I have asked this question on math.stackexchange, however, have not got any answer. Therefore, I suspect that this system of ordinary differential equations cannot be solved analytically. But I still have some hope that it is possible to show that $a(u)\sim -1/u$ for $u\rightarrow \infty$. The system arises in physics, approximate treatment of the Hubbard model using some newly developed formalism.

$$2 a'(u)=u x'(u),$$ $$4 a u x'(u)-4 y'(u)=-2 x(u) \big[u a'(u)+a(u)\big],$$ $$4 a u y'(u)-4 x'(u)=\big[u\lambda'(a)a'(u)+\lambda(a)\big]-2y(u)\big[u a'(u)+a(u)\big].$$ with initial conditions $$x(0)=y(0)=0, a(0)=-1/2,$$ and $\lambda(a)=a^3+\frac14 a$.

It is easy to find a series solution in the vicinity of $u=0$. It is also possible to integrate one of the equations. But the problem is in the infinity point $u\rightarrow \infty$.

To my astonishment, a simpler approximation results in a system of 2 ODEs of similar structure, which indeed permits analytic solution.

*Edit*

In the meantime I managed to integrate 2 equations out of 3 and got: $$ y(u)=\frac{1}{2} u a(u) x(u)+\frac{a(u)^2}{2}-\frac{1}{8},\\ 2 \Lambda (a(u))-u x(u) \lambda (a(u))+4 u a(u) x(u) y(u)-2 \left(x(u)^2+y(u)^2\right)=\frac{3}{32}, $$ where $\Lambda (a)=\frac{1}{8} a^2 \left(2 a^2+1\right)$. But I still do not know how to get asymptotics. The remaining equations are $$ 2 a'(u)=u x'(u),\\ 4 a(u)^2 \left(3 u^2 x(u)^2+1\right)=4 x(u) \big(u a(u)+4 x(u)\big)+1. $$