"\\frac{dp}{dt} = 0.5p - 330"
(i) Population at any time is given by
"\\int {dp} = \\int (0.5p-330 )dt"
"\\int \\frac{dp}{ (0.5p-330 )} = \\int dt"
solving integral, we get
"2log( 0.5p-330 ) = t + c" , where c is integral constant. . . . . . . . . . (a)
for finding the time for extinction of the population,
"2[log (0.5p-330 )]_{620}^{0} = t|_0^t"
it will give,
"t = 5.61 months" (approx)
(ii) Using equation (a) to find the initial population so that it become extinct in one year
Equation will be
"2log[0.5p-330 ]_p^0 = t|_0^{12}"
solving this equation for p, we get "p=658.36 \\approx" "658."
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