I'm looking for functions that satisfy:

$\displaystyle \frac{dy}{dx}=\frac{y}{x}k$

(k being a constant)

So far I am aware that $\displaystyle y=ax^k$ works, but are there any other functional forms out there that would also work?

Thanks

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- Jan 9th 2011, 11:19 AM #1

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## dy is to dx as y is to x

I'm looking for functions that satisfy:

$\displaystyle \frac{dy}{dx}=\frac{y}{x}k$

(k being a constant)

So far I am aware that $\displaystyle y=ax^k$ works, but are there any other functional forms out there that would also work?

Thanks

- Jan 9th 2011, 11:27 AM #2

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- Jan 9th 2011, 01:43 PM #3

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So, integrating $\displaystyle \frac{dy}{dx}=\frac{y}{x} k$ I get $\displaystyle y=ky\ln{x}+c$

Solving for y I get: $\displaystyle y=\frac{c}{1-k\ln{x}}$

This yields the following relationship:

$\displaystyle \frac{dy}{dx}=\frac{y^2}{x}\frac{k}{c}$

Which is tantalizingly close to the proportion I was looking for, but not quite.

I don't mind the "c" being in there, but is there some way to get the y to not be squared?

- Jan 9th 2011, 01:48 PM #4

- Jan 9th 2011, 02:47 PM #5

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NO, no, no! That's not what "separating" means. You cannot treat the "y" on the right as if it were a constant while integrating dy on the left.

Instead you need to actually**separate**x and y.

$\displaystyle \frac{dy}{y}= \frac{k}{x} dx$

Now, integrate both sides to get $\displaystyle ln(|y|)= k ln(|x|)+ C= ln(|x|^k)+ C$ and take the exponential of both sides: $\displaystyle y= |x|^ke^C= c|x|^k$ where $\displaystyle c= e^C$.

This yields the following relationship:

$\displaystyle \frac{dy}{dx}=\frac{y^2}{x}\frac{k}{c}$

Which is tantalizingly close to the proportion I was looking for, but not quite.

I don't mind the "c" being in there, but is there some way to get the y to not be squared?

- Jan 13th 2011, 10:12 AM #6

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- Jan 13th 2011, 10:51 AM #7

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- Jan 25th 2011, 05:20 PM #8

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Hey guys, just one last question:

When taking the integral $\displaystyle k\int \frac{1}{x}\: dx$ why doesn't the k get factored out?

I.e., to my naive mind it should be:

$\displaystyle k\int \frac{1}{x}\: dx=k(\ln{x}+C)$

Where you guys have:

$\displaystyle k\int \frac{1}{x}\: dx=k\ln{x}+C$

Thanks

- Jan 25th 2011, 06:53 PM #9
Notice that $\displaystyle \displaystyle k(\ln{x} + C) = k\ln{x} + kC = k\ln{x} + C'$ where $\displaystyle \displaystyle C'$ is some other constant.

So it really doesn't matter how you write it.

BTW I expect you need modulus signs inside the logarithm...

- Jan 27th 2011, 09:09 AM #10

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