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First order DE, using integrating factor

  1. Jan 20, 2010 #1
    Hi,

    I tried to solve this by using the integrating factor technique
    [tex]\begin{cases}
    dy/dt +10y = 1 \\
    y(1/10) = 2/10
    \end{cases}[/tex]

    So [tex]p(x) = 10t \rightarrow e^{10t}[/tex]

    [tex]e^{10t} \cdot \frac{dy}{dt} + e^{10t} \cdot 10y = e^{10t}[/tex]

    This part is confusing to me, i have two different variables y and t.

    How do i integrate the left side?
     
  2. jcsd
  3. Jan 20, 2010 #2

    vela

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    The independent variable in this problem is t. It's playing the role x would usually play.

    The point of the integrating factor is to turn the lefthand side into a derivative. In your case, you have

    [tex]e^{10t} \cdot \frac{dy}{dt} + e^{10t} \cdot 10y = (e^{10t} y)'[/tex]

    You can differentiate [itex]e^{10t} y[/itex] to verify that's the case. So integrating the lefthand side is trivial because you're just integrating a derivative.
     
  4. Jan 20, 2010 #3
    Ok,

    So after integrating both sides, i have
    [tex]e^{10t}\cdot y = \frac{1}{10} e^{10t} +C[/tex]

    After inserting applying the initial conditions i get:

    [tex] e^{10t} \cdot \frac{1}{10} = \frac{1}{10} e^{10t} +C[/tex]

    This gives that the e terms cancel and C = 2/10 but that's incorrect.

    Hm, what am i missing here?
     
  5. Jan 20, 2010 #4

    vela

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    You can solve for y to get

    [tex]y=\frac{1}{10}+C e^{-10t}[/tex]

    A few mistakes in what you did: you didn't substitute a value in for t, and you plugged in the wrong initial value for y. Also, you apparently did the algebra wrong because the exponentials can't cancel.
     
  6. Jan 22, 2010 #5
    Hm, after plugging in the right value for y i get, like you.

    [tex]e^1\cdot \frac{2}{10} = \frac{e^1}{10} + C[/tex] which clearly gives C = e/10

    But how did you get [tex]e^{-10t}[/tex] ?

    That must be from dividing by 2e^10 but why does the variable t reappear when you already plugged in 1/10 ?

    Sorry for the ignorance, i suck.
     
  7. Jan 22, 2010 #6

    vela

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    I divided both sides of the equation by [itex]e^{10t}[/itex] to solve for y. All you have to do now is take the value of C you found and plug it back into the solution.
     
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