http://homework.math.ucsb.edu/webwor...36b80d48b1.png

Does anyone know how to find the general solution for the above ODE?

I've tried an integrating factor and I can't seem to get anywhere with it. Any pointers would be much appreciated!

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- Jun 6th 2010, 04:27 PMcdlegendaryODE
http://homework.math.ucsb.edu/webwor...36b80d48b1.png

Does anyone know how to find the general solution for the above ODE?

I've tried an integrating factor and I can't seem to get anywhere with it. Any pointers would be much appreciated! - Jun 6th 2010, 04:33 PMTheEmptySet
- Jun 6th 2010, 04:57 PMcdlegendary
$\displaystyle ln(y) = \int -e^t/(1+e^t) dt$

$\displaystyle ln(y) = -ln(1+e^t) + c$

$\displaystyle y = -e^{1+e^t} * e^c$

$\displaystyle y = -Ce^{1+e^t}$

Alright, so I've integrated the equation and solved for y. It doesn't seem to be correct though. Is there any error in my calculations? - Jun 6th 2010, 05:04 PMTheEmptySet
You have made an algebra error

This is correct but to simplify calculations add $\displaystyle \ln(c)=c$

for the arbitary constant

$\displaystyle ln(y) = -ln(1+e^t) + \ln(c) \iff \ln(y)=\ln\left( \frac{1}{1+e^{t}}\right)+\ln(c)=\ln\left( \frac{c}{1+e^{t}} \right)$

$\displaystyle y=\frac{c}{1+e^{t}}$