Is it possible to solve $\frac{dy}{dx}=y^x$? I was trying to solve the equation below just for curiosity,
$\frac{dy}{dx}=y^x$
I found that i could only equate $y^x$ with $e^{-i^2x\ln(y)}=\cos(ix\ln(y))-i\sin(ix\ln(y))$, so the original equation became,
$\frac{dy}{dx}+i\sin(ix\ln(y))=\cos(ix\ln(y))$
After I arrived here I couldn't continue.
Can someome solve this problem, please?
 A: Unable to proceed analytically I used numerical integration after logarithmic differentiation.
Take logs and differentiate
$$ \log y' = x \log y,\; \dfrac{y^{''}}{y'}= \log y + \dfrac{x y{'}}{y}$$
The ode is (primes differentiation with respect to $x$ )
$$ \dfrac{d^2y}{dx^2}= y'\;\big[ \log \;y + \dfrac{xy^{'}}{y} \big] $$
with BC for two curves: $( x=0, y'=1, y_i= (5,9,18,25)\;) $
The Boundary condition $ x=0, y'=1 $  for all $y_i$ compels adoption of ode to be of next higher order i.e., second order.

Curve terminates at $y_{max}$  due to log singularity.
$y>0$ always. Negative $x-$ axis exists, but not computed /plotted.
EDIT1:
Integrand replotted after minor error (of no concern here). The computation errors are checked and seen to plot to zero (brown line plot) residue using second order ode.
EDIT2:
This morning even after diligent checks on same program could find no error so far but surprisingly got slightly different $y$ values. I am reporting the same along with the code and the previous graph for anyone to take a look even if qualitatively and checking-wise nothing has really changed.
NDSolve[{Y''[x]==Y[x]^x*(Log[Y[x]]+x 
Y[x]^x/Y[x]),Y[0]==5,Y'[0]==1},Y,{x,0,1.715028}];
y[u_]=Y[u]/.First[%];
Res[u_]:=y'[u]-y[u]^u;
g5=Plot[{y[x],Res[x]},{x,0,1.715027}]
FindRoot[y[xu]==20.,{xu,1.3}]

The four curves have BC:
$$ (y_i,x_{max})=\big[(5,1.715028),(9,1.66261),(18,1.609705),(25,1.58755)\big]$$
EDIT3:
The matter of Output Difference is not acceptable. So it has been posed at the StackExchange Mathematica site with link to this problem in this site ...analysis on-going.
A: Perhaps an interesting point about the ode is the essential singularity at $(0,0)$ which separates solutions into domains.  As mentioned by @yves-daoust, $y(x)\equiv 0$, $x>0$ is a solution. If instead you take an initial condition $y(0)=\epsilon$ for some $\epsilon>0$ the solution explodes in finite positive time. Furthermore, the explosion time has a well-defined linit as $\epsilon\rightarrow 0$.  Numerically I find that the max time before explosion is $x_{\rm max}=2.10958799...$
You may find this by integrating the reciprocal ode:
$$ \frac{dx}{dy} = y^{-x} $$
from $0^+$ to $\infty$, with initial condition $x(0^+)=0$.
