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Starting initial value problems when terms are expressed only in x

  1. Sep 17, 2011 #1
    How should I go about starting the following two problems

    solve the following initial value problems

    x''-4x=0 ; x(0)=1, x'(0)=0
    and
    x''-4x'+4x=0 ; x(0)=1, x'(0)=0

    Should i use a general solution with the basis m^2-4=0 for problem one and m^2-4x+4=0 for the second?
     
  2. jcsd
  3. Sep 17, 2011 #2

    ehild

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    Yes, if you meant the solutions in form exp(mt), and m^2-4m+4=0 for the second problem.


    ehild
     
  4. Sep 17, 2011 #3
    Thanks ehild, do I then use x=1 and and t=0 to find constant A and x'=1 and t=0 to find constant B for both questions?
     
  5. Sep 17, 2011 #4

    ehild

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    The general solution is x=A exp(m1t)+B exp(m2t). Use t=0 x=1 to get one equation for A and B, and t=0, x'=0 in x'(t) to get the other equation for A and B. You get two equations, with two unknown, solve for A and B.

    ehild
     
  6. Sep 18, 2011 #5
    Working through my first problem x''-4x=0 ; x(0)=1 x'(0)=0

    characteristic equation is m^2-4=0

    solutions m=+/-2i

    so complex solution (e^2it, e^-2it)

    since (cos2t, sin2t)

    x(t)=Acos2t-Bsin2t

    For x(0)=1

    1=A-0
    so
    A=1

    x'(t)= -2Asin2t-2Bcos2t

    for x'(0)=0

    0=0-2B
    so
    B=0

    therefore

    x(t)=cos2t

    does this seem correct?
     
  7. Sep 18, 2011 #6

    ehild

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    The solutions of the characteristic equation m^2-4=0 are real m=±2.

    ehild
     
  8. Sep 18, 2011 #7
    Im sorry ehild im confused by this, does this mean that my A and B constants are correct making x(t) = cos2t a solution?
     
  9. Sep 18, 2011 #8

    ehild

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    cos(2t) and sin(2t) are not solutions.

    ehild
     
  10. Sep 18, 2011 #9

    HallsofIvy

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    If one solution to the characteristic equation of a linear differential equation with constant, real, coeficients is the complex number a+ bi, then [itex]y= e^{at}cos(bt)[/itex] and [itex]e^{at}sin(bt)[/itex] are solutions.

    In particular, if the real number a (+ 0i) is a root then a solution is [itex]y= e^{at}[/itex].

    If the imaginary number 0+ bi is a root then two solutions are cos(bt) and sin(bt). (Of course, complex, including imaginary, roots of an equation with real coefficients always come in conjugate pairs: [itex]a\pm bi[/itex].)

    Which case applies here?
     
  11. Sep 18, 2011 #10
    the real number is a root, so my solution must be in the form y= e^{at}[/itex]. ?
     
  12. Sep 18, 2011 #11
    the real number is a root, so my solution must be in the form y= e^{at}[/itex]. ?
     
  13. Sep 18, 2011 #12

    ehild

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    Yes, and you have two values for the exponent, the solutions of your characteristic equation m^2-4=0. You have written that already, but you said that the solutions are 2i and -2i, which is wrong. m^2-4=0 is equivalent m^2=4. What are the roots?

    ehild
     
  14. Sep 18, 2011 #13
    The solutions of the characteristic equation m^2-4=0 are real m=±2.

    does the above not mean that -2 and 2 are roots?
    So that then m1 and m2 are -2and 2?
     
    Last edited: Sep 18, 2011
  15. Sep 18, 2011 #14

    ehild

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    Yes, 2 and -2 are the roots.

    You got the characteristic equation originally by assuming that the solution of the differential equation x''-4x=0 is of the form x(t)=emt.
    The second derivative is x''(t)=m2emt.
    Substituting x(t) and x''(t) back into the DE, you get

    m2emt-4emt=0,

    factoring out the exponent :

    (m2-4)emt=0, which can be true for all t if

    m2-4 =0.

    This is true for m=2 and m=-2, so the solutions of the DE can be written really in the form of x(t)=emt, and we have two independent solutions, x1(t)=e2t and x2(t)=e-2t.

    The general solution is a linear combination of these two functions,

    x(t)=Ae2t+Be-2t.

    Find A and B so that x(0)=1 and x'(0)=0.

    ehild
     
  16. Sep 18, 2011 #15
    I think ehild I have finally arrived at the correct solution

    when x(t) = Ae^2t + Be^-2t

    x(0)=1

    Ae^2t + Be^-2t=1

    and x'(0)=0

    2Ae^2t - 2Be^2t =0

    so with simultaneous equations A = 1/2 AND B = 1/2

    I arrive at

    x(t)=1/2 e^2t + 1/2 e^-2t

    I hope you can tell me this is correct?
     
  17. Sep 18, 2011 #16

    ehild

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    Yes, it is correct. Nice work! Now solve the other problem.

    ehild
     
  18. Sep 18, 2011 #17
    Thank you ehild. I believe I have now managed the other problem, can you confirm the solution to be x(t) = e^2t - 2te^2t

    I really appreciate all the help you have given me.
     
    Last edited: Sep 18, 2011
  19. Sep 19, 2011 #18

    ehild

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    Good, it is the solution! But use parentheses in the exponent!

    ehild
     
  20. Sep 29, 2011 #19
    Hi,

    I too am trying to solve the second of your problems.

    I have used the characteristic equation to find ou that m = 2

    I have then used the general solution:

    x(t) = Ae^m0t + Bte^m0t

    This has given me the following:

    x(0) = Ae^2t + Bte^2t = 1

    Ae^2(0) + B(0)e^2(0) = 1

    A(1) + B(0)(1) = 1

    A + 0 = 1

    Therfore A = 1

    For B:

    x'(0) = 2Ae^2t + 2Bte^2t = 0

    This is where I am getting confused. Plugging in the 0 value for t:

    2Ae^2(0) + 2B(0)e^2(0) = 0

    2A(1) + 2B(0)(1) = 0

    As A = 1

    2(1)(1) + 2B(0)(1) = 0

    2 + 2B(0)(1) = 0

    2B(0)(1) = -2

    Now from your final answer I see that A = 1 and B = -2 to give x(t) = e^2t - 2te^2t

    When I multiply out the RHS of the above equation does B remain even though it has been multiplied by zero? I cant see this as in the equation to solve A as 1 the B side equals 0 and the sum is the same. I am sure I am missing something obvious but the more I look at it the more clouded it becomes!
     
  21. Sep 29, 2011 #20

    ehild

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    x(t) = Ae^(2t) + Bte^(2t)

    The second term is a product of t and e^(2t), apply the product rule!

    x'(t) = 2Ae^(2t) + 2Bte^(2t)+Be^(2t).

    ehild
     
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