So BD/AB=CE/AC.

So y/5=x/3,

Therefore, the analytic formula of y's culvert about x is: y=5x/3.

The domain is: 0 less than x less than 5.

(2) Solution: Let the centers of the circles with BD and CE as diameters be O 1 and O2, respectively.

Then the connecting line O65438+o02 is the center line of the trapezoidal DBCE,

In triangle ABC, we can get from cosine theorem:

BC squared =AB squared +AC squared -2 times AB times AC times cosA.

= 25+9-30 times 3/ 10

=25,

BC=5，

Because the circle O 1 is circumscribed with the circle O2,

So O 1O2= sum of radii of two circles =? It can't be solved anymore.