Differential equations question 2?

Hello

I only have 1 attempt left.

Wil you please help me?

http://tinypic.com/r/2ez76sx/5

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  • 7 years ago
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    separation of variables

    dy/dt = y(8 - y)

    1/(8 - y) (1/y) dy = 1 dt

    convert 1/(8 - y) (1/y) to partial fractions

    1/(8 - y) (1/y) = A/(8 - y) + B/y

    1 = Ay + B(8 - y)

    1 = Ay + 8B - By

    B = 1/8

    A = B

    A = 1/8

    1/(8 - y) (1/y) = 1/8 [ 1/(8 - y) + 1/y ]

    1/8 ∫ [ 1/(8 - y) + 1/y ] = ∫1 dt

    ∫ [ 1/(8 - y) + 1/y ] = ∫8 dt

    -ln(8 - y) + ln(y) = 8t + c

    ln [ y/(8 - y) ] = 8t + c

    y/(8 - y) = e^(8t) e^c

    y (0) = 1

    1/(8 - 1) = e^(0) e^c

    e^c = 1/7

    y/(8 - y) = (1/7)e^(8t)

    y = (8/7)e^(8t) - (y/7)e^(8t)

    y + (y/7)e^(8t) = (8/7)e^(8t)

    y[ 7 + e^(8t) ] = 8e^(8t)

    y = 8e^(8t) / [ 7 + e^(8t) ]

    ===================

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