1. ## Help with this D.E please

1/v dv=-ke^(s/h) ds

Where k is a constant

Im trying to get v as a function of s by sep. of variables and am fairly confused by the algebra. Any help would be appreciated.

2. Originally Posted by msu_15
1/v dv=-ke^(s/h) ds

Where k is a constant

Im trying to get v as a function of s by sep. of variables and am fairly confused by the algebra. Any help would be appreciated.
What is h?

If it is just another constant then your equation is ready to be integrated...

$\int \frac{1}{v} dv = -k \int e^{\frac{s}{h}}ds$

3. Originally Posted by msu_15
1/v dv=-ke^(s/h) ds

Where k is a constant

Im trying to get v as a function of s by sep. of variables and am fairly confused by the algebra. Any help would be appreciated.
The variables are already separated in this case. My solution only works if k and h are constants too.

Integrating : $\int \frac{dv}{v} = \int -ke^{(s/h)} ds$

$ln|v| = -ke^{\frac{s}{h}} \times \frac{h}{s} + C$

Can you solve it from there? I tried going further but could only get to s^s =f(v)

4. h is a constant as well. my bad