in $x^y=y^x$ I know that all rational solutions are of the form :
$$\begin{align*}x &= (1+1/u)^u \\ y &= (1+1/u)^{u+1}\end{align*} \ \ u =1 ,2 , ...$$
let $0<x<y$ and $x,y$ be irrational number . now what ? Does it exist ? if yes How is that ? i want all solutions in irrational number(generaly real number) .
From your equation,$$x^y=y^x,$$ by dividing both sides by $x^x$ you'll arrive at $$\frac{x^y}{x^x}=x^{y-x}=x^{x(y/x-1)}=\frac{y^x}{x^x}=\left(\frac{y}x\right)^x.$$ Under your assumption $x>0$, raising both sides to power $1/x$ gives $$x^{y/x-1}=\frac{y}x.$$ You assume $y>x$, so you have $$\frac{y}x=1+\dfrac1u\tag1$$ with some $u>0$, and you obtain $$x^{1/u}=1+\frac1u.$$ Raising both sides to power $u$ implies $$x=\left(1+\frac1u\right)^u\tag2.$$ Combining (1) and (2) you get $$y=\left(1+\frac1u\right)^{u+1}\tag3.$$ So you see that (2) and (3) with some real $u>0$ give all real solutions $0<x<y$ of your equation, indeed.