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# question on diff eqautions! watch

1. Hello,

Is my method for this right?

In an industrial process, the mass of a chemical, m kg, produced after t hours is modelled by the differential equation:
dm/ dt = ke^-t (1+m) (1-m)

Solve this diff equation given that when t = 0, m = 0 and that the initial rate at which the chemical is produced is 0.5 kg per hour.

A:

1/2 ln (m+1) -1/2 ln (1-m) = -0.5e^-t + 0.5

thanks! :0
2. (Original post by laurawoods)
Hello,

Is my method for this right?

In an industrial process, the mass of a chemical, m kg, produced after t hours is modelled by the differential equation:
dm/ dt = ke^-t (1+m) (1-m)

Solve this diff equation given that when t = 0, m = 0 and that the initial rate at which the chemical is produced is 0.5 kg per hour.

A:

1/2 ln (m+1) -1/2 ln (1-m) = -0.5e^-t + 0.5

thanks! :0
Yes that's right. You could combine the logarithms.

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Updated: April 1, 2013
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