the rate of increase in the number of people,x, who own a alaptop is proportional to the product of x and N-x where N is the total polpulation. Initially x=1/10N and after one week x=1/4N. Form a differential equation from information above and show that after t weeks x/N-x= 3^t-2

### Step 1: Set Up the Differential Equation

Given that the rate of increase in the number of people \( x \) who own a laptop is proportional to the product of \( x \) and \( N - x \), we can express this relationship mathematically as:

\[
\frac{dx}{dt} = kx(N - x)
\]

where \( k \) is a constant of proportionality, and \( N \) is the total population.

### Step 2: Initial Conditions

The problem provides initial conditions:
- Initially, \( x = \frac{1}{10}N \).
- After one week (\( t = 1 \)), \( x = \frac{1}{4}N \).

### Step 3: Solve the Differential Equation

To solve the differential equation, first, let's rewrite it by separating the variables:

\[
\frac{dx}{x(N - x)} = k \, dt
\]

Next, we perform partial fraction decomposition on the left-hand side:

\[
\frac{1}{x(N - x)} = \frac{1}{N} \left( \frac{1}{x} + \frac{1}{N - x} \right)
\]

So the equation becomes:

\[
\frac{1}{N} \left( \frac{1}{x} + \frac{1}{N - x} \right) dx = k \, dt
\]

Integrating both sides with respect to their respective variables:

\[
\int \left( \frac{1}{x} + \frac{1}{N - x} \right) dx = kN \int dt
\]

The integral of the left-hand side is:

\[
\ln |x| - \ln |N - x| = kNt + C
\]

This can be simplified to:

\[
\ln \left( \frac{x}{N - x} \right) = kNt + C
\]

Exponentiating both sides to remove the natural logarithm:

\[
\frac{x}{N - x} = e^{kNt + C} = Ae^{kNt}
\]

where \( A = e^C \) is a constant.

### Step 4: Apply Initial Conditions

At \( t = 0 \), \( x = \frac{1}{10}N \):

\[
\frac{\frac{1}{10}N}{N - \frac{1}{10}N} = A
\]

Simplifying:

\[
\frac{\frac{1}{10}N}{\frac{9}{10}N} = A \implies A = \frac{1}{9}
\]

So the equation becomes:

\[
\frac{x}{N - x} = \frac{1}{9} e^{kNt}
\]

After one week (\( t = 1 \)), \( x = \frac{1}{4}N \):

\[
\frac{\frac{1}{4}N}{N - \frac{1}{4}N} = \frac{1}{9}e^{kN}
\]

Simplifying:

\[
\frac{\frac{1}{4}N}{\frac{3}{4}N} = \frac{1}{3} = \frac{1}{9}e^{kN} \implies e^{kN} = 3
\]

Thus, \( kN = \ln 3 \), and the equation becomes:

\[
\frac{x}{N - x} = \frac{1}{9} \cdot 3^t
\]

This simplifies to:

\[
\frac{x}{N - x} = 3^{t-2}
\]

This is the required solution.

### Conclusion

The differential equation formed is:

\[
\frac{dx}{dt} = kx(N - x)
\]

And we have shown that:

\[
\frac{x}{N - x} = 3^{t - 2}
\]

after \( t \) weeks.

---

Would you like further details on any part of the solution? 

### Relative Questions:
1. What is the significance of the constant \( k \) in the differential equation?
2. How would the solution change if the initial condition was different?
3. Can the general solution be applied to other proportional growth models?
4. How does the assumption of proportionality affect the outcome?
5. What mathematical techniques are used to solve such differential equations?
6. What would happen if \( N \) were to change over time?
7. How can you interpret the solution \( \frac{x}{N-x} = 3^{t-2} \) in real-world terms?
8. What other real-life scenarios can be modeled by similar differential equations?

### Tip:
When dealing with proportional relationships in differential equations, always remember to check the boundary conditions, as they can provide essential constants for your general solution.

Learn how to form and solve a differential equation for proportional growth in a population scenario. This problem covers the mathematical modeling of growth rates and initial conditions, demonstrating how to derive and solve the equation step by step.

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