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Stumped- DE problem

  • Thread starter hola
  • Start date
  • #1
38
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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.
 

Answers and Replies

  • #2
321
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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.
 
  • #3
38
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That's the problem..... I can't integrate it.
 
  • #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.
 
  • #5
Tide
Science Advisor
Homework Helper
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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.
 
  • #6
38
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I got some really messy answer.... is that ok?
 
  • #7
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
That depends on the answer! :biggrin:

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

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