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 AMrainerdy 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 AMwonderboy1953
- Jan 9th 2011, 01:43 PMrainer
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 PMe^(i*pi)
I get $\displaystyle |y| = e^{k\ln|x| + C} = e^{\ln|x^k|} \cdot e^C = Px^k \implies y = \pm P|x|^k$ where P and k are constants for the integral which is what you got in the OP

- Jan 9th 2011, 02:47 PMHallsofIvy
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 $\displaystyle c= e^C$.

Quote:

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 AMrainer
- Jan 13th 2011, 10:51 AMHallsofIvy
Good response! By the way, I note that I took the logarithm incorrectly on the right and have edited my previous response.

- Jan 25th 2011, 05:20 PMrainer
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 PMProve It
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 AMrainer
oh yeah. Duh!