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Homework Help: Stumped- DE problem

  1. Sep 27, 2005 #1
    I am stumped..... here is the problem:
    Solve the DE using the following:
    L and R are constants

    [tex]L\frac{di}{dt} + Ri = E(t)[/tex]

    [tex]i(0) = i_0[/tex]

    [tex]E(t) = E_0*sin(wt)[/tex]

    Here is my work so far:

    I got the integrating factor to become [tex]e^{Rt/L}[/tex]. But now:

    [tex]\frac{d(e^{\frac{Rt}{L}}*i)}{dt} = e^{\frac{Rt}{L}}\frac{E_0}{L}*sin(wt)[/tex]

    But I am stuck from there. Help would be appreciated.
     
  2. jcsd
  3. Sep 27, 2005 #2
    Since you have

    [tex]\mu(t) = e^{\frac {R}{L}t}[/tex]

    and you also have

    [tex](\mu(t)i)' = \mu(t) \frac {E_0}{L} \sin (\omega t)[/tex]

    You can integrate both sides and divide by [itex]\mu(t)[/itex]

    Hence
    [tex]i(t) = \frac {\int \mu(s) \frac {E_0}{L} \sin (\omega s) ds}{\mu(t)}[/tex]

    I switched the t to a s in the numerator to avoid confusion. After you integrate the numerator, you can replace the s with a t.
     
  4. Sep 27, 2005 #3
    That's the problem..... I can't integrate it.
     
  5. Sep 27, 2005 #4

    Tom Mattson

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    Integrate it by parts. Let [itex]u=\sin(\omega t)[/itex] and let [itex]dv=exp\left(\frac{Rt}{L})[/itex].

    You'll have to integrate by parts twice and then algebraically solve for the integral. This integral actually pops up all the time in second order dynamic systems.
     
  6. Sep 27, 2005 #5

    Tide

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    You can either integrate by parts or, if you're comfortable with complex analysis, you can note that

    [tex]\int e^{at} \sin \omega t dt = Im \int e^{(a + i \omega) t} dt[/tex]

    and extract the imaginary part after performing the integration.
     
  7. Sep 27, 2005 #6
    I got some really messy answer.... is that ok?
     
  8. Sep 27, 2005 #7

    Tom Mattson

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    That depends on the answer! :biggrin:

    Why don't you post what you did so we can see it?
     
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