# Thread: Another Seperable Differential EQ

1. ## Another Seperable Differential EQ

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

I moved the $\displaystyle e^{t+z}$ to the right side of the equation and came out with.

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

Am I right so far, before I take the integral of both sides?

And can the integral of $\displaystyle \frac{1}{e^z}$ be taken without using parts?

2. Originally Posted by xwanderingpoetx
$\displaystyle \frac{dz}{dt} + e^{t+z} = 0$

I moved the $\displaystyle e^{t+z}$ to the right side of the equation and came out with.

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

Am I right so far, before I take the integral of both sides?

And can the integral of $\displaystyle \frac{1}{e^z}$ be taken without using parts?
$\displaystyle \frac{dz}{dt} = -e^{t+z}$

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

$\displaystyle e^{-z} \, dz = -e^t \, dt$

$\displaystyle -e^{-z} = C - e^t$

$\displaystyle e^{-z} = e^t + C$

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

$\displaystyle z = \ln\left(\frac{1}{e^t + C}\right)$

3. I just have one question, why did you take the reciprocal at the end when you divided out the -1?

4. Originally Posted by xwanderingpoetx
I just have one question, why did you take the reciprocal at the end when you divided out the -1?
power rule for exponents ...

$\displaystyle -\ln{a} = \ln{a^{-1}} = \ln\left(\frac{1}{a}\right)$

5. Wow, I completely blanked on that. Thank you =D