solve the initial value problem of dp/dt=10p(1-p) when p(0)=0.1

Question

when i tried this i found that the constant, C,was equal to 0.09 when t=0 and so ended up with |p-p^2|=0.09e^t.   I'm not sure what to next though, any ideas would be greatlyappreciated. Thanks

Answer 1

The variables can be separated: dp/(10p(1-p))=dt. This can be written 10*integral((1/p)+(1/(1-p))dp=t+k, by splitting into partial fractions.

10ln(p)-10ln(1-p)=t+k⇒10ln(p/(1-p))=t+k or ln(p/(1-p))=0.1t+0.1k. Therefore, p/(1-p)=e^(0.1t+0.1k) (or ae^0.1t where a=e^0.1k).

Applying initial conditions, p=0.1 when t=0, 0.1/0.9=e^0.1k=a. So a=1/9.

p/(1-p)=e^(0.1t)/9 which can be converted: p=e^(0.1t)/9-pe^(0.1t)/9, p=e^(0.1t)/9)/(1+e^(0.1t)/9).

This can also be written: p(t)=e^0.1t/(9+e^0.1t). Check: when t=0 p(0)=1/10=0.1.