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Math Help - Partial fraction decomposition.

  1. #1
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    Question Partial fraction decomposition.

    Decompose into partial fractions: \frac{x}{x^4-a^4}
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    Quote Originally Posted by ChristopherDunn View Post
    Decompose into partial fractions: \frac{x}{x^4-a^4}
    \frac{x}{x^4 - a^4} = \frac{x}{(x^2)^2 - (a^2)^2}

     = \frac{x}{(x^2 - a^2)(x^2 + a^2)}

     = \frac{x}{(x - a)(x + a)(x^2 + a^2)}.


    Now try using partial fractions with

    \frac{A}{x - a} + \frac{B}{x + a} + \frac{Cx + D}{x^2 + a^2} = \frac{x}{(x - a)(x + a)(x^2 + a^2)}.
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    1. \frac{A}{x - a} + \frac{B}{x + a} + \frac{Cx + D}{x^2 + a^2} = \frac{x}{(x - a)(x + a)(x^2 + a^2)}
    2. x=A(x+a)(x^2+a^2)+B(x-a)(x^2+a^2)+(Cx+D)(x-a)(x+a)

    let x=a;
    1. a=A(a+a)(a^2+a^2)
    2. a=4a^3A
    3. A=\frac{1}{4a^2}

    let x=-a;
    1. -a=B(-a-a)((-a)^2+a^2)
    2. -a=-4a^3B
    3. B=\frac{1}{4a^2}

    How do I find C and D?
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  4. #4
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    Quote Originally Posted by ChristopherDunn View Post
    1. \frac{A}{x - a} + \frac{B}{x + a} + \frac{Cx + D}{x^2 + a^2} = \frac{x}{(x - a)(x + a)(x^2 + a^2)}
    2. x=A(x+a)(x^2+a^2)+B(x-a)(x^2+a^2)+(Cx+D)(x-a)(x+a)

    let x=a;
    1. a=A(a+a)(a^2+a^2)
    2. a=4a^3A
    3. A=\frac{1}{4a^2}

    let x=-a;
    1. -a=B(-a-a)((-a)^2+a^2)
    2. -a=-4a^3B
    3. B=\frac{1}{4a^2}

    How do I find C and D?
    Let x = 0 to find D.

    Then you can use your information about A, B, D to find C.
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  5. #5
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    That'll do it -- thanks!
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