My Math Forum Separation of Variables

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 September 22nd, 2010, 02:21 AM #1 Member   Joined: Jul 2010 Posts: 34 Thanks: 0 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
 September 22nd, 2010, 03:40 AM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 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).
 September 22nd, 2010, 05:27 AM #3 Member   Joined: Jul 2010 Posts: 34 Thanks: 0 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.
 September 22nd, 2010, 11:28 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs 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)
 September 23rd, 2010, 12:58 AM #5 Member   Joined: Jul 2010 Posts: 34 Thanks: 0 Re: Separation of Variables Thanks heaps Mark.

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