Question Solved1 Answer Question 25 4 Points Solve the problem. The size of a population p as a function of time t is given by op -0.02p + 9t, Solve for p as a function of t if the initial population is 10,000. A p =-450t - 22,500 + 32,500e -0.02t B p=-450t - 22,500 + 32,500e 0.02t p=-450t + 32,500e 0.02t D p=- 22,500 + 32,500e -0.02t

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Transcribed Image Text: Question 25 4 Points Solve the problem. The size of a population p as a function of time t is given by op -0.02p + 9t, Solve for p as a function of t if the initial population is 10,000. A p =-450t - 22,500 + 32,500e -0.02t B p=-450t - 22,500 + 32,500e 0.02t p=-450t + 32,500e 0.02t D p=- 22,500 + 32,500e -0.02t
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Transcribed Image Text: Question 25 4 Points Solve the problem. The size of a population p as a function of time t is given by op -0.02p + 9t, Solve for p as a function of t if the initial population is 10,000. A p =-450t - 22,500 + 32,500e -0.02t B p=-450t - 22,500 + 32,500e 0.02t p=-450t + 32,500e 0.02t D p=- 22,500 + 32,500e -0.02t
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{:[(dp)/(dt)=0.02 p+9t],[=>(dp)/(dt)-0.02 p=9t]:}This is a linear 1 st order D.E." Integrating factor "=e^(-0.02 int dt)=e^(-0.02 t)". "multiplying both sides irth I.F we get.{:[e^(-0.02 t)(dp)/(dt)-0.02e^(-0.02 t)*p=9te^(-0.02 t)],[=>quad int d(e^(-0.02 t)*p)=int9te^(-0.02 t)*dt". "],[=>p*e^(-0.02 ... See the full answer