# Separating Variables

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• 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

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

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

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

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

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

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

You should find $c = -2$

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

Separate:

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

Integrate:

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

e to both sides:

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

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

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

Now, use your IC to solve for C1
• Apr 1st 2014, 09:07 PM
DanielPeyton
Re: Separating Variables
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