# Thread: Another Seperable Differential EQ

1. ## Another Seperable Differential EQ

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 ...

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

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