# Separating Variables

• Dec 24th 2009, 11:55 AM
matlabnoob
Separating Variables
Hi everyone...

Im stuck on this question (and its xmas eve (Crying)i dont want to be spending hours on this what appears to be a very simple problem....aaah! )

so in my lecture notes, i have the equation,

dy/dt = y+1/t-1 given the boundary condition y=1 at t=0

separating variables gives,

integral( dy/y+1) = integral (dt/y-1)
ln(y+1) = ln(t-1) + lnc - c

OR (now this is where im lost..how did they get c - bc :confused: and how on earth did c equal -2??? you'll see...)

y+1/t-1 = c - bc tells us that c = -2

y = 2(1-t)-1
• Dec 24th 2009, 12:06 PM
pickslides
Quote:

Originally Posted by matlabnoob

integral( dy/y+1) = integral (dt/y-1)
ln(y+1) = ln(t-1) + lnc - c

You could say

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

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

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

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

$\displaystyle y = c(t-1)-1$

Now use your I.C. $\displaystyle y=1$ at $\displaystyle t=0$ to find $\displaystyle c$.

You should find $\displaystyle c = -2$

Quote:

Originally Posted by matlabnoob

OR (now this is where im lost..how did they get c - bc :confused: and how on earth did c equal -2??? you'll see...)

Where did $\displaystyle b$ come from?
• Dec 24th 2009, 12:11 PM
galactus
$\displaystyle \frac{dy}{dt}=\frac{y+1}{t-1}$

Separate:

$\displaystyle \frac{dy}{y+1}=\frac{dt}{t-1}$

Integrate:

$\displaystyle ln(y+1)=ln(t-1)+C$

e to both sides:

$\displaystyle y+1=e^{ln(t-1)+C}=e^{C}(t-1)$

But $\displaystyle e^{C}$ is a constant we can rename and call $\displaystyle C_{1}$

$\displaystyle y=C_{1}(t-1)-1$

Now, use your IC to solve for C1
• Apr 1st 2014, 09:07 PM
DanielPeyton
Re: Separating Variables
As wells immune balancing and the body Raspberry Ultra Drops and that was really the evolution of his particular uh... program the original ail dot go repair developed by the late doctor heard other without them is now available north America the treatment procedure two thousand.