# Thread: dy is to dx as y is to x

1. ## dy is to dx as y is to x

I'm looking for functions that satisfy:

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

(k being a constant)

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

Thanks

2. Originally Posted by rainer
I'm looking for functions that satisfy:

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

(k being a constant)

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

Thanks
This is a separable equation, so you would integrate (and don't forget the constant of integration).

3. Originally Posted by wonderboy1953
This is a separable equation, so you would integrate (and don't forget the constant of integration).
So, integrating $\frac{dy}{dx}=\frac{y}{x} k$ I get $y=ky\ln{x}+c$

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

This yields the following relationship:

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

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?

4. I get $|y| = e^{k\ln|x| + C} = e^{\ln|x^k|} \cdot e^C = Px^k \implies y = \pm P|x|^k$ where P and k are constants for the integral which is what you got in the OP

5. Originally Posted by rainer
So, integrating $\frac{dy}{dx}=\frac{y}{x} k$ I get $y=ky\ln{x}+c$

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

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

This yields the following relationship:

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

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?
Yes, integrate correctly!

6. Originally Posted by HallsofIvy
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...

...integrate correctly!
Oops!

7. Good response! By the way, I note that I took the logarithm incorrectly on the right and have edited my previous response.

8. Hey guys, just one last question:

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

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

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

Where you guys have:

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

Thanks

9. Notice that $\displaystyle k(\ln{x} + C) = k\ln{x} + kC = k\ln{x} + C'$ where $\displaystyle C'$ is some other constant.

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

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

10. oh yeah. Duh!