To find the equation of the graph passing through the point (9,1), we need to solve the given differential equation:

`y' = y/(2x)` .

First, rewrite it as

`(dy)/(dx) = (y)/(2x)` . This equation can be solved by the method of separating variables.

Multiply by dx and divide by y:

`(dy)/y = (dx)/(2x)` . Now we can integrate both sides:

`lny = 1/2lnx+C = lnx^(1/2)+C` , where C is an arbitrary constant.

Rewriting this in exponential form results in

`y = e^(ln(x^(1/2)) + C) = e^C*x^(1/2)` .

Since the graph of this equation passes through the point (9,1), we can find C:

`1 = e^C*9^(1/2)`

`e^C = 1/3`

`C = ln(1/3) = -ln3` .

**So the equation of the graph passing through the point (9,1) with the given slope is**

`y(x) = 1/3x^(1/2) = 1/3sqrt(x)` .

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