Another Seperable Differential EQ

**xwanderingpoetx** · April 9th 2009, 01:46 PM

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

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?

**skeeter** · April 9th 2009, 01:53 PM

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?

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

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

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

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

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

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

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

**xwanderingpoetx** · April 9th 2009, 02:03 PM

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

**skeeter** · April 9th 2009, 02:05 PM

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 ...

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

**xwanderingpoetx** · April 9th 2009, 02:07 PM

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

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