© All Rights reserved @ LearnWithDash
Step-by-step Solution
Step 1: Write the differential equation
The problem gives the following first-order differential equation for the population p(t) of mice:
$ \frac{dp(t)}{dt} = 0.5\,p(t) \;-\; 450 $
Step 2: Rewrite in a more convenient form
It is often helpful to make the coefficient in front of p(t) clear. Since 0.5 = 1/2, we can write:
$ \frac{dp(t)}{dt} = \frac{1}{2}\,p(t) \;-\; 450 \;=\; \frac{p(t)-900}{2}\,.
$
Step 3: Separate variables
Bring all terms involving p(t) to one side and dt to the other side. Observe that:
$ \frac{dp(t)}{dt} \;=\; \frac{p(t) - 900}{2} \quad \Longrightarrow \quad 2\,\frac{dp(t)}{p(t) - 900} = dt.
$
To make the integral more direct, rewrite p(t) - 900 = -(900 - p(t)). Then:
$ 2\,\frac{dp(t)}{900 - p(t)} = -\,dt.
$
Step 4: Integrate both sides
Let $ u = 900 - p(t) $. Therefore $ du = -dp(t) $, and the integral becomes:
$ -2 \int \frac{dp(t)}{900 - p(t)} = \int dt
$
Substituting $ du = -dp(t)$ gives:
$ 2 \int \frac{du}{u} = \int dt.
$
The integral of $1/u$ is $ \ln |u| $, so we have:
$ 2 \ln|u| = t + C
$
where C is the constant of integration.
Step 5: Express the general solution and apply initial condition
Substituting back $u = 900 - p(t)$:
$ 2 \ln|900 - p(t)| = t + C.
$
Exponentiating both sides:
$ |900 - p(t)| = e^{\frac{t + C}{2}} = e^{\frac{t}{2}} \cdot e^{\frac{C}{2}}.
$
Let $ K = e^{\frac{C}{2}} $. Then:
$ 900 - p(t) = K\,e^{\frac{t}{2}}.
$
Using the initial condition $ p(0)=850 $ at $ t = 0 $:
$ 900 - 850 = K \,\cdot e^0 \quad \Longrightarrow \quad 50 = K.
$
Hence, the particular solution is:
$ 900 - p(t) = 50\,e^{\frac{t}{2}} \quad \Longrightarrow \quad p(t) = 900 - 50\,e^{\tfrac{t}{2}}.
$
Step 6: Find the time t when p(t) becomes zero
We need to find $ t_1 $ such that $ p(t_1) = 0 $:
$ 0 \;=\; 900 - 50\,e^{\frac{t_1}{2}}.
$
So,
$ 50\,e^{\frac{t_1}{2}} = 900 \quad \Longrightarrow \quad e^{\frac{t_1}{2}} = \frac{900}{50} = 18.
$
Taking the natural logarithm on both sides,
$ \frac{t_1}{2} = \ln(18) \quad \Longrightarrow \quad t_1 = 2\,\ln(18).
$
Step 7: Final answer
The time at which the population becomes zero is:
$ \boxed{2 \ln(18)}.
