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Differential Eq. Initial Value Problem (separable/partial fractions/hyperbolic)

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data

    dx/dt = 2*(1-x^2) with x(0)=0

    Obtain the analytic solution to the IVP. Then, obtain the terms in the Taylor series up to and including the t^5 term.

    Note: should be possible to express x(t) in terms of hyperbolic functions.

    3. The attempt at a solution

    some of my steps;

    dx/(1-x^2) = 2dt

    *partial fractions + integrated*

    0.5[ ln|-x+1| + ln|x+1| ] = 2t + C

    -x^2 +1 = A*exp(4t) , A>=0

    x(t)=sqrt(1-exp(4t))

    im not sure if this is right at all, as its simply been a while since ive looked at DE's. If it is right, i cant seem to figure out what combination of hyperbolic functions it relates to! any help would be much appreciated!
     
  2. jcsd
  3. Mar 9, 2010 #2

    Mark44

    Staff: Mentor

    The work to here is fine, but the line below doesn't take into account the absolute value. Using the initial condition, you should have |1 - x2| = e4t.

    For what x is the expression 1 - x2 nonnegative? Negative?
    It seems to me that there is another part of this problem: to assume a solution in the form of a Maclaurin series, and to find its coefficients up to the x5 term. You're assuming a solution of the form x(t) = a0 + a1t + ... + a5t5 + ... + antn + ...

    Basically, what you want to do is to substitute this into dx/dt + 2x2 - 2 = 0, and equate coefficients.
     
  4. Mar 9, 2010 #3
    i thought because i conditioned A to be greater than or equal to zero, the RHS would stay positive, and i could drop the absolutes on the LHS... is there something wrong with my logic there?

    all i really need to find is a suitable x(t) before i start subbing things into a maclaurin series... the hint said i should be able to write x(t) in terms of hyperbolic functions, and because none are springing out at me im getting the feeling im wrong.
     
  5. Mar 9, 2010 #4

    Mark44

    Staff: Mentor

    Yes, there is. Use your initial condition to find A. You can't just arbitrarily say that A needs to be positive. Assuming that you find that A is positive (I did), you can't just get rid of the absolute values. Without them, 1 - x^2 is negative for some x and positive for others.
     
  6. Mar 9, 2010 #5
    But because x is the dependant variable, we don't have to worry about crazy x values that cause errors... as all values of t (-inf, inf) will make x behave nicely. If the initial values didn't agree with the restraints on A then id agree there's a problem, but because they do I'm still not hugely convinced I'm wrong.

    looking at 1-x2 = Ae4t , A>=0

    x(0)=0 implies A=1

    therefore x(t)=sqrt(1-exp(4t))

    I know I must be sounding like a stubborn idiot, I apologize. I've got class tomorrow so I can run it through with the lecturer and post the answer.
     
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