ODE

• Jun 6th 2010, 05:27 PM
cdlegendary
ODE
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, 05:33 PM
TheEmptySet
Quote:

Originally Posted by cdlegendary
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!

The equation is first order linear and seperable

$(1+e^{t})\frac{dy}{dt}=-e^{t}y \iff \frac{dy}{y}=-\frac{e^{t}}{1+e^{t}}dt$

Now just integrate...
• Jun 6th 2010, 05:57 PM
cdlegendary
$ln(y) = \int -e^t/(1+e^t) dt$

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

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

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

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, 06:04 PM
TheEmptySet
Quote:

Originally Posted by cdlegendary
$ln(y) = \int -e^t/(1+e^t) dt$

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

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

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

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?

You have made an algebra error

This is correct but to simplify calculations add $\ln(c)=c$
for the arbitary constant
$ln(y) = -ln(1+e^t) + \ln(c) \iff \ln(y)=\ln\left( \frac{1}{1+e^{t}}\right)+\ln(c)=\ln\left( \frac{c}{1+e^{t}} \right)$

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