The population P = P(t) at time ‘t’ of a certain spectes follows the differential equation dPdt=0.5P-450. If P(0) – 850, then the time at which population becomes zero is:

Given dp(t)dt−12p(t)=−450I.F= e∫−12 dt=e−t2Solution isp(t)(I.F)=∫(−450)(I.F) dtp(t)e−t2=−450∫e−t2dt=−450e−t2(−12)+c⇒ p(t)e−t2=900 e−t2+cPut t=0⇒ p(0) e0=900 e0+c⇒850−900=c⇒c=−50p(t)=900−50 et2If p(t)=0⇒ 0= 900−50 et2⇒ 90050=et2⇒ 18= et2⇒ log 18 = t2⇒ t= 2 log 18

The population P = P(t) at time ‘t’ of a certain spectes follows the differential equation dPdt=0.5P-450. If P(0) – 850, then the time at which population becomes zero is:

Given dp(t)dt−12p(t)=−450I.F= e∫−12 dt=e−t2Solution isp(t)(I.F)=∫(−450)(I.F) dtp(t)e−t2=−450∫e−t2dt=−450e−t2(−12)+c⇒ p(t)e−t2=900 e−t2+cPut t=0⇒ p(0) e0=900 e0+c⇒850−900=c⇒c=−50p(t)=900−50 et2If p(t)=0⇒ 0= 900−50 et2⇒ 90050=et2⇒ 18= et2⇒ log 18 = t2⇒ t= 2 log 18

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The population P = P(t) at time ‘t’ of a certain spectes follows the differential equation $\frac{dP}{dt} = 0.5 P - 450 .$ If P(0) – 850, then the time at which population becomes zero is:

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$\log_{e} 18$

$2 \log_{e} 18$

$\log_{e} 9$

$\frac{1}{2} \log_{e} 18$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n \frac{d p (t)}{d t} - \frac{1}{2} p (t) = - 450 \\ I . F = e^{\int \frac{- 1}{2} d t} = e^{\frac{- t}{2}} \\ S o l u t i o n i s \\ p (t) (I . F) = \int (- 450) (I . F) d t \\ p (t) e^{\frac{- t}{2}} = - 450 \int e^{\frac{- t}{2}} d t \\ = - 450 \frac{e^{\frac{- t}{2}}}{(\frac{- 1}{2})} + c \end{array}$

$\begin{array}{l} \Rightarrow p (t) e^{\frac{- t}{2}} = 900 e^{\frac{- t}{2}} + c \\ P u t t = 0 \\ \Rightarrow p (0) e^{0} = 900 e^{0} + c \\ \Rightarrow 850 - 900 = c \\ \Rightarrow c = - 50 \\ p (t) = 900 - 50 e^{\frac{t}{2}} \end{array}$

$\begin{array}{l} I f p (t) = 0 \\ \Rightarrow 0 = 900 - 50 e^{\frac{t}{2}} \\ \Rightarrow \frac{900}{50} = e^{\frac{t}{2}} \\ \Rightarrow 18 = e^{\frac{t}{2}} \\ \Rightarrow \log 18 = \frac{t}{2} \\ \Rightarrow t = 2 \log 18 \end{array}$