write and find the general solution of the differential equation that models the verbal statement. evaluate…

write and find the general solution of the differential equation that models the verbal statement. evaluate the solution at the specified value of the independent variable. the rate of change of p is proportional to p. when t = 0, p = 4,000 and when t = 1, p = 3,400. what is the value of p when t = 4? write the differential equation. (use k for the constant of proportionality.) dp/dt = solve the differential equation. p = evaluate the solution at the specified value of the independent variable. (round your answer to three decimal places.)
Answer
Explanation:
Step1: Write the differential equation
Since the rate of change of $P$ is proportional to $P$, we have $\frac{dP}{dt}=kP$.
Step2: Solve the differential equation
Separate variables: $\frac{dP}{P}=kdt$. Integrate both sides: $\int\frac{dP}{P}=\int kdt$. We get $\ln|P| = kt + C$. Exponentiating both sides gives $P = Ce^{kt}$. Using the initial - condition when $t = 0$, $P=4000$. Substituting into $P = Ce^{kt}$, we have $4000 = Ce^{k\times0}$, so $C = 4000$. Then $P = 4000e^{kt}$. When $t = 1$, $P = 3400$. Substitute into $P = 4000e^{kt}$: $3400=4000e^{k\times1}$. Then $e^{k}=\frac{3400}{4000}=0.85$, so $k=\ln(0.85)$. So the general solution is $P = 4000e^{t\ln(0.85)}=4000(0.85)^{t}$.
Step3: Evaluate the solution at $t = 4$
Substitute $t = 4$ into $P = 4000(0.85)^{t}$: $P = 4000\times(0.85)^{4}=4000\times0.52200625 = 2088.025$.
Answer:
$\frac{dP}{dt}=kP$ $P = 4000(0.85)^{t}$ $2088.025$