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First-Order Differential Equation

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data


    Find x(t) given the following:
    x(0)=2
    [tex]dx/dt+4x=10e^{-2t}[/tex]
    [tex]e^{4t}(dx/dt+4x)=10e^{-2t}e^{4t}[/tex]
    [tex]d/dt(e^{4t}x)=10e^{2t}[/tex]
    [tex]\int{d(e^{4t}x)}=\int{10e^{2t}dt}[/tex]
    [tex]e^{4t}x+C=\int{10e^{2t}}[/tex]
    [tex]e^{4t}x=\int{10e^{2t}}+C_1[/tex]
    [tex]x(t)=\int{10e^{-2t}}+e^{-4t}C_1[/tex]
    From here I get stuck. I know I can integrate it, but then I have another constant?
     
  2. jcsd
  3. Jan 10, 2012 #2

    Dick

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    No, no. Your last line should read,

    [tex]x(t)=e^{-4t}\int{10e^{2t}}+e^{-4t}C_1[/tex]

    You can't just move the e^(-4t) inside the integral like that. Now sure, integrate it. If you add another constant, does it matter?
     
  4. Jan 10, 2012 #3
    [tex]x(t)=e^{-4t}\int{10e^{2t}}+e^{-4t}C_1[/tex]
    [tex]x(t)=e^{-4t}(5e^{2t}+C_2)+e^{-4t}C_1[/tex]

    Then, since
    [tex]x(0)=2[/tex]
    we have
    [tex]-3=C_2+C_1[/tex]
    Now if I combine these constant to make a new one, that doesn't really help me out in regards to the equation does it?
     
  5. Jan 10, 2012 #4

    SammyS

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    Why not integrate the right hand side ?
     
  6. Jan 10, 2012 #5
    @ Sammy:
    I was doing that before, but I could not seem to get the answer. I just tried again given the information Dick presented and got the right answer. Thanks!
     
  7. Jan 10, 2012 #6

    Dick

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    [tex]x(t)=e^{-4t}(5e^{2t}+C_2)+e^{-4t}C_1[/tex]

    is the same as

    [tex]x(t)=5e^{-2t}+e^{-4t}(C_2+C_1)[/tex]

    You may as well just call C=C1+C2. The extra constant doesn't do anything different than the first one did.
     
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