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

Question

Let $f(x)=|\log_e x|-|x-1|+5$. Statement 1: $f(x)$ is differentiable for all $x\in(0,\infty)$
Statement 2: $f(x)$ is increasing in $(1,\infty)$
Statement 3: $f(x)$ is decreasing in $(0,1)$
Which of the following is correct?

Answer 1

Step 1: Variable Separation
Rearrange the differential equation to isolate variables: \[ \frac{dP}{P} = k \, dt. \]
Step 2: Integration
Integrate both sides of the equation: \[ \int \frac{1}{P} \, dP = \int k \, dt. \] This yields: \[ \ln P = kt + C, \] where $ C $ represents the constant of integration.
Step 3: Exponential Form
Exponentiate both sides to remove the natural logarithm:
\[ P = e^{kt + C} = e^C \cdot e^{kt}. \] Redefine $ e^C $ as a new constant, $ C_1 $:
\[ P = C_1 e^{kt}. \]

Answer 2

Step 1: Apply the general solution. The general solution is $ P = C_1 e^{kt}. $ Given that $ P = 1000 $ at $ t = 0 $, we have $ 1000 = C_1 e^{k(0)} $, which simplifies to $ C_1 = 1000 $. The equation is then $ P = 1000 e^{kt}. $
Step 2: Determine $ k $ by substituting values. At $ t = 1 $, $ P = 2000 $. This yields $ 2000 = 1000 e^{k(1)} $, so $ 2 = e^k $. Taking the natural logarithm of both sides gives $ k = \ln 2 $.

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

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Solution and Explanation

If the population of bacteria is 1000 at \( t = 0 \), and 2000 at \( t = 1 \), find the value of \( k \).

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Solution and Explanation

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