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)` .