Question

The population ( p ) of the city at time ( t ) is given by ( \frac{dp}{dt} = \frac{p}{2} - 100 ). If initial population is 100 then ( p = )

• A. ( 200 + 100e^{t/2} )
• B. ( 200 - 100e^{t/2} )
• C. ( 300 - 100e^{t/2} )
• D. ( 300 + 100e^{t/2} )

Hint:

Differential equations of the form $\frac{dy}{dx} = ay + b$ always have exponential solutions.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This is a first-order differential equation modeling population change.
Step 3: Detailed Explanation:
\[ \frac{dp}{dt} = \frac{p - 200}{2} \]
Using variable separable method:
\[ \int \frac{dp}{p - 200} = \int \frac{1}{2} dt \]
\[ \ln|p - 200| = \frac{t}{2} + C \]
\[ p - 200 = K e^{t/2} \]
Initial condition: At \( t = 0, p = 100 \).
\[ 100 - 200 = K e^0 \implies K = -100 \]
Substituting \( K \) back:
\[ p - 200 = -100 e^{t/2} \]
\[ p = 200 - 100 e^{t/2} \]
Step 4: Final Answer:
The population equation is \( p = 200 - 100 e^{t/2} \).