`(9, 1) , y' = y/(2x)` Find an equation of the graph that passes through the given point and has the given slope.

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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