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Math Help - Laplace Transforms and differential equations

  1. #1
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    Laplace Transforms and differential equations

    I am new to Laplace transforms and have started from the beginning with learning it.
    I am troubled by the process of solving differential equations.

    I have the problem:

    dx/dt -4x = 8

    I have rearranged to get sx - 2 - 4x = 8/s

    now not really sure where to go from here. Can someone please show me?
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  2. #2
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    No need for Laplace Transforms, this is first order linear, so you can use the integrating factor method...

    \frac{dx}{dt} - 4x = 8

    The integrating factor is e^{\int{-4\,dt}} = e^{-4t}, so multiplying through by the integrating factor gives

    e^{-4t}\frac{dx}{dt} - 4x\,e^{-4t} = 8e^{-4t}

    \frac{d}{dt}(x\,e^{-4t}) = 8e^{-4t}

    x\,e^{-4t} = \int{8e^{-4t}\,dt}

    x\,e^{-4t} = -2e^{-4t} + C

    x = -2 + Ce^{4t}.
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  3. #3
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    Quote Originally Posted by harrisonjane70 View Post
    I am new to Laplace transforms and have started from the beginning with learning it.
    I am troubled by the process of solving differential equations.

    I have the problem:

    dx/dt -4x = 8

    I have rearranged to get sx - 2 - 4x = 8/s

    now not really sure where to go from here. Can someone please show me?
    Use an Upper case X(s) to denote the LT of x(t).

    Then you have:

    X(s) (s-4)=\dfrac{8}{s}+2

    so:

    X(s)=\dfrac{8+2s}{s(s-4)}

    Which you find the ILT of by the usual methods (partial fractions and table look up)

    CB
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  4. #4
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    how about dx/dt + 2x = 10e^3t

    rearranged to X(s)(s + 2) - 6 = 10 / s - 3
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    Again, first order linear. No need for Laplace Transforms...

    The integrating factor is e^{\int{2\,dt}} = e^{2t}.

    So multiplying through by the integrating factor gives

    e^{2t}\,\frac{dx}{dt} + 2x\,e^{2t} = 10e^{5t}

    \frac{d}{dt}(x\,e^{2t}) = 10e^{5t}

    x\,e^{2t} = \int{10e^{5t}\,dt}

    x\,e^{2t} = 2e^{5t} + C

    x = 2e^{3t} + Ce^{-2t}.
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  6. #6
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    Is it possible to see it with Laplace Transforms? I'm looking at examples in a text book.
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  7. #7
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    transposing to obtain X somehow gets (6s - 8) / (s + 2)(s - 3)???

    Can someone explain why the 10 has become -8?
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  8. #8
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    Prove It, he's meant to do it by applying LT according his(er) professor.

    i've found many threads here when people say "integrate by using trig. substitution," then i appear and say "no need for trig. sub." but then one realises that they are learning how to use a new technique on solving a problem.
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  9. #9
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    yeah I'm going into year 2 Electrical engineering in october and I will be studying this in one of my modules along with fourier series and matrix algebra. At the moment I'm just looking at the material trying to see what Im up against as Im not the most mathematical person.
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by harrisonjane70 View Post
    transposing to obtain X somehow gets (6s - 8) / (s + 2)(s - 3)???

    Can someone explain why the 10 has become -8?
    It is called algebra:

    X(s)(s+2) - 6=\dfrac{10}{s-3}

    becomes:

    X(s)(s+2)=\dfrac{10}{s-3}+6=\dfrac{10+6(s-3)}{s-3}

    etc

    CB
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  11. #11
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    This question was asked by the same person using a different screen name here: http://www.mathhelpforum.com/math-he...ms-153657.html

    Interesting that the same responses have been given ....

    Thread closed.
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