Solveeit Logo
Login

Question

Mathematics Question on Differential equations

Let the population of rabbits surviving at a time t be governed by the differential equation dp(t)/dt=(1/2)p(t)−200.dp(t)/dt = (1/2) p(t) - 200. If p(0)=100, p(0) = 100, then p(t) equals

A

600−500 et/2 600 - 500 \,e^{t/2}

B

400−300 e−t/2400 - 300\, e^{-t/2}

C

400−300 et/2400 - 300 \, e^{t/2 }

D

300−200 e−t/2.300 - 200 \, e^{-t/2.}

Answer

400−300 et/2400 - 300 \, e^{t/2 }

Explanation

Solution

Since dpdt−12p(t)=−200\frac{dp}{dt} - \frac{1}{2} p\left(t\right) = -200 is lmear in yy ∴I.F.=e∫12dt=e−t2\therefore I.F. = e^{\int \frac{1}{2} dt} = e^{-\frac{t}{2}} ∴\therefore role is p⋅e−t2=∫−200⋅(e−t2)dt+Cp\cdot e^{\frac{-t}{2}} = \int -200\cdot\left(e^{-\frac{t}{2}}\right)dt + C =−200⋅e−t/2−1/2+C= -200\cdot \frac{e^{-t/ 2}}{-1/ 2}+C =400 e−t/2+C= 400 \,e^{-t/2} + C Since p(0)=100p\left(0\right) = 100 ∴100 e0=400 e0+C\therefore 100 \,e^{0} = 400\, e^{0} + C ⇒100=400+C\Rightarrow 100 = 400 + C ⇒C=−300\Rightarrow C = - 300 ∴p(t)=400−300 et/2\therefore p\left(t\right) = 400 - 300 \,e^{t/2}