Question

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

& A.400-300{{e}^{t}} \\\ & B.300-200{{e}^{t}} \\\ & C.600-500{{e}^{t}} \\\ & D.400-300{{e}^{t}} \\\ \end{aligned}$$

Answer 1

To solve this question, first of all try to bring terms having P on one side of the equation and other terms on the other side of the equation. Now, we can easily integrate and for that we will use two formulae of integration given as
$\int{\dfrac{1}{x}dx=\log \left( x \right)}\text{ and }\int{1dx}=x$

Complete step by step answer:
Given that, $\dfrac{dp\left( t \right)}{dt}=\dfrac{1}{2}p\left( t \right)-200$
Which can be written as $\dfrac{dp\left( t \right)}{dt}=\dfrac{p\left( t \right)-400}{2}$
Now, making all term of P (t) at one side and all left terms on the other side, we get:
$\dfrac{dp\left( t \right)}{p\left( t \right)-400}=\dfrac{1}{2}dt$
Now, both sides of the above equation have 'd' type terms, so we will integrate.
Integrating both sides, we get:
$\int{\dfrac{dp\left( t \right)}{p\left( t \right)-400}}=\int{\dfrac{1}{2}dt}$
We have, $\int{\dfrac{1}{x}dx=\log \left( x \right)}\text{ and }\int{1dx}=x$
Here, let $x=p(t)=400$ then $\int{\dfrac{dp(t)}{p(t)-400}}=\log \left| p(t)-400 \right|$
This is done by using $\int{\dfrac{1}{x}dx=\log \left| x \right|}$
$\Rightarrow \log \left| p(t)-400 \right|=\dfrac{1}{2}t+C\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
Where C is the constant of integration.
Given in the question is that P (0) = 100
Means at t = 0 value of P = 100
So, substituting t = 0 and P (0) = 100 in above equation (i), we get

& \log \left| 100-400 \right|=\dfrac{1}{2}0+C \\\ & \Rightarrow C=\log \left| -300 \right| \\\ & \Rightarrow C=\log \left| 300 \right| \\\ \end{aligned}$$ As mode value is always positive. Now, substituting value of C as $$C=\log \left| 300 \right|$$ in equation (i), we get: $$\log \left| p-400 \right|=\dfrac{1}{2}t+\log \left| 300 \right|$$ Subtracting $\log \left| 300 \right|$ both sides, we get: $$\log \left| p-400 \right|-\log \left| 300 \right|=\dfrac{1}{2}t$$ Using property of log which says $$\log \left( \dfrac{m}{n} \right)=\log m-\log n$$ in above and using $$m=p-400\text{ and n=300}$$ we get $$\log \left| \dfrac{p-400}{300} \right|=\dfrac{1}{2}t\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}$$ Again, we will use a property of log which is as below: $${{e}^{\log x}}=x$$ Applying exponential e on both sides of equation (ii) we get: $${{e}^{\log \left| \dfrac{p-400}{300} \right|}}={{e}^{\dfrac{1}{2}t}}$$ Using $${{e}^{\log x}}=x$$ in above, we get: $$\left| \dfrac{p-400}{300} \right|={{e}^{\dfrac{1}{2}t}}$$ Multiplying 300 on both sides, we get: $$\left| p-400 \right|=300{{e}^{\dfrac{1}{2}t}}$$ Here, $\left| 300 \right|=300$ as mode of a number is always positive. We have, $$\left| p-400 \right|=300{{e}^{\dfrac{1}{2}t}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}$$ Opening the mode here and considering both positive and negative value as $\left| x \right|=\pm x$ $$\Rightarrow \left| p-400 \right|=+\left( p-400 \right)\text{ and }\left| p-400 \right|=-\left( p-400 \right)$$ Both are possible. Case I- Let $$\left| p-400 \right|=+\left( p-400 \right)$$ Then, as it is given in the question that P (0) = 100 $$\Rightarrow \left( 100-400 \right)=-300$$ Then, if Case I is possible then LHS of equation (iii) is negative at t = 0 but at t = 0 RHS of equation (iii) is $$300{{e}^{\dfrac{1}{2}0}}=300\text{ as }{{\text{e}}^{0}}=1$$ which is positive, a contradiction. Hence, Case I is wrong. Consider, Case II - when $$\begin{aligned} & \left| p-400 \right|=-\left( p-400 \right) \\\ & \Rightarrow \left| p-400 \right|=400-p \\\ \end{aligned}$$ Here, at P (0) = 100 both LHS and RHS of equation (iii) are positive and equal. As $$400-100=300\text{ and 300}{{\text{e}}^{0}}=300$$ So, Case II is correct and $$\left| p-400 \right|=400-p$$ Using this in equation (iii) we get $$\begin{aligned} & 400-p=300{{e}^{\dfrac{1}{2}t}} \\\ & p=-300{{e}^{\dfrac{1}{2}t}}+400 \\\ \end{aligned}$$ Take $\dfrac{1}{2}t=2{{t}^{1}}$ then $$p=400-300{{e}^{{{t}^{1}}}}=400-300{{e}^{t}}$$ which is option D. **So, the correct answer is “Option D”.** **Note:** The biggest possibility of mistake a student can commit here in this question is, considering $\left| p-400 \right|=p-400$ without checking whether positive or negative value is possible or not. Always, first check which of the positive or negative values are not giving any contradiction then use it. Also, a possibility of confusion can be taking $\dfrac{1}{2}t={{t}^{1}}$ at last. That step is correct because, integration value does not change if a constant is multiplied to it. So, taking $\dfrac{1}{2}t={{t}^{1}}$ both are correct.