Let the population of rabbits surviving at ti | vedclass

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(C) Given the differential equation: $\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200 = \frac{p(t) - 400}{2}$.Separating the variables,we get: $\int \frac{dp

Vedclass · Accepted Answer

$400 - 300e^{t/2}$

Vedclass · Answer

(C) Given the differential equation: $\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200 = \frac{p(t) - 400}{2}$.Separating the variables,we get: $\int \frac{dp(t)}{p(t) - 400} = \int \frac{1}{2} dt$.Integrating both sides: $\ln |p(t) - 400| = \frac{t}{2} + C$.This implies $|p(t) - 400| = e^{C} \cdot e^{t/2}$,or $p(t) - 400 = Ke^{t/2}$ where $K = \pm e^C$.Using the initial condition $p(0) = 100$: $100 - 400 = Ke^0 \implies K = -300$.Substituting $K$ back into the equation: $p(t) - 400 = -300e^{t/2}$.Therefore,$p(t) = 400 - 300e^{t/2}$.

Let the population of rabbits surviving at time $t$ be governed by the differential equation $\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200$. If $p(0) = 100$,then $p(t)$ equals:

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