"\\frac{dP}{dt}=\\frac{P}{100}-\\frac{P^2}{10^8}\\\\\n10^8\\frac{dP}{10^6P-P^2}=dt"
"10^8\\int{\\frac{dP}{10^6P-P^2}}=\\int{dt}"
"100ln\\frac{P}{P-10^6}=t+C_1"
"\\frac{P}{P-10^6}=Ce^{\\frac{t}{100}}"
Given that in the year 1980, t=0 and "x(0)=10^5" we get "C=-\\frac{1}{9}"
Then, "\\frac{P}{P-10^6}=-\\frac{1}{9}e^{\\frac{t}{100}}"
"P(t)=\\frac{10^6e^{\\frac{t}{100}}}{e^{\\frac{t}{100}}+9}"
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