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Math Help - Solving a second order DE using the method of Laplace Transforms

  1. #1
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    Solving a second order DE using the method of Laplace Transforms

    can anyone help me to solve this problem [using laplace Transforms] ...just wanna know if we have the same answers..i have difficulty in answering the partial fraction of this prob...3/8 d^2x/dt^2 + 24x =24sin4t where: x(0)=1/3 and x'(0)= 0
    Last edited by mr fantastic; January 11th 2009 at 12:48 PM. Reason: A new question moved from an old existing thread - I added the Laplace Transform requirement
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  2. #2
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    Quote Originally Posted by mhyann View Post
    can anyone help me to solve this problem [using laplace Transforms] ...just wanna know if we have the same answers..i have difficulty in answering the partial fraction of this prob...3/8 d^2x/dt^2 + 24x =24sin4t where: x(0)=1/3 and x'(0)= 0
    If you have an answer I assume you've done the calculation. If you post your calculation and answer someone will be able to check it.
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    Quote Originally Posted by mhyann View Post
    can anyone help me to solve this problem [using laplace Transforms] ...just wanna know if we have the same answers..i have difficulty in answering the partial fraction of this prob...3/8 d^2x/dt^2 + 24x =24sin4t where: x(0)=1/3 and x'(0)= 0

    \frac{3}{8}x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right) where x\left( 0 \right) = \frac{1}{3} and x'\left( 0 \right) = 0.

    First, simplifying the equation: x''\left( t \right) + 64x\left( t \right) = 64\sin \left( {4t} \right).

    \sin \left( {4t} \right)\underset{\raise0.3em\hbox{$\smash{\scripts  criptstyle\cdot}$}}{ \doteq } \frac{4}{{{s^2} + 16}}

    x\left( t \right)\underset{\raise0.3em\hbox{$\smash{\scripts  criptstyle\cdot}$}}{ \doteq } \overline {x\left( s \right)}

    x''\left( t \right)\underset{\raise0.3em\hbox{$\smash{\scripts  criptstyle\cdot}$}}{ \doteq } {s^2}\overline {x\left( s \right)}  - x\left( 0 \right)s - x'\left( 0 \right) = {s^2}\overline {x\left( s \right)}  - \frac{s}{3}

    So, we have

    {s^2}\overline {x\left( s \right)}  - \frac{s}{3} + 64\overline {x\left( s \right)}  = \frac{{256}}{{{s^2} + 16}} \Rightarrow

    \Rightarrow \left( {{s^2} + 64} \right)\overline {x\left( s \right)}  = \frac{{256}}{{{s^2} + 16}} + \frac{s}{3} \Rightarrow

    \Rightarrow \overline {x\left( s \right)}  = \frac{s}{{3\left( {{s^2} + 64} \right)}} + \frac{{256}}{{\left( {{s^2} + 16} \right)\left( {{s^2} + 64} \right)}}

    \frac{s}{{3\left( {{s^2} + 64} \right)}}\underset{\raise0.3em\hbox{$\smash{\scrip  tscriptstyle\cdot}$}}{ \doteq } \frac{1}{3}\cos \left( {8t} \right)

    \frac{{256}}{{\left( {{s^2} + 16} \right)\left( {{s^2} + 64} \right)}} = \frac{{16}}{{3\left( {{s^2} + 16} \right)}} - \frac{{16}}{{3\left( {{s^2} + 64} \right)}} =

    = \frac{4}{3}\frac{4}{{{s^2} + 16}} - \frac{2}{3}\frac{8}{{{s^2} + 64}}\underset{\raise0.3em\hbox{$\smash{\scriptscri  ptstyle\cdot}$}}{ \doteq } \frac{4}{3}\sin \left( {4t} \right) - \frac{2}{3}\sin \left( {8t} \right)

    \overline {x\left( s \right)}  {\underset{\raise0.3em\hbox{$\smash{\scriptscripts  tyle\cdot}$}}{ \doteq } } \frac{1}{3}\cos \left( {8t} \right) + \frac{4}{3}\sin \left( {4t} \right) - \frac{2}{3}\sin \left( {8t} \right)

    Finally, we have

    \boxed{x\left( t \right) = \frac{1}{3}\cos \left( {8t} \right) + \frac{4}{3}\sin \left( {4t} \right) - \frac{2}{3}\sin \left( {8t} \right).}
    Last edited by DeMath; January 12th 2009 at 05:24 AM.
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    thanks for helping me

    but its not 64 its 24sin4(t)..anyway i will follow the concept.

    muchas gracias to u man...
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by mhyann View Post
    but its not 64 its 24sin4(t)..anyway i will follow the concept.
    In your first post you wrote is not clear:

    thise \frac{3}{8}x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right) or thise x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right)???

    If it was thise x''\left( t \right) + 24x\left( t \right) = 24\sin \left( {4t} \right), then your answer must be x\left( t \right) = \frac{1}{3}\cos \left( {2\sqrt 6 t} \right) - \sqrt 6 \sin \left( {2\sqrt 6 t} \right) + 3\sin \left( {4t} \right).
    Last edited by DeMath; January 11th 2009 at 07:12 PM.
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  6. #6
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    the first one with 3/8

    i get u now.u multiplied the 8 to 24 then divided by 3 that's why it comes up to 64.maybe you we're right....maybe i was wrong in distributing the term/numbers.but we almost have the same answer...hehehe ive got another mistakes in exam.i'll keep working on this promise...
    thank u for helping me dude!
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