}$
Let $u=\ln x$ ,
Then $x=e^u$
$\dfrac{dx}{dt}=e^u\dfrac{du}{dt}$
$\therefore e^u\dfrac{du}{dt}=e^{tu}$
$\dfrac{du}{dt}=e^{(t-1)u}$
Let $v=t-1$ ,
Then $\dfrac{du}{dt}=\dfrac{du}{dv}\dfrac{dv}{dt}=\dfrac{du}{dv}$
$\therefore\dfrac{du}{dv}=e^{vu}$
Let $w=vu$ ,
Then $u=\dfrac{w}{v}$
$\dfrac{du}{dv}=\dfrac{1}{v}\dfrac{dw}{dv}-\dfrac{w}{v^2}$
$\therefore\dfrac{1}{v}\dfrac{dw}{dv}-\dfrac{w}{v^2}=e^w$
$\dfrac{1}{v}\dfrac{dw}{dv}=e^w+\dfrac{w}{v^2}$
$(e^wv^2+w)\dfrac{dv}{dw}=v$
Let $z=v^2$ ,
Then $\dfrac{dz}{dw}=2v\dfrac{dv}{dw}$
$\therefore\dfrac{e^wv^2+w}{2v}\dfrac{dz}{dw}=v$
$(v^2+we^{-w})\dfrac{dz}{dw}=2e^{-w}v^2$
$(z+we^{-w})\dfrac{dz}{dw}=2e^{-w}z$
Let $s=z+we^{-w}$ ,
Then $z=s-we^{-w}$
$\dfrac{dz}{dw}=\dfrac{ds}{dw}+(w-1)e^{-w}$
$\therefore s\dfrac{ds}{dw}+(w-1)e^{-w}s=2e^{-w}(s-we^{-w})$
$s\dfrac{ds}{dw}=(3-w)e^{-w}s-2we^{-2w}$
This belongs to an Abel equation of the second kind.
