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Another DE problem i've encoutered...

ok i got the differential equation:

[tex]\frac{dx}{dt}=k(x)(1-x)[/tex] (x is the proportion of the population)

[tex]\int \frac{dx}{x(1-x)} = \int k dt[/tex]

and change it into partial fraction:

[tex]\int [\frac {1}{x} + \frac {1}{1-x}]dx = \int k dt[/tex]

[tex]ln|x|-ln|1-x|=kt+c[/tex]

After i substitute the t=0, x=0.5, and got

[tex]ln|\frac{0.5}{1-0.5}|=k \cdot 0 + c[/tex]

[tex]ln 1 = c[/tex]

[tex]c = 0[/tex]

[tex]\therefore ln|\frac {x}{1-x}|=kt [/tex]

The problem is when i tried to find the constant k, as i substituted t=24 and x=1, i got:

[tex]ln|\frac{1}{0}|=24k[/tex]

So, how to solve [tex]\frac{1}{0}=e^{24k}[/tex]?

:uhh: thanks for your time here.

**A contagious disease spreads at a rate directly proportional to the product of the number of population infected and the remaining population that is not infected. Initially, one-half of the population is infected and if the rate of infection is kept constant, the whole population will be infected in 24 days. Find the proportion of the population that will be infected after 12 days.**ok i got the differential equation:

[tex]\frac{dx}{dt}=k(x)(1-x)[/tex] (x is the proportion of the population)

[tex]\int \frac{dx}{x(1-x)} = \int k dt[/tex]

and change it into partial fraction:

[tex]\int [\frac {1}{x} + \frac {1}{1-x}]dx = \int k dt[/tex]

[tex]ln|x|-ln|1-x|=kt+c[/tex]

After i substitute the t=0, x=0.5, and got

[tex]ln|\frac{0.5}{1-0.5}|=k \cdot 0 + c[/tex]

[tex]ln 1 = c[/tex]

[tex]c = 0[/tex]

[tex]\therefore ln|\frac {x}{1-x}|=kt [/tex]

The problem is when i tried to find the constant k, as i substituted t=24 and x=1, i got:

[tex]ln|\frac{1}{0}|=24k[/tex]

So, how to solve [tex]\frac{1}{0}=e^{24k}[/tex]?

:uhh: thanks for your time here.

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