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Let the population of rabbits surviving at a time t be governed by the differential equation

$\frac{dp\left ( t \right )}{dt}= \frac{1}{2}p\left ( t \right )-200$ . If $p\left ( 0 \right )= 100,then\: p\left ( t \right )$ equal to:

Option 1)
$600-500\: e^{t/2}$

Option 2)
$400-300\: e^{-t/2}$

Option 3)
$400-300\: e^{t/2}$

Option 4)
$300-200\: e^{-t/2}$

Answers (1)

best_answer

As we learnt in

Solution of Differential Equation -

$\frac{\mathrm{d}y }{\mathrm{d} x} =f\left ( ax+by+c \right )$

put

$Z =ax+by+c$

- wherein

Equation with convert to

$\int \frac{dz}{bf\left ( z \right )+a} =x+c$

$\int \frac{dp(t)}{\frac{1}{2}p(t)-200}=\int dt$

$\Rightarrow 2\int_{100}^{p(t)} \frac{dp(t)}{p(t)-400}=\int_{t=0}^{t=t} dt$

$\Rightarrow log|p(t)-400|\int_{100}^{p(t)}=\frac{t}{2}\left.\begin{matrix} & \\ & \end{matrix}\right|_{0}^{t}=\frac{t}{2}$

$\therefore \log \left (400 -p(t) \right ) - \log 300 = \frac{t}{2}$

$\therefore \log \frac{(400 -p(t))}{300}= \frac{t}{2}$

$\therefore 400 -p(t)=e^{t/2}\times 300$

$\therefore p(t)=400 -300 \ e^{t/2}$

Option 1)

$600-500\: e^{t/2}$

This option is incorrect.

Option 2)

$400-300\: e^{-t/2}$

This option is incorrect.

Option 3)

$400-300\: e^{t/2}$

This option is correct.

Option 4)

$300-200\: e^{-t/2}$

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