How to find an equation of the graph that passes through the point (9,1) and has a slope of #y'=y/(2x)#?

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Answer 1 · 2018-02-13T15:00:35

# y=sqrt(x)/3 #

We have:

# dy/dx = y/(2x) #

Which is a First Order Separable Ordinary Differential Equation, so we can arrange as follows:

# 1/y \ dy/dx = 1/(2x) #

And we can "separate the variables" to get:

# int \ 1/y \ dy = 1/2 \ int \ 1/x \ dx #

Both integrals are standard results, so integrating we get:

# ln |y| = 1/2ln|x| + C #

We know that #y=1# when #x=9#, so:

# ln 1 = 1/2ln9 + C #
# :. 0 = ln3 + C #
# :. C = -ln3#

So the solution is:

# ln |y| = 1/2ln|x| -ln3 #
# \ \ \ \ \ \ \ = ln(sqrt(|x|)/3) #

And if we assume that #x gt # we have:

# y=sqrt(x)/3 #

Verification

1) #x=9 => y = sqrt(9)/3 = 1#
2) #y'=1/2(x^(-1/2)/3) = 1/2 sqrt(x)/(3x) = (y)/(2x) \ \ \ # QED