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Math Help - powers

  1. #1
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    powers

    Very quickly, if your taking powers across the other side eg deifferential equations, how does it work again, im having a blank!!!

    eg
    solving for y

    -y^-5 = 6x+C

    becomes

    y = (6x+C)-1/5

    IS this right??

    If it is can anyone explain the thoery behind it???
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  2. #2
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    ah man i think i wrote the question out wrong...

    can anyone help me find the solution for the separable equation
    6y^6

    I already know the answer

    for y(0)=3
    (1/243 - 30x)^-(1/5)

    but can anyone help me through the stages?
    I just need explicit help with this solution, I keep going wrong with the basic Algebra i think
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  3. #3
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    It should be y=(6x+C)^-(1/5).

    Just multiply the exponent by its inverse (like if the exponent is 2, multiply both sides of the equation by 1/2).
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  4. #4
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    Quote Originally Posted by dankelly07 View Post
    can anyone help me find the solution for the separable equation
    6y^6
    This is not an equation. What is your original equation?

    -Dan
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  5. #5
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    sorry for the delayed reply,

    The equation I asked to solve was

    dy/dx = 6y^6

    find the general solution
    w/ condition y(0) = 3
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  6. #6
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    Quote Originally Posted by dankelly07 View Post
    sorry for the delayed reply,

    The equation I asked to solve was

    dy/dx = 6y^6

    find the general solution
    w/ condition y(0) = 3
    \frac{dx}{dy} = \frac{1}{6} \frac{1}{y^6} = \frac{1}{6} y^{-6}.

    Therefore:

    x = \frac{1}{6} \left( \frac{-1}{5} \right) y^{-5} + C

    = -\frac{1}{30} \frac{1}{y^5} + C

    \Rightarrow 30(x - C) = \frac{1}{y^5}.

    You know that when x = 0, y = 3:

    -30C = \frac{1}{3^5} = \frac{1}{243}.

    Therefore 30x - \frac{1}{243} = \frac{1}{y^5}.

    You can finish it off if you want to make y the subject .....
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  7. #7
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    Yeah thats gr8, I cant see where I was going wrong, I'm v bad at factorising!
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