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Math Help - LaPlace transforms!

  1. #1
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    LaPlace transforms!

    Solve the following Laplace Transform:

    x````(t) + x```(t) = sint where: x(0)=x`(0)=x```(0) = 0 and x``(0)=2

    I believe I manipulated it correctly to the following:

    X(s) = 1/(s^4+s^3)(s^2+1) + 2/s^3

    Now I do not know how to manipulate this using partial fractions method. The denominator of the 1st RHS term is confusing me. Is there anyone that could help?
    Last edited by tactical; May 1st 2011 at 08:09 PM.
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  2. #2
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    Quote Originally Posted by tactical View Post
    Solve the following Laplace Transform:

    x````(t) + x```(t) = sint where: x(0)=x`(0)=x```(0) = 0 and x``(0)=2

    I believe I manipulated it correctly to the following:

    X(s) = 1/(s^4)(s^2+1) + 2/s^3

    Now I do not know how to manipulate this using partial fractions method. The denominator of the 1st RHS term is confusing me. Is there anyone that could help?
    You need denominators of s, s^2, s^3, and s^4...

    http://tinyurl.com/tacticalNewbie
    Last edited by TheChaz; May 1st 2011 at 07:50 PM. Reason: wolfram link = fail
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  3. #3
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    Quote Originally Posted by TheChaz View Post
    You need denominators of s, s^2, s^3, and s^4...

    http://tinyurl.com/tacticalNewbie
    Well.....i guess I had a typo :/ the equation should be

    X(s) = 1/(s^4+s^3)(s^2+1) + 2/s^3 which simplifies to;

    X(s) = 1/(s^3)(s+1)(s^2+1) + 2/s^3

    Now, do i need to find a common denominator to do partial fractions?

    Man, It's been a long day
    Last edited by tactical; May 1st 2011 at 08:13 PM.
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by tactical View Post
    Solve the following Laplace Transform:

    x````(t) + x```(t) = sint where: x(0)=x`(0)=x```(0) = 0 and x``(0)=2

    I believe I manipulated it correctly to the following:

    X(s) = 1/(s^4)(s^2+1) + 2/s^3

    Now I do not know how to manipulate this using partial fractions method. The denominator of the 1st RHS term is confusing me. Is there anyone that could help?
    s^4X-s^3(0)-s^2(0)-s(2)-0+s^3X-s^2(0)-s(0)-2=\frac{1}{s^2+1}

    s^3(s+1)X=2s+2+\frac{1}{s^2+1} \iff X=\frac{2}{s^3}+\frac{1}{s^3(s+1)(s^2+1)}=\frac{2s  ^2+2s^2+2s+3}{s^3(s+1)(s^2+1)}

    So the partial fractions will look like

    \frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s  +1}+\frac{Es+F}{s^2+1}=\frac{2s^2+2s^2+2s+3}{s^3(s  +1)(s^2+1)}

    Multiplying out and solving gives

    A=0,B=-1,C=3,D=-\frac{1}{2},E=\frac{1}{2},F=\frac{1}{2}
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    Quote Originally Posted by TheEmptySet View Post
    s^4X-s^3(0)-s^2(0)-s(2)-0+s^3X-s^2(0)-s(0)-2=\frac{1}{s^2+1}

    s^3(s+1)X=2s+2+\frac{1}{s^2+1} \iff X=\frac{2}{s^3}+\frac{1}{s^3(s+1)(s^2+1)}=\frac{2s  ^2+2s^2+2s+3}{s^3(s+1)(s^2+1)}

    So the partial fractions will look like

    \frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s  +1}+\frac{Es+F}{s^2+1}=\frac{2s^2+2s^2+2s+3}{s^3(s  +1)(s^2+1)}

    Multiplying out and solving gives

    A=0,B=-1,C=3,D=-\frac{1}{2},E=\frac{1}{2},F=\frac{1}{2}
    Life Saver......

    Now going back into the time domain I would get

    (-t)+(3t^2)-(.5e^-t) +(?) I know this is someway to algebraically manipulate the last term, but im 3 semesters out of my Diff EQ class and forgot how. Is there anyway you could explain what to do?

    Edit: can I do s/s^2+1 + 1/s^2+1 to get

    (-t) + (3t^2) - (.5e^-t) + (.5sint) +(.5cost)
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  6. #6
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    Okay, I understand that above. I have one last question, I'm working on another problem and I get my equation in the s domain to be the following:
    x(s) = \frac{1}{s}  -  \frac{s+3}{(s^2+3s+1)}

    How would I go about algebraically manipulating the second part of the RHS to get it back into the time domain?
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by tactical View Post
    Okay, I understand that above. I have one last question, I'm working on another problem and I get my equation in the s domain to be the following:
    x(s) = \frac{1}{s}  -  \frac{s+3}{(s^2+3s+1)}

    How would I go about algebraically manipulating the second part of the RHS to get it back into the time domain?
    If you complete the square in the denominator you get

    -\frac{s+3}{(s+\frac{3}{2})^2-\frac{5}{4}}=-\frac{s+\frac{3}{2}}{(s+\frac{3}{2})^2-\frac{5}{4}}-\frac{\frac{3}{2}}{(s+\frac{3}{2})^2-\frac{5}{4}}

    Now we can use the the fact that

    \mathcal{L}\{e^{at}f(t) \}=F(s-a)

    So the first half is

    -\frac{s+\frac{3}{2}}{(s+\frac{3}{2})^2-\frac{5}{4}} = -\frac{w}{w^2-\frac{5}{4}}\bigg|_{w \to s+\frac{3}{2}}=-e^{-\frac{3}{2}t}\cosh\left(\frac{\sqrt{5}}{2}t\right)

    The 2nd one is similar you can convert the hyperbolics back to exponentials if you wish.
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  8. #8
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    Quote Originally Posted by tactical View Post
    Okay, I understand that above. I have one last question, I'm working on another problem and I get my equation in the s domain to be the following:
    x(s) = \frac{1}{s}  -  \frac{s+3}{(s^2+3s+1)}

    How would I go about algebraically manipulating the second part of the RHS to get it back into the time domain?
    Next time, please do not start a new question in the middle of a thread. Start a new problem in a new thread. Thank you.
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