1. ## ODE

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!

2. Originally Posted by cdlegendary

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

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

Now just integrate...

3. $\displaystyle ln(y) = \int -e^t/(1+e^t) dt$

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

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

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

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?

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

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

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

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

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?
You have made an algebra error

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

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