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

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