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Brushing up on the basics of diff eq

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data
    y(x) = C1sin(2x) + C2cos(2x)

    2. Relevant equations
    y(∏/8) = 0


    3. The attempt at a solution
    C1(1/2)(√2) + C2(1/2)(√2) + 1 = 0


    The book jumps to the equation below and I'm having trouble figuring out how they got there.
    C1 + C2 = -√2
     
  2. jcsd
  3. Dec 30, 2011 #2

    SammyS

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    What is the dif. eq. you're trying to solve?

    Multiply your "attempt at a solution" by √2 .
     
  4. Dec 30, 2011 #3
    Determine C1 and C2 so that y(x) = C1sin(2x) + C2cos(2x) will satisfy the initial conditions y(∏/8) = 0 and y'(∏/8) = √2
     
  5. Dec 30, 2011 #4

    SammyS

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    Multiply your "attempt at a solution" in your original post by √2 .
     
  6. Dec 30, 2011 #5
    I guess it's safe to assume that C1/2 = C1 and C2/2 = C2 considering they're both unknowns at this point?
     
  7. Dec 30, 2011 #6

    SammyS

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    That doesn't make sense, unless you are asking if you can replace C1/2 by C1 and C2/2 by C2 .

    If [itex]C_1+C_2=-\sqrt{2}\,,[/itex] then you can replace [itex]C_2[/itex] with [itex]-C_1-\sqrt{2}[/itex]
     
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