# simple integral separation problem

• Feb 18th 2011, 10:38 PM
rainer
simple integral separation problem
If I know that:

$\displaystyle \frac{dk}{dl}=\frac{w}{v}$

$\displaystyle dk=\frac{w}{v}dl$

$\displaystyle \int 1 dk=\frac{w}{v}\int1dl$

$\displaystyle \frac{k}{l}=\frac{w}{v}$

?

Thanks
• Feb 18th 2011, 10:41 PM
Prove It
What variable are $\displaystyle \displaystyle w$ and $\displaystyle \displaystyle v$ functions of?
• Feb 18th 2011, 10:49 PM
CaptainBlack
Quote:

Originally Posted by rainer
If I know that:

$\displaystyle \frac{dk}{dl}=\frac{w}{v}$

$\displaystyle dk=\frac{w}{v}dl$

$\displaystyle \int 1 dk=\frac{w}{v}\int1dl$

$\displaystyle \frac{k}{l}=\frac{w}{v}$

?

Thanks

You are treating w/v as though it is a constant, and you have also lost a constant of integration.

CB
• Feb 18th 2011, 10:56 PM
rainer
Yes, w/v is a constant.

Where am I missing a constant of integration? I thought the constant of integratoin on one side cancels out with the constant of integration on the other side.
• Feb 18th 2011, 11:57 PM
CaptainBlack
Quote:

Originally Posted by rainer
Yes, w/v is a constant.

Where am I missing a constant of integration? I thought the constant of integratoin on one side cancels out with the constant of integration on the other side.

What you have is:

$\displaystyle \displaystyle \dfrac{dk}{dl}=a$

This has solution

$\displaystyle k(l)=al+c$

Or to put it another way, you would have two arbitrary constants doing it your way. The sum and difference of arbitrary constants are both another arbitrary constant. That is the two constants are not equal.

CB
• Feb 18th 2011, 11:59 PM
CaptainBlack
Quote:

Originally Posted by rainer
Yes, w/v is a constant.

Where am I missing a constant of integration? I thought the constant of integratoin on one side cancels out with the constant of integration on the other side.

In that case; this is an exercise in the use of the fundamental theorem of calculus not in differential equations (so don't try to separate the variables).

CB