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    Can anyone help me see where I have gone wrong with this question?

    Mould is spreading through a piece of cheese of mass 1 kg in such a way that at a time t days, x kg of the cheese has become mouldy where

    dx/dt = 2x(1-x)

    Given that initially 0.01 kg of the cheese is mouldy, show that x = (e^2t)/(99+e^2t)

    I end up with the solution x = (e^2t)/([1/A] + e^2t), but 1/A evalutes to 100 as 1/0.01 and not 99?

    Thank you for your help
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    (Original post by Dapperblook22)
    Can anyone help me see where I have gone wrong with this question?

    Mould is spreading through a piece of cheese of mass 1 kg in such a way that at a time t days, x kg of the cheese has become mouldy where

    dx/dt = 2x(1-x)

    Given that initially 0.01 kg of the cheese is mouldy, show that x = (e^2t)/(99+e^2t)

    I end up with the solution x = (e^2t)/([1/A] + e^2t), but 1/A evalutes to 100 as 1/0.01 and not 99?

    Thank you for your help
    Just use your initial condition

    x(t)=\dfrac{e^{2t}}{\frac{1}{A} + e^{2t}}

    x(0)=0.01 \text{kg}=\dfrac{1}{\frac{1}{A}+  1}

    Rearrange for A or 1/A and plug back in to your expression for x(t)


    It might help you if you write your general expression in terms of B=1/A rather than 1/A. You can do this because A is just some arbitrary constant, so rewriting it in terms of some other arbitrary constant doesn't change anything.
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    (Original post by Implication)
    Just use your initial condition

    x(t)=\dfrac{e^{2t}}{\frac{1}{A} + e^{2t}}

    x(0)=0.01 \text{kg}=\dfrac{1}{\frac{1}{A}+  1}

    Rearrange for A or 1/A and plug back in to your expression for x(t)


    It might help you if you write your general expression in terms of B=1/A rather than 1/A. You can do this because A is just some arbitrary constant, so rewriting it in terms of some other arbitrary constant doesn't change anything.
    Thank you so much, I see the proof now
 
 
 
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