The population of sheep on a farm is 
given by dN
dt = k(N − 1800), where N is 
the number of sheep. If there are initially 
3000 sheep and after 3 years there are 
3400 sheep, find:
a the number of sheep on the farm after 
5 years
b when the sheep population will reach 
8000

We have the differential equation for the population of sheep given by:

\[
\frac{dN}{dt} = k(N - 1800)
\]

where \( N(t) \) is the number of sheep at time \( t \), and \( k \) is a constant. 

### Step 1: Solve the differential equation

The equation is separable, so we can rewrite it as:

\[
\frac{dN}{N - 1800} = k \, dt
\]

Integrating both sides:

\[
\int \frac{1}{N - 1800} \, dN = \int k \, dt
\]

The left-hand side integrates to:

\[
\ln|N - 1800| = kt + C
\]

Exponentiating both sides:

\[
N - 1800 = e^{kt + C} = Ae^{kt}
\]

where \( A = e^C \) is a constant. So the general solution is:

\[
N(t) = 1800 + Ae^{kt}
\]

### Step 2: Determine the constants \( A \) and \( k \)

We are given that initially (when \( t = 0 \)), \( N(0) = 3000 \). Plugging in these values:

\[
3000 = 1800 + A \cdot e^{0} \implies A = 1200
\]

Thus, the equation becomes:

\[
N(t) = 1800 + 1200e^{kt}
\]

We are also given that after 3 years (\( t = 3 \)), the population \( N(3) = 3400 \). Plugging in these values:

\[
3400 = 1800 + 1200e^{3k}
\]

Solving for \( k \):

\[
1600 = 1200e^{3k}
\]

\[
\frac{4}{3} = e^{3k}
\]

\[
k = \frac{1}{3} \ln\left(\frac{4}{3}\right)
\]

### Step 3: Calculate the number of sheep after 5 years

Now that we have \( k \), we can find \( N(5) \):

\[
N(5) = 1800 + 1200e^{5k}
\]

Substituting \( k = \frac{1}{3} \ln\left(\frac{4}{3}\right) \):

\[
N(5) = 1800 + 1200 \left(\frac{4}{3}\right)^{5/3}
\]

We can calculate this numerically:

\[
\left(\frac{4}{3}\right)^{5/3} \approx 1.5874
\]

\[
N(5) = 1800 + 1200 \times 1.5874 \approx 1800 + 1904.88 \approx 3704.88
\]

So the number of sheep after 5 years is approximately **3705**.

### Step 4: Determine when the population will reach 8000

We need to find \( t \) such that \( N(t) = 8000 \):

\[
8000 = 1800 + 1200e^{kt}
\]

\[
6200 = 1200e^{kt}
\]

\[
\frac{31}{6} = e^{kt}
\]

Taking the natural logarithm:

\[
kt = \ln\left(\frac{31}{6}\right)
\]

Substituting \( k = \frac{1}{3} \ln\left(\frac{4}{3}\right) \):

\[
t = \frac{3 \ln\left(\frac{31}{6}\right)}{\ln\left(\frac{4}{3}\right)}
\]

Calculating this numerically:

\[
t \approx \frac{3 \times 1.8326}{0.2877} \approx 19.11 \text{ years}
\]

So the population will reach 8000 sheep in approximately **19.1 years**.

---

Do you want more details or have any questions? Here are some related questions:

1. How would the population change if the initial number of sheep was different?
2. What happens to the population if the growth constant \( k \) changes?
3. How does the carrying capacity of 1800 affect the population dynamics?
4. What would the equation look like if the population was decreasing?
5. How does the population model compare to exponential growth without limits?

**Tip:** When dealing with population growth models, always check if there's a carrying capacity or a threshold that limits growth.

Explore the population growth of sheep using differential equations and exponential growth models. Solve for the number of sheep after 5 years and when the population will reach 8000. Understand the dynamics of population growth with a detailed step-by-step solution. Suitable for advanced high school students and learners interested in applied mathematics.

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