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Solving equation simultaneously for constant

  1. Jul 15, 2012 #1
    This is for 2nd order homogenous differential equation but that isn't what needs to be solved. What I need help with is solving simultaneously-- something that I am terrible with.

    [itex]y=c_1e^t+c_2e^{-t}[/itex]
    With t = 0, y = 2 we have [itex]c_1 + c_2 = 2[/itex]

    Upon differentiation of y, we obtain: [itex]y'=c_1e^t-c_2e^{-t}[/itex]

    With t = 0, y' = -1 [itex]c_1 - c_2 = -1[/itex]

    Now the book says that upon solving these two equations simultaneously, [itex] c_1 = 1/2 & c_2 = 3/2[/itex]

    So I try to verify the result, I set up the two equations to solve by elimination

    [itex]c_1 + c_2 = 2[/itex]
    [itex]c_1 - c_2 = -1[/itex]
    c_2 cancels out & we obtain c_1 = 1 which is the wrong conclusion. I can reason out that c_1 should be 1/2 & c_2 should be 3/2, but first and foremost I want to know why solving by elimination gives the wrong answer and what method would be the correct one.
     
  2. jcsd
  3. Jul 15, 2012 #2

    SammyS

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    If
    [itex]c_1 + c_2 = 2[/itex]
    [itex]c_1 - c_2 = -1[/itex]​
    Then adding the two equations gives
    [itex]2c_1=1\ [/itex]​
     
  4. Jul 15, 2012 #3
    Oh--duh. That was probably the easiest problem you've had to solve to date :rofl: .

    1 + 1 ≠ 1 .. maybe it is time I get some sleep. :grumpy:
     
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