# First order differential

1. Apr 28, 2015

### Marcis Rancans

I dont understand this first order differential equation:
How is it possible to get an exponent as answer?

Last edited by a moderator: May 7, 2017
2. Apr 28, 2015

### SteamKing

Staff Emeritus
What if Eq. (2) were re-written as follows:

dV/dt = -(E/R) * V

then, separating the variables:

dV/V = -(E/R) dt

Care to take it from here?

3. Apr 28, 2015

### Marcis Rancans

I would like to see what comes up next! :)

4. Apr 28, 2015

### SteamKing

Staff Emeritus
Why can't you solve this DE now? All you have to do is integrate both sides of the equation.

5. Apr 28, 2015

### Marcis Rancans

I have no idea how to get constant in front of exponent.

EDIT: Forgot to add "-C" in exponent function which I got drom integration.

Last edited by a moderator: May 7, 2017
6. Apr 28, 2015

### SteamKing

Staff Emeritus
If you have ln V = Mess, what do you do to the LHS of the equation to get V? i.e., how is the natural log of a number related to the constant e?

Last edited by a moderator: May 7, 2017
7. Apr 28, 2015

### Marcis Rancans

Isn't that correct what I wrote? V is e^(MESS).

8. Apr 28, 2015

### SteamKing

Staff Emeritus
It is. I'm sorry for not fully recognizing that.

However, when you did the integration to obtain ln V, you forgot to include the constant of integration on the RHS, thus:

∫ dV/V = ∫ (-E/R) dt

ln V = (-E/R)*t + C

Let's say at t = 0, V = V0, then

ln V0 = C, so

ln V = (-E/R)*t + ln V0

exponentiating both sides gives:

eln V = e[(-E/R)*t + ln V0]

which can be simplified:

V = e(-E/R)*t * eln V0 {using the law of exponents}

V = V0 * e(-E/R)*t

if T = R/E, then

V = V0 * e(-t / T)

9. Apr 28, 2015

### Marcis Rancans

Thanks for clear explanation! Only thing I dont understand why constant appears only on one side of equation?

10. Apr 28, 2015

### SteamKing

Staff Emeritus
Technically, integrating both sides of the equation will result in a constant of integration for each integral:

∫ dV/V = ∫ (-E/R) dt

ln V + C1 = (-E/R) * t + C2

The two separate constants of integration can be combined into one constant:

ln V = (-E/R) * t + C2 - C1 = (-E/R) * t + C, where C = C2 - C1

and the solution proceeds as described in Post #8.