September 22nd, 2010, 01:21 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(x2) My working dx/dt = t(x2) = f(t)g(x) where f(t) = t and g(x) = x2 dividing both sides by g(x)gives (1/(x2)dx/dt = t integrating I get ?1/(x2) dx/dt dt = ?t dt simplifying ?1/(x2) dx = ?t dt so ln(x2)=(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, 02:40 AM  #2 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Re: Separation of Variables
dx/dt = t(x2) (dx/dt) * dt = dx = t(x2)dt dx/(x2) = 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, 04:27 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, 10:28 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,167 Thanks: 472 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 22nd, 2010, 11:58 PM  #5 
Member Joined: Jul 2010 Posts: 34 Thanks: 0  Re: Separation of Variables
Thanks heaps Mark.


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