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Math Help - help with partial fractions

  1. #91
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    I got

    s = -1

    1 = -A + B -5C + 5/2

    s = 2

    1 = 8A + 4B + 4C + 1

    then not sure what to do from here.
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  2. #92
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    I don't agree with either one of those equations. Try again, and double-check your work.
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  3. #93
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    S = -1

    1 = -5a + 5/2 - c + d

    s = = 2

    1 = 4a + 1 + 8c + 4d
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  4. #94
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    There you go. Now you have three linear equations with three unknowns. Pick your favorite solution method and fire away. (I'd recommend Gaussian elimination with back substitution - there is no more efficient exact method known to man.)
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  5. #95
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    I have tried cramers rule to solve the simultaneous equation but it doesnt work because you have to multiply by 0.
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  6. #96
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    So your system appears to be this:

    \begin{bmatrix}1&1&1\\-5&-1&1\\4&8&4\end{bmatrix}\begin{bmatrix}A\\C\\D\end{  bmatrix}=\begin{bmatrix}1/2\\-3/2\\0\end{bmatrix}.

    Correct? Show me what work you've done.
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  7. #97
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    Ive gone through it and all the values = 1/2.

    When \dfrac{1/2s+1/2}{(s-1)^2+1)}. I have to put that in laplace transform form so this should = 1/2e^tcost. Just in the answers its says -1/2
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  8. #98
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    A=C=D=1/2 doesn't even satisfy A+C+D=1/2. Therefore, your solution is incorrect. Show me some work, here, and I can show you where you're going wrong. If you don't show me work, I can't help you.
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  9. #99
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    I got det A = \begin{bmatrix}1/2&1&1\\-3/2&-1&1\\0&8&4\end{bmatrix}.= (1/2)(-1)(4)+(1)(1)(0)+(1)(-3/2)(8)-(1)(-1)(0)-(1)(-3/2)(4)-(1/2)(1)(8)
    divide by \begin{bmatrix}1&1&1\\-5&-1&1\\4&8&4\end{bmatrix}. = (1)(-1)(4)+(1)(1)(4)+(1)(-5)(8)-(1)(-1)(4)-(1)(-5)(4)-(1)(1)(8) = -12/-24

    det C = \begin{bmatrix}1/2&1&1\\-3/2&-5&1\\0&4&4\end{bmatrix}.= (1/2)(-5)(4)+(1)(1)(0)+(1)(-3/2)(4)-(1)(-5)(0)-(1)(-3/2)(4)-(1/2)(1)(4)
    divide by \begin{bmatrix}1&1&1\\-5&-1&1\\4&8&4\end{bmatrix}.= (1)(-1)(4)+(1)(1)(4)+(1)(-5)(8)-(1)(-1)(4)-(1)(-5)(4)-(1)(1)(8) = -12/-24

    det D = \begin{bmatrix}1/2&1&1\\-3/2&-5&-1\\0&4&8\end{bmatrix}.= (1/2)(-5)(8)+(1)(-1)(0)+(1)(-3/2)(4)-(1)(-5)(0)-(1)(-3/2)(8)-(1/2)(-1)(4)
    divide by \begin{bmatrix}1&1&1\\-5&-1&1\\4&8&4\end{bmatrix}.= (1)(-1)(4)+(1)(1)(4)+(1)(-5)(8)-(1)(-1)(4)-(1)(-5)(4)-(1)(1)(8) = -12/-24

    also the answers say the laplace transforms are -1/2e^tcost + 1/2 + 1/2t
    Last edited by hunterage2000; November 12th 2010 at 08:05 AM.
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  10. #100
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    You're not computing the correct determinants. You should have

    \displaystyle A=\frac{\left|\begin{matrix}<br />
1/2&1&1\\<br />
-3/2&-1&1\\<br />
0&8&4<br />
\end{matrix}\right|}{\left|\begin{matrix}<br />
1&1&1\\<br />
-5&-1&1\\<br />
4&8&4<br />
\end{matrix}\right|},<br />
C=\frac{\left|\begin{matrix}<br />
1&1/2&1\\<br />
-5&-3/2&1\\<br />
4&0&4<br />
\end{matrix}\right|}{\left|\begin{matrix}<br />
1&1&1\\<br />
-5&-1&1\\<br />
4&8&4<br />
\end{matrix}\right|},<br />
D=\frac{\left|\begin{matrix}<br />
1&1&1/2\\<br />
-5&-1&-3/2\\<br />
4&8&0<br />
\end{matrix}\right|}{\left|\begin{matrix}<br />
1&1&1\\<br />
-5&-1&1\\<br />
4&8&4<br />
\end{matrix}\right|}.<br />

    What does that give you?
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  11. #101
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    cheers mate thats correct.


    Moderator edit: New question deleted. Ask new questions in a new thread, as explained in the forum rules stuck at every subforum.
    Last edited by mr fantastic; November 12th 2010 at 01:46 PM.
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