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Differential Equation, Solve for function

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Given

    [itex]\frac{dx}{dt} + ax = Asin(ωt), x(0) = b[/itex]

    Solve for [itex]x(t)[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I take the Laplace transform of both sides and get

    [itex]sX(s) - x(0) + aX(s) = \frac{Aω}{s^{2} + ω^{2}}[/itex]
    [itex]X(s) = \frac{b}{s + a} + \frac{Aω}{(s^{2} + ω^{2})(s+1)}[/itex]

    The solution then state that the expression above is equal to

    [itex](b + \frac{ωA}{a^{2}+ω^{2}})\frac{1}{s + a} + \frac{aA}{a^{2} + ω^{2}}\frac{ω}{s^{2} + ω^{2}} - \frac{ωA}{a^{2} + ω^{2}}\frac{s}{s^{2} + ω^{2}}[/itex]

    I don't know how the two expressions are equal to each other or were [itex]\frac{aA}{a^{2} + ω^{2}}\frac{ω}{s^{2} + ω^{2}} - \frac{ωA}{a^{2} + ω^{2}}\frac{s}{s^{2} + ω^{2}}[/itex] came from. I believe that [itex]\frac{b}{s + a} + \frac{Aω}{(s^{2} + ω^{2})(s+a)} = (b + \frac{ωA}{a^{2}+ω^{2}})\frac{1}{s + a}[/itex] as it's just factoring. So then if this is true than [itex]\frac{aA}{a^{2} + ω^{2}}\frac{ω}{s^{2} + ω^{2}} = \frac{ωA}{a^{2} + ω^{2}}\frac{s}{s^{2} + ω^{2}}[/itex], which I don't see how that's true.

    Thanks for any help.
     
  2. jcsd
  3. Jan 21, 2014 #2

    LCKurtz

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    I think you meant ##s+a## where I have corrected in red. I didn't bother to work through your next calculations. This is going to be messy even without any mistakes. What the author surely did was use the standard procedure of partial fractions to break up that last fraction:$$
    \frac{Aω}{(s^2 + ω^2)(s+a)}= \frac C {s+a} + \frac{Ds+E}{s^2+\omega^2}$$Use your calculus partial fraction methods to solve for ##C,D,E##. They will be a bit messy and if neither you nor the author made any mistakes it should work. That last fraction is easy enough to take the inverse transform. I would inverse just as it stands first, then substitute the values for ##C,D##, and ##E## in your answer.

    I should comment that LaPlace transforms is an terrible way to work this problem. I'm assuming you were asked to do it that way or I would have a rather different hint.
     
    Last edited: Jan 21, 2014
  4. Jan 22, 2014 #3
    How would you recommend solving this problem then, if there's a much easier way to solve this than I would like to know how that can be done. Resolving that into partial fractions is very difficult to do.
     
  5. Jan 22, 2014 #4
    Another way of inverting the second term is to use convolution, especially with the s+a term there in the denominator.
     
  6. Jan 23, 2014 #5

    pasmith

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    The method of undertermined coefficients suggests a solution of the form
    [tex]
    x(t) = Be^{-at} + C\cos \omega t + D \sin \omega t
    [/tex]
    for some constants [itex]B[/itex], [itex]C[/itex] and [itex]D[/itex].
     
  7. Jan 23, 2014 #6

    LCKurtz

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    It's a constant coefficient equation. I would have recommended the same as pasmith did above.
     
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