The population p(t) at time t of a certain mouse species satisfies the differential equation `d (p(t))/(dt)=0. 5 p(t)-450` If `p(0)""=""850` , then the time at which the population becomes zero is ............................................................... (1) 2 ln 18 (2) ln 9 (3) `1/2` In 18 (4) ln 18

Options

A. `1/2 ln 18`
B. `ln 18`
C. `2 ln 18`
D. `ln 9`

Answer

To solve the given differential equation for the population of the mouse species, we will follow these steps:

### Step 1: Write the differential equation
The given differential equation is:
\[ \frac{dp(t)}{dt} = 0.5 p(t) - 450 \]

### Step 2: Rearrange the equation
We can rearrange the equation as follows:
\[ \frac{dp(t)}{dt} = 0.5 p(t) - 450 \]
This can be rewritten as:
\[ \frac{dp(t)}{dt} = \frac{1}{2} p(t) - 450 \]

### Step 3: Separate variables
To separate the variables, we can rewrite the equation:
\[ \frac{dp(t)}{p(t) - 900} = \frac{1}{2} dt \]

### Step 4: Integrate both sides
Now we integrate both sides:
\[ \int \frac{1}{p(t) - 900} \, dp(t) = \int \frac{1}{2} \, dt \]
The left side integrates to:
\[ \ln |p(t) - 900| = \frac{1}{2} t + C \]

### Step 5: Solve for the constant C using the initial condition
We know that \( p(0) = 850 \). Substituting \( t = 0 \) and \( p(0) = 850 \) into the equation gives:
\[ \ln |850 - 900| = \frac{1}{2} \cdot 0 + C \]
This simplifies to:
\[ \ln | -50 | = C \implies C = \ln 50 \]

### Step 6: Substitute C back into the equation
Now we substitute \( C \) back into the equation:
\[ \ln |p(t) - 900| = \frac{1}{2} t + \ln 50 \]

### Step 7: Exponentiate both sides
Exponentiating both sides gives:
\[ |p(t) - 900| = 50 e^{\frac{1}{2} t} \]
Since \( p(t) < 900 \) for our case, we can write:
\[ 900 - p(t) = 50 e^{\frac{1}{2} t} \]

### Step 8: Solve for p(t)
Rearranging gives:
\[ p(t) = 900 - 50 e^{\frac{1}{2} t} \]

### Step 9: Find the time when the population becomes zero
To find the time when the population becomes zero, set \( p(t) = 0 \):
\[ 0 = 900 - 50 e^{\frac{1}{2} t} \]
Solving for \( e^{\frac{1}{2} t} \):
\[ 50 e^{\frac{1}{2} t} = 900 \implies e^{\frac{1}{2} t} = \frac{900}{50} = 18 \]

### Step 10: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \frac{1}{2} t = \ln 18 \implies t = 2 \ln 18 \]

### Conclusion
The time at which the population becomes zero is:
\[ \boxed{2 \ln 18} \]