# 1st order diff equ-separation of variables EASY!!!

• Jan 30th 2010, 08:51 PM
mxrider777
1st order diff equ-separation of variables EASY!!!
Okay lets see if someone on here can help me. This problem looks pretty easy:
dy/dt=1(1-y)
I see that its an autonomous equation that will allow for separation of variables, which I do. I separate and integrate accordingly which yields me:
t+c=ln(y)-ln(y-1)+c
now I raise each side exponentially to get rid of the ln's:
e^(t+c)=e^(ln(y))-e^(ln(y-1))+e^c
this is where I think I am making a mistake. There must be some law of logarithms that I don't see because the answer turns out to be (according to the back of my book):
y(t)=ce^t/(ce^t+1)
I can only come up with the ce^t and the rest is wrong. Can someone fill me in to what I need to do/what I am doing wrong?
• Jan 30th 2010, 08:59 PM
VonNemo19
Quote:

Originally Posted by mxrider777
Okay lets see if someone on here can help me. This problem looks pretty easy:
dy/dt=1(1-y)
I see that its an autonomous equation that will allow for separation of variables, which I do. I separate and integrate accordingly which yields me:
t+c=ln(y)-ln(y-1)+c
now I raise each side exponentially to get rid of the ln's:
e^(t+c)=e^(ln(y))-e^(ln(y-1))+e^c
this is where I think I am making a mistake. There must be some law of logarithms that I don't see because the answer turns out to be (according to the back of my book):
y(t)=ce^t/(ce^t+1)
I can only come up with the ce^t and the rest is wrong. Can someone fill me in to what I need to do/what I am doing wrong?

$\displaystyle \frac{1}{1-y}dy=(1)dt$

$\displaystyle \int\frac{1}{1-y}dy=\int(1)dt$.

Can you finish?

EDIT: That wasn't youre trouble. Sorry, misread.
• Jan 30th 2010, 09:52 PM
Jhevon
Quote:

Originally Posted by mxrider777
Okay lets see if someone on here can help me. This problem looks pretty easy:
dy/dt=1(1-y)
I see that its an autonomous equation that will allow for separation of variables, which I do. I separate and integrate accordingly which yields me:
t+c=ln(y)-ln(y-1)+c
now I raise each side exponentially to get rid of the ln's:
e^(t+c)=e^(ln(y))-e^(ln(y-1))+e^c
this is where I think I am making a mistake. There must be some law of logarithms that I don't see because the answer turns out to be (according to the back of my book):
y(t)=ce^t/(ce^t+1)
I can only come up with the ce^t and the rest is wrong. Can someone fill me in to what I need to do/what I am doing wrong?

check what you wrote. the solution you claim the book has is not the solution to your problem. the solution to $\displaystyle \frac {dy}{dt} = 1 - y$ is $\displaystyle y = 1 - Ce^{-t}$ NOT $\displaystyle y = \frac {Ce^t}{Ce^t + 1}$
• Jan 31st 2010, 06:08 AM
Coomast
just for information:

$\displaystyle y=\frac{Ce^t}{Ce^t+1}$ is the solution of $\displaystyle \frac{dy}{dt}=y(1-y)$.

Maybe a typo in the book?

Coomast