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# Differential equations question 2?

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- King LeoLv 77 years agoFavorite Answer
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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