Differential Equations 3 Question 3
3. Let the population of rabbits surviving at a time $\frac{d p(t)}{d x}=\frac{1}{2} p(t)-200$. If $p(0)=100$, then $p(t)$ is equal to
(a) $400-300 e^{\frac{1}{2}}$
(b) $300-200 e^{-\frac{t}{2}}$
(c) $600-500 e^{\frac{1}{2}}$
(d) $400-300 e^{-\frac{t}{2}}$
(2014 Main)
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Answer:
Correct Answer: 3. (a)
Solution:
- Given, differential equation is $\frac{d p}{d t}-\frac{1}{2} p(t)=-200$ is a linear differential equation.
Here,
$$ \begin{aligned} p(t) & =\frac{-1}{2}, Q(t)=-200 \\ IF & =e^{\int-\frac{1}{2} d t}=e^{-\frac{t}{2}} \end{aligned} $$
Hence, solution is
$$ \begin{aligned} p(t) \cdot IF & =\int Q(t) \cdot IF d t \\ p(t) \cdot e^{-\frac{t}{2}} & =\int-200 \cdot e^{-\frac{t}{2}} d t \\ p(t) \cdot e^{-\frac{t}{2}} & =400 e^{-\frac{t}{2}}+K \\ p(t) & =400+k e^{-1 / 2} \\ \text { If } p(0)=100, \text { then } k & =-300 \\ \Rightarrow \quad p(t) & =400-300 e^{\frac{t}{2}} \end{aligned} $$
