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Diff eq and circuits problem

  1. Jan 16, 2005 #1
    dv(t)/dt + 200v(t) = 10 cos(100t)
    v(0)=0

    find v(t)
    i know v(t) is going to have an exponential and sinusoid component

    i also know that 200*ke^-st=-ske^-st
    so, s = -200. i dont know how to find k.
    i also cant figure out how the sinusoid part is going to be. the equations from my circuits book dont look anything like what i need.
     
  2. jcsd
  3. Jan 16, 2005 #2

    dextercioby

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    Take the particular solution of the nonhomogenous LODE as a combination
    [tex] v_{p}(t)=A\sin 100t+B\cos 100t [/tex]

    Plug it in the equation and find A and B.Then find the constant of integration (the one which resulted when integrating the homogenous LODE) by asking that
    [tex] v(0)=v_{hom}(0)+v_{p}(0)=0 [/tex]

    Daniel.
     
  4. Jan 16, 2005 #3
    how do i find the k for the exponential portion?
     
  5. Jan 16, 2005 #4

    Curious3141

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    This is a first order differential equation that you can solve by multiplying by an integrating factor.

    In an equation of the form :

    [tex]v'(t) + f(t)v = g(t)[/tex]

    you can multiply both sides by [tex]e^{\int{f(t)}dt[/tex] (this is called an integrating factor)

    to get

    [tex]\frac{d}{dt} ({v(t)e^{\int{f(t)}dt}) = g(t)e^{\int{f(t)}dt[/tex]

    which can then be solved by integrating both sides wrt [tex]t[/tex].

    What is [tex]f(t)[/tex] in this case ? What is the corresponding value of [tex]e^{\int{f(t)}dt[/tex] ?
     
    Last edited: Jan 16, 2005
  6. Jan 16, 2005 #5

    dextercioby

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    Separate variables and integrate.
    [tex] \frac{dv(t)}{dt}+200v(t)=0 [/tex]

    Daniel.
     
  7. Jan 16, 2005 #6
    oh man, im a complete bonehead. i know how to do these.
    first-order linear differential equation, for some reason it didnt click that that was what it was.
     
  8. Jan 16, 2005 #7

    dextercioby

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    Look at the bright side...You finally got the "click"... :tongue2:

    Daniel.
     
  9. Jan 16, 2005 #8
    the final form of this equation has an integral, this cant be right.

    heres what im getting

    integrating factor = e^-200t

    so,e^200t [tex]int{(e^-200t)(10cos(100t))}dt[/tex]
    that means integration by parts. as far as i can tell, im doing everything right.
     
    Last edited: Jan 16, 2005
  10. Jan 16, 2005 #9

    dextercioby

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    Nope,his version is right,too.It's an alternative.Obviously will lead to the same result like mine.

    Daniel.
     
  11. Jan 16, 2005 #10

    Curious3141

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    Tell me wherever you're lost in the following :

    In this case,

    [tex]f(t) = 200[/tex]

    [tex]e^{\int{f(t)}dt = e^{200t}[/tex]

    You now have the differential equation :

    [tex]\frac{d}{dt} (v(t)e^{200t}) = 10 \cos(100t)e^{200t}[/tex]

    Integration will give you :

    [tex]v(t)e^{200t} = \int{10 \cos(100t)e^{200t}}dt + constant[/tex]

    The easiest way to evaluate the integral on the RHS is to make the substitution [tex]x = 100t[/tex] then integrate by parts. You'll have to integrate by parts twice to solve this one. Don't forget to solve for the constant by setting [tex]v = 0[/tex] at [tex]t = 0[/tex]

    The final expression is :

    [tex]v(t) = \frac{1}{50}(\sin(100t) + 2\cos(100t) - 2e^{-200t})[/tex]
     
    Last edited: Jan 16, 2005
  12. Jan 16, 2005 #11
    im not getting this stuff to work out like the book does. the answer in the back of the book has a exp part, a sin part, and a cos part. but im getting left with integrals in my final equations.

    and when i try to separate the variables and solve for k, the variables im separating for are v and t. i end up with t= -ln200v.
     
  13. Jan 16, 2005 #12

    Curious3141

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    Is my answer the same as the book answer (it should be, because it satisfies the given d.e.) ?
     
  14. Jan 16, 2005 #13
    okay, i follow all that. i was getting lost at the part where both sides are integrated.
    i was setting it up wrong.
    thanks
     
  15. Jan 16, 2005 #14

    Curious3141

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  16. Jan 16, 2005 #15
    your answer works with the books.
     
  17. Jan 16, 2005 #16
    okay, im still having a problem working through this on my own.

    i have [tex] v(t) = 10\sin(100t) + 20\cos(100t) + 20\int{cos(100t)e^{200t}}dt[/tex]
    how do i get rid of the integral?
     
  18. Jan 16, 2005 #17

    Curious3141

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    I'll go through the integration step by step. It's a little tedious, so I hope you're appreciative. :tongue2:

    You want to work out [tex]\int{10 \cos(100t)e^{200t}}dt = 10\int{\cos(100t)e^{200t}}dt[/tex]

    Let [tex]I = \int{\cos(100t)e^{200t}}dt[/tex]

    Substitute [tex]100t = x[/tex]

    [tex]I = \int{\cos(x)e^{2x}}dt = \frac{1}{100}\int{\cos(x)e^{2x}}dx[/tex]

    Now let [tex]F = \int{\cos(x)e^{2x}}dx[/tex]

    Integration by parts, [tex]\int udw + \int wdu = uw[/tex]

    First let [tex]u = e^{2x}, dw = \cos x[/tex]

    [tex]\int{e^{2x}\cos{x}}dx + 2\int{\sin x}{e^{2x}}dx = (e^{2x})(\sin x)[/tex] ---eqn (1)

    From the LHS of eqn (1), we need to find [tex]G = \int{(\sin x)e^{2x}}dx[/tex]

    Integrate again by parts. This time let [tex]u = e^{2x}, dw = \sin x[/tex]

    Then [tex]\int{(\sin x)e^{2x}}dx + 2\int{(-\cos x) (e^{2x})}dx = -e^{2x}\cos x[/tex]

    Therefore [tex]G = -e^{2x}\cos x + 2\int{(\cos x)(e^{2x})}dx[/tex] ---eqn(2)

    Put the expression for [itex]G[/itex] from eqn (2) back into eqn (1) and solve for F, then use that to find I, put it all back in the original equation, and you have your solution.

    As I said, tedious. :rofl: I just hope I haven't slipped up somewhere in my LaTEX.
     
    Last edited: Jan 16, 2005
  19. Jan 16, 2005 #18

    dextercioby

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    That's why i still recommend my version.

    Daniel.
     
  20. Jan 16, 2005 #19
    geez, that seems complicated.
    thanks for going into such detail. that explains it for me.
     
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