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Initial value problem

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data
    How can I find a general solution to this problem:

    5y" + 9y' - 9y = 0


    2. Relevant equations



    3. The attempt at a solution

    I am confused with second order different equations in it
     
  2. jcsd
  3. Sep 28, 2009 #2

    CompuChip

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    Homework Helper

    Usually with this type of equations (homogeneous ordinary differential equations) the easiest way is to plug in a trial solution [itex]y(t) = e^{\lambda t}[/itex], then solve the resulting polynomial for [itex]\lambda[/itex] and make a linear combination to get the general solution.

    (For non-homogeneous equations, add to that a particular solution).
     
  4. Sep 28, 2009 #3
    can't I just integrate twice and get the answer? how can I solve this using the method of undeterminant coefficients
     
  5. Sep 28, 2009 #4

    Mark44

    Staff: Mentor

    No. You can integrate y'' twice, but how are you going to integrate y' and y twice?

    CompuChip has given you some good advice. Try it out.
     
  6. Sep 28, 2009 #5
    I don't get what he meant by make a linear combination
     
  7. Sep 28, 2009 #6

    Mark44

    Staff: Mentor

    Assume that the solution is of the form y = ert. Calculate the first and second derivatives of this function and plug them into your differential equation. After you factor out ert, you will have a quadratic that you can solve for r. (CompuChip used lambda, but the idea is the same.) You should get two values for r--r1 and r2, meaning that y1 = er1t and y2 = er2t are solutions of this DE. The general solution is all possible linear combinations of these two functions. IOW, y = c1er1t + c2er2t.

    A linear combination of functions is the sum of multiples of those functions.

    Clear?
     
  8. Sep 30, 2009 #7
    what if it only has one root, say (y-3)^2, so the only root is y = 3
     
    Last edited: Sep 30, 2009
  9. Sep 30, 2009 #8

    Mark44

    Staff: Mentor

    Then your solutions would be e3t and te3t. Your general solution would be the linear combinations of these two functions.

    IOW, yg = c1e3t + c2te3t.
     
  10. Sep 30, 2009 #9
    if it's y^2 + 25y , then the two roots are y = 0 and y = -25. so yg = c1 + c2e^(-25t).
     
  11. Oct 1, 2009 #10

    CompuChip

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    Yes. You can understand why you only get one solution which is non-trivial (i.e. not equal to a constant number everywhere): if r = 0 is a solution to the polynomial, then that means that your equation must have been of the form
    a y'' + b y' = 0
    so without a term in y.
    In that case you can view it as a (simpler) first order equation for v = y':
    a v' + b v = 0
    which of course only has one independent solution.

    In general, if your equation is nth order (i.e. the highest derivative of y that occurs is the nth derivative) you get n independent solutions which you should add (indeed, when you plug in the exponential solution, you get a polynomial of degree n). You should be advised, however, that the roots of the polynomial are not always real, like for example in y'' + y = 0.
     
  12. Oct 1, 2009 #11
    well if have initial value y(0) = 2 and y'(0)= 5

    then what I get is

    y' = -25c2e^(-25t)
    5 = -25c2
    so c2 = -1/5

    and c1 is equal to 11/5

    is the true? reason I asked is because the stupid webassign won't accept this answer
     
  13. Oct 1, 2009 #12

    Mark44

    Staff: Mentor

    Are you talking about this initial value problem? (This is a different problem from the one you first posted in this thread.)
    y'' + 25y = 0, y(0) = 2, y'(0) = 5

    If so, your solution looks fine to me. Maybe the webassign thing is looking for decimal values instead of fractions.
     
  14. Oct 1, 2009 #13
    yes this is a different problem, thanks though for confirming and all the help
     
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