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2nd Order Differential Equation Problem

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Let y be the solution of the initial value problem:
    y'' + 2y' + 2y = 0 , y(0) = 0 , y'(0) = 5

    The maximum value of y over 0 less than or equal to x less than infinity is ??.

    3. The attempt at a solution
    r = -2 +/- i

    I solved it:

    y = c1 e^-2x cos(x) + c2 e^-2x sin(x)
    y' = -2c1 e^-2x cos(x) - c1 e^-2x sin(x) - 2 c2 e^-2x sin(x) + c2 e^-2x cos(x)

    y = c1 = 0
    y' = -2c1 + c2 = 5 => c2 = 5

    y = 5 e^-2t sin(x)

    But I am terrible at trig. What is the maximum value of y? I got (x = -3pi/2 so therefore max y is 5 e^(6pi/2) but that is incorrect. I have one more attempt and don't want to lose out on this question.

    Thanks.
     
    Last edited: Nov 19, 2009
  2. jcsd
  3. Nov 19, 2009 #2

    Mark44

    Staff: Mentor

    The roots of your characteristic equation are wrong. They should be r = -1 +/- i.
     
  4. Nov 19, 2009 #3
    Yeah, I missed that but doesn't change my answer much.

    r = -1 +/- i

    I solved it:

    y = c1 e^-x cos(x) + c2 e^-x sin(x)
    y' = -c1 e^-x cos(x) - c1 e^-x sin(x) -c2 e^-x sin(x) + c2 e^-x cos(x)

    y = c1 = 0
    y' = -c1 + c2 = 5 => c2 = 5

    y = 5 e^-x sin(x)

    But I still can't find the max value of y. Help please?

    I think max of y is 5 e^(3pi/2) where x = -3pi/2. I don't want to exhaust my last attempt. Any confirmations?
     
  5. Nov 19, 2009 #4

    Mark44

    Staff: Mentor

    How do you normally find the maximum value of a function? You probably spent the good part of a quarter or semester doing that.
     
  6. Nov 19, 2009 #5
    I really don't know, my professor moves way to fast and doesn't give enough examples. I also completely forgot about that. If you show me how to do it instead of giving hints, I am sure I will learn it better that way. I would really appreciate that. Thanks.
     
  7. Nov 19, 2009 #6

    Mark44

    Staff: Mentor

    And I'm sure that you will learn it better if you dig for it than if I just tell you. Crack your calculus book open and see if it has anything about finding the largest value of a function.
     
  8. Nov 19, 2009 #7
    I am sorry, but I really don't know. I have been looking for awhile and I can't seem to find it. Did I do the differential equation correct though?
     
  9. Nov 19, 2009 #8

    Mark44

    Staff: Mentor

    Yes, the solution of the initial value problem is y(x) = 5e-xsin(x).

    Any maximum or minimum will come at an endpoint of the interval in question or at a point at which the derivative is zero. I truly hope that rings a bell with you.
     
  10. Nov 20, 2009 #9
    Yeah, I found out that you set the derivative to zero but as I said before I am terrible at trigonmetry. So:

    y'=5e^-x (cos(x)- sin(x))
    0=5e^-x (cos(x)- sin(x))
    5e^-x cos(x) = 5e^-x sin(x)
    cos(x) = sin(x)

    Whats next?
     
  11. Nov 20, 2009 #10
    try solving for x? (dividing throughout by cos(x))
     
  12. Nov 20, 2009 #11
    So, 0 = sin(x)/cos(x) = tan(x)? so tan (0)=0 so the max value of y is 0???
     
  13. Nov 20, 2009 #12
    cos(x)/cos(x) does not equal 0....

    what is 1/1, 5/5, pi/pi ?
     
  14. Nov 20, 2009 #13
    Oh yeah, did the algebra wrong...

    So, sinx/cosx = 1 = tanx
    x=45?

    so I should input the answer as:
    y = 5 e^(-45) sin(45)?
     
  15. Nov 20, 2009 #14
    i think you need to put it in radians, but your method is correct
     
  16. Nov 20, 2009 #15
    Yes!!

    I got it. Thanks Mark and Chewy. Yeah, I forgot to set my calculator back to radians, thanks for the help guys.
     
  17. Nov 20, 2009 #16

    Mark44

    Staff: Mentor

    There are an infinite number of solutions to the equation tan x = 1 on the interval [0, infinity). You have found only one of them. All you know is that dy/dx = 0 at x = pi/4. You don't know whether you have a maximum, minimum, or neither at this point.
     
  18. Nov 20, 2009 #17

    Mark44

    Staff: Mentor

    Some thoughts on several of your comments...
    You've mentioned a couple of times that you're terrible at trigonometry, you seem to have some trouble with algebra, and you didn't remember that the derivative of a function could be used to find the maximum and minimum values of that function. Now you are apparently in a class in differential equations, where knowledge of these areas is assumed. To be weak in all these areas is definitely not a recipe for success in differential equations.

    I don't know what classes you are planning to take in the future, but you probably are going to have a much harder time of it until you can get up to speed in the areas where you are weak. I can't imagine that you are having an easy time of it in your present class.
     
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