dy is to dx as y is to x

**rainer** · January 9th 2011, 11:19 AM

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

**wonderboy1953** · January 9th 2011, 11:27 AM

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).

**rainer** · January 9th 2011, 01:43 PM

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?

**e^(i*pi)** · January 9th 2011, 01:48 PM

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

**HallsofIvy** · January 9th 2011, 02:47 PM

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!

**rainer** · January 13th 2011, 10:12 AM

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!

**HallsofIvy** · January 13th 2011, 10:51 AM

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

**rainer** · January 25th 2011, 05:20 PM

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

**Prove It** · January 25th 2011, 06:53 PM

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...

**rainer** · January 27th 2011, 09:09 AM

oh yeah. Duh!

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