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Homework Help: Differential Equation by Separation of Variables

  1. Mar 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation: [itex]y' + cos(x)y = cos(x)[/itex]

    3. The attempt at a solution
    [itex]y' = cos(x)(1 -y)[/itex]
    [itex]\frac{dy}{1-y} = cos(x) dx[/itex]
    [itex]-ln|1-y| = sin(x) + C[/itex]
    [itex]\frac{1}{1-y} = e^{sin(x)} + C[/itex]
    [itex]1-y = \frac{1}{e^{sin(x)} + C}[/itex]
    [itex]y = \frac{-1}{e^{sin(x)} + C} + 1[/itex]

    Have I done this correctly?
     
  2. jcsd
  3. Mar 30, 2012 #2

    Dick

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    No. e^(sin(x)+C) isn't the same as e^sin(x)+C.
     
  4. Mar 30, 2012 #3
    So I know [itex]e^{sin(x) + C} = e^{sin(x)}e^C[/itex]. But since [itex]e^C[/itex] is just a constant, can't I just compress it into the constant C?
     
  5. Mar 30, 2012 #4

    Dick

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    Yes, but that makes it C*e^(sin(x)), doesn't it?
     
  6. Mar 30, 2012 #5
    so I should write,

    [itex]\frac{1}{1-y} = e^{sin(x)}C[/itex]
    [itex]1-y = \frac{1}{e^{sin(x)}C}[/itex], and C is totally arbitrary so write,
    [itex]y = \frac{C}{e^{sin(x)}} + 1[/itex]
     
  7. Mar 30, 2012 #6

    Dick

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    I think that works.
     
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