Question

Obtain the general solution of the differential equation: \[ \frac{dP}{dt} = kP, \] and express it as an exponential function of \( t \).

Answer 1

Step 1: Separate the variables 
Rearrange the differential equation: \[ \frac{dP}{P} = k \, dt. \]
Step 2: Integrate both sides
Integrate with respect to their respective variables: \[ \int \frac{1}{P} \, dP = \int k \, dt. \] \[ \ln P = kt + C, \] where \( C \) is the constant of integration. 
Step 3: Express the solution in exponential form
Exponentiate both sides to eliminate the natural logarithm: 
\[ P = e^{kt + C} = e^C \cdot e^{kt}. \] Let \( e^C = C_1 \) (a new constant): 
\[ P = C_1 e^{kt}. \]

Answer 2

Step 1: Use the general solution
From the general solution: \[ P = C_1 e^{kt}. \] At \( t = 0 \), \( P = 1000 \): \[ 1000 = C_1 e^{k(0)} \implies C_1 = 1000. \] Thus, the equation becomes: \[ P = 1000 e^{kt}. \] 
Step 2: Substitute the values to find  \( k \), At \( t = 1 \), \( P = 2000 \):
 \[ 2000 = 1000 e^{k(1)} \implies 2 = e^k. \] 
Take the natural logarithm of both sides: \[ k = \ln 2. \]