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Math Help - partial fraction

  1. #1
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    partial fraction

    can anybody help solve this?

    x= 1/(s^2(s+2))

    i need to find the partial fractions of the RHS of the equation.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by al2308 View Post
    can anybody help solve this?

    x= 1/(s^2(s+2))

    i need to find the partial fractions of the RHS of the equation.
    \frac 1{s^2(s + 2)} = \frac As + \frac B{s^2} + \frac C{s + 2}

    can you continue?
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  3. #3
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    I got that far and then i got A=1/2, thats where i'm stuck from
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  4. #4
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    (No partial fractions foundation.)

    \begin{aligned}<br />
  \frac{1}<br />
{{s^2 (s + 2)}} &= \frac{4}<br />
{{4s^2 (s + 2)}} \hfill \\<br />
   &= \frac{{4 - s^2  + s^2 }}<br />
{{4s^2 (s + 2)}} \hfill \\<br />
   &= \frac{{(2 - s)(s + 2) + s^2 }}<br />
{{4s^2 (s + 2)}} \hfill \\<br />
   &= \frac{1}<br />
{4}\left( {\frac{{2 - s}}<br />
{{s^2 }} + \frac{1}<br />
{{s + 2}}} \right).<br />
\end{aligned}
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jhevon View Post
    \frac 1{s^2(s + 2)} = \frac As + \frac B{s^2} + \frac C{s + 2}
    just to finish this method up, since Krizalid showed you another.

    \Rightarrow 1 = As(s + 2) + B(s + 2) + Cs^2 .............(1)

    plug in s = 0 in (1), we get:

    1 = 2B \implies \boxed{B = \frac 12}

    plug in s = -2 in (1), we get:

    1 = 4C \implies \boxed{C = \frac 14}

    can we plug anything else in to eliminate B and C but not A? Nah, we will have to do A the hard way, but it will be easier considering we know B and C

    expand (1):

    1 = As^2 + 2As + Bs + 2B + Cs^2

    group like powers of s:

    1 = (A + C)s^2 + (2A + B)s + 2B

    now the coefficients of each power of s on the right must be the same as the coefficient of that same power on the left.

    thus we have the system:

    A + C = 0

    2A + B = 0

    2B = 1

    and we could have easily found each variable from this system to begin with (but i wanted to show you the two methods, you usually need only one). but now let's just find A. we can find A by plugging in the value we know for C in the A + C = 0 equation
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