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Let the population of rabbits surviving at a time $'t'$ be governed by the differential equation $\large\frac{dp(t)}{dt}=\frac{1}{2}$$p(t)-200$ <br> If p(0)= 100, then what is the value of p(t)?

$\begin{array}{1 1}(A)\;400-300e^{\Large\frac{t}{2}}\\(B)\;300-200e^{\Large\frac{-t}{2}}\\(C)\;600-500e^{\Large\frac{t}{2}}\\(D)\;400-300e^{\Large\frac{-t}{2}}\end{array} $

1 Answer

Let p(t) = y
$\frac{dy}{dt} = \frac{1}{2}t - 200$
$\implies \frac{dy}{dt} - \frac{1}{2} t = -200$
General solution for the above linear equation is
$\begin{align*} y e ^{\int p dx} = \int Q e ^{\int p dx} dx + c\end{align*}$
where $p = \frac{-1}{2}; \; \; \; Q =-200 $
$\begin{align*} y e^{\int-\frac{1}{2} dt} &= \int Q e^{\int -\frac{1}{2 }dt} dt + c \\ \implies y e^{\frac{-1}{2} t}& = -200 \int e^{-\frac{1}{2} dt} + c \\ \implies y e^{-\frac{1}{2} t}& = -200 \frac{e ^{-\frac{1}{2} t}}{-\frac{1}{2}} + c \\ y e^{-\frac{1}{2}t} & = 400 e^{-\frac{1}{2} t} + c \end{align*}$
dividing throughout by $e^{-\frac{1}{2} t}$
$y = 400 + ce^{\frac{t}{2}}$
when $t=0$, $y = 100$
$\implies 100 = 400 + c e^0$
$\implies c = -300$
$\therefore y = 400 - 300 e^{\frac{t}{2}}$
answered Apr 7, 2014 by sreemathi.v
edited Nov 3 by priyanka.c
 

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