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Math Help - Laplace transform

  1. #1
    Junior Member Singular's Avatar
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    Laplace transform

    Using Laplace Transform I was asked to solved this eq.

    <br />
\frac{{d^2 y}}<br />
{{dx^2 }} + 4\frac{{dy}}<br />
{{dx}} + 13y = 145\cos 2t<br />
<br />
{\text{       }};y(0) = 9<br />
and <br />
\left. {\frac{{dy}}<br />
{{dx}}} \right|_{t = 0}  = 19<br />


    I've try it but I got Y(s) that I couldn't find its inverse ... help

    thank you
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by Singular View Post
    Using Laplace Transform I was asked to solved this eq.

    <br />
\frac{{d^2 y}}<br />
{{dx^2 }} + 4\frac{{dy}}<br />
{{dx}} + 13y = 145\cos 2t<br />
<br />
{\text{       }};y(0) = 9<br />
and <br />
\left. {\frac{{dy}}<br />
{{dx}}} \right|_{t = 0}  = 19<br />


    I've try it but I got Y(s) that I couldn't find its inverse ... help

    thank you
    Can you show your work - or at least your answer for Y(s) - to save re-inventing the wheel .....
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  3. #3
    Junior Member Singular's Avatar
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    OK wait .... I'll find it first....I forgot where I put it ....
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  4. #4
    Junior Member Singular's Avatar
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    Here is my attempt to solve it ...

    <br />
\displaylines{<br />
  L(y^{''} ) + 4[L(y')] + 13[L(y)] = 145\cos 2t \cr <br />
  s^2 Y - sy(0) - y'(0) + 4sY - 4y(0) + 13Y = 145\cos 2t \cr <br />
  s^2 Y - 9s - 19 - 4sY - 36 + 13Y = 145\cos 2t \cr <br />
  (s^2  - 4s + 13)Y - 9s - 55 = 145\cos 2t \cr <br />
  Y = \frac{{145\cos 2t + 9s + 55}}{{s^2  - 4s + 13}} \cr}<br /> <br /> <br />

    After taht I stuck , I don't know what I should do then ...
    Anyone can HElp ThankS
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  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Singular View Post
    Here is my attempt to solve it ...

    <br />
\displaylines{<br />
  L(y^{''} ) + 4[L(y')] + 13[L(y)] = 145\cos 2t \cr <br />
  s^2 Y - sy(0) - y'(0) + 4sY - 4y(0) + 13Y = 145\cos 2t \cr <br />
  s^2 Y - 9s - 19 - 4sY - 36 + 13Y = 145\cos 2t \cr <br />
  (s^2  - 4s + 13)Y - 9s - 55 = 145\cos 2t \cr <br />
  Y = \frac{{145\cos 2t + 9s + 55}}{{s^2  - 4s + 13}} \cr}<br /> <br /> <br />

    After taht I stuck , I don't know what I should do then ...
    Anyone can HElp ThankS
    The laplace transform of L(145 \cos(2t))=145 \frac{s}{s^2+4}

    You need to transform the right hand side
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  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    I ended up with

    s^2Y-9s-19+4[sY-9]+13Y=145 \frac{s}{s^2+4}

    [s^2+4s+13]Y=9s+55+145\frac{s}{s^2+4}

    [(s+2)^2+9]Y=9(s+2)+37+145\frac{s}{s^2+4}

    Y=9\frac{s+2}{[(s+2)^2+9]} +\frac{37}{[(s+2)^2+9]}+145\frac{s}{(s^2+4)[(s+2)^2+9]}

    by partial fractions of the last one we get

    Y=9\frac{s+2}{[(s+2)^2+9]} +\frac{37}{[(s+2)^2+9]}+145\left[\frac{1}{145}\frac{9s+16}{s^2+4}-\frac{1}{145}\frac{9s+52}{(s+2)^2+9} \right]

    Combinging like terms we get...


    Y=\frac{3}{[(s+2)^2+9]}+9\frac{s}{s^2+4}+8\frac{2}{s^2+4}

    Taking the inverse transfrom we get...

    y(t)=e^{-2t}\sin(3t)+9\cos(2t)+8\sin(2t)
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