First order DE, using integrating factor

  1. Hi,

    I tried to solve this by using the integrating factor technique
    dy/dt +10y = 1 \\
    y(1/10) = 2/10

    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. vela

    vela 12,505
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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. 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. vela

    vela 12,505
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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. 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. vela

    vela 12,505
    Staff Emeritus
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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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