1. ## First Order ODE

$\frac{dz}{dt} + 3e^{t+z} = 0$

Here's what I have done:

$\frac{dz}{dt} = -3e^{t+z}$

$\ln(dz) - \ln(dt) = \ln(-3) + ln(e^{t+z})$

Am I approaching this correctly? Something is obviously wrong since the ln(-3) is undefined, which makes me think that I am doing this wrong.

Thanks!

2. ## re: First Order ODE

I see it like this

$\frac{dz}{dt} = -3e^{t+z}$

$\frac{dz}{dt} = -3e^{t}e^{z}$

$e^{-z}~dz= -3e^{t}~dt$

$\int e^{-z}~dz= -3 \int e^{t}~dt$

etc..

3. ## re: First Order ODE

Thanks for that. This is my very first differential equation (my deflowering, so to speak), so this is all new to me.

4. ## re: First Order ODE

But wait...

Carrying on from where you left off:

$e^{-z} = -3e^t + C$

$-z = \ln(-3e^t + C)$

I'm back to my same dilemma. I have the logarithm of a negative number. How do I handle that? Thanks again.

5. ## re: First Order ODE

Originally Posted by joatmon

$-z = \ln(-3e^t + C)$

I'm back to my same dilemma. I have the logarithm of a negative number. How do I handle that?
But do you? consider the case where t=0 and C=4. Is the logarithm negative?

6. ## re: First Order ODE

I thought of that, too, that the constant can always push the equation positive. Where I am confused is that when I run this problem through Wolfram (or my calculator) to check my work, it shows the answer as $\ln(3e^t + C)$. I'm just trying to understand whether I am getting the right answer or not, and I have two sources (that I trust more than my own work) that have a slightly different answer.

Any thoughts? Thanks again.

7. ## Re: First Order ODE

Originally Posted by joatmon
But wait...

Carrying on from where you left off:

$e^{-z} = -3e^t + C$

$-z = \ln(-3e^t + C)$

I'm back to my same dilemma. I have the logarithm of a negative number. How do I handle that? Thanks again.
I don't quite get the same result. I have

$-e^{-z}=-3e^{t}+C,$ and so

$e^{-z}=3e^{t}+C,$ where I've redefined my constant. Then

$-z=\ln(3e^{t}+C),$ and hence

$z=-\ln(3e^{t}+C).$