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September 22nd, 2010, 01:21 AM   #1
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Separation of Variables

Okay,
So I'm attempting to find the general solution to dx/dt = t(x-2)
My working
dx/dt = t(x-2)
= f(t)g(x)
where f(t) = t
and g(x) = x-2
dividing both sides by g(x)gives
(1/(x-2)dx/dt = t
integrating I get
?1/(x-2) dx/dt dt = ?t dt
simplifying
?1/(x-2) dx = ?t dt
so
ln(|x-2|)=(t^2)/2
solving explicitly I get
x = 2 + e^(t^2/2) or x = 2 - e^(t^2/2)
However my textbook give the answer x = 2 + Ae^(t^2/2)
My question is where does the "A" come from and how does it negate the "2 -"
Thanks
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September 22nd, 2010, 02:40 AM   #2
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Re: Separation of Variables

dx/dt = t(x-2)
(dx/dt) * dt = dx = t(x-2)dt
dx/(x-2) = tdt
ln(x - 2) = 1/2 * t^2 + C_1 C_1 is a constant. Also I did not use absolute value because if 2 - x was a possibility, then the integrand would've been 1 / (2 - x) in the first place.
e^(1/2 * t^2 + C_1) = x - 2
x = e^(1/2 * t^2 + C_1) + 2 = 2 + e^(C_1) * e^(1/2 * t^2) e^(C_1) is a constant.
Therefore x = 2 + Ce^(1/2 * t^2).
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September 22nd, 2010, 04:27 AM   #3
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Re: Separation of Variables

Okay, Thanks Heaps
So the next part of the questions is "find the particular solution that to satisfy condition x(0)=5.
Can you offer some insight into this? I don't even know where to begin.
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September 22nd, 2010, 10:28 AM   #4
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Re: Separation of Variables

x = 2 + Ae^(t^2/2), x(0) = 5

What you do is let t = 0, x = 5, and solve for A in the general solution:

5 = 2 + Ae^(0^2/2) = 2 + A

A = 3

Thus, the particular solution satisfying x(0) = 5 is given by:

x(t) = 2 + 3e^(t^2/2)
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September 22nd, 2010, 11:58 PM   #5
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Re: Separation of Variables

Thanks heaps Mark.
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