Given that S(t) + I(t) = N_0, and I(0) = 1, can you find S(t) and I(t)

Given the additional information:

- \( S(t) + I(t) = N_0 \) (which tells us that the total population \( N_0 \) is constant over time)
- \( I(0) = 1 \)

We can find the specific solutions for \( S(t) \) and \( I(t) \).

### Step 1: Use \( S(t) + I(t) = N_0 \)
Since \( S(t) + I(t) = N_0 \), we have:
\[
S(t) = N_0 - I(t)
\]
We substitute this into the differential equation for \( \frac{dI}{dt} \):

\[
\frac{dI}{dt} = \beta S I = \beta (N_0 - I) I
\]

### Step 2: Solve the differential equation
The equation becomes:
\[
\frac{dI}{dt} = \beta N_0 I - \beta I^2
\]
This can be rewritten as:
\[
\frac{dI}{I(N_0 - I)} = \beta dt
\]

### Step 3: Perform partial fraction decomposition
We decompose \( \frac{1}{I(N_0 - I)} \) as:
\[
\frac{1}{I(N_0 - I)} = \frac{1}{N_0} \left(\frac{1}{I} + \frac{1}{N_0 - I}\right)
\]
Thus, the equation becomes:
\[
\frac{1}{N_0} \left(\frac{1}{I} + \frac{1}{N_0 - I}\right) dI = \beta dt
\]
Integrating both sides:
\[
\frac{1}{N_0} \left(\ln|I| - \ln|N_0 - I|\right) = \beta t + C_1
\]
\[
\ln\left(\frac{I}{N_0 - I}\right) = \beta N_0 t + C_2
\]
where \( C_2 \) is a constant of integration.

### Step 4: Express \( I(t) \)
Exponentiating both sides:
\[
\frac{I}{N_0 - I} = A e^{\beta N_0 t}
\]
where \( A = e^{C_2} \) is a constant.

Rearranging to solve for \( I(t) \):
\[
I(t) = \frac{A N_0 e^{\beta N_0 t}}{1 + A e^{\beta N_0 t}}
\]

### Step 5: Determine the constant \( A \) using the initial condition \( I(0) = 1 \)
At \( t = 0 \):
\[
I(0) = \frac{A N_0}{1 + A} = 1
\]
So,
\[
A N_0 = 1 + A
\]
\[
A(N_0 - 1) = 1
\]
\[
A = \frac{1}{N_0 - 1}
\]

### Step 6: Write the final expressions for \( S(t) \) and \( I(t) \)
Now, substituting \( A \) back into the equation for \( I(t) \):
\[
I(t) = \frac{\frac{N_0}{N_0 - 1} e^{\beta N_0 t}}{1 + \frac{1}{N_0 - 1} e^{\beta N_0 t}}
\]
Simplifying:
\[
I(t) = \frac{N_0 e^{\beta N_0 t}}{N_0 - 1 + e^{\beta N_0 t}}
\]

And since \( S(t) = N_0 - I(t) \):
\[
S(t) = \frac{N_0 (N_0 - 1)}{N_0 - 1 + e^{\beta N_0 t}}
\]

### Final Answer:

\[
I(t) = \frac{N_0 e^{\beta N_0 t}}{N_0 - 1 + e^{\beta N_0 t}}
\]
\[
S(t) = \frac{N_0 (N_0 - 1)}{N_0 - 1 + e^{\beta N_0 t}}
\]

Would you like more details on this derivation or have any questions?

Here are some related questions:
1. How do the initial conditions affect the spread of \( I(t) \) over time?
2. What happens to \( I(t) \) as \( t \) approaches infinity?
3. How would a change in \( \beta \) affect the dynamics of \( S(t) \) and \( I(t) \)?
4. Can this model be applied to real-world epidemic modeling?
5. How would you interpret the relationship between \( S(t) \) and \( I(t) \) at any time \( t \)?

**Tip:** When dealing with differential equations in models, always check how initial conditions influence the long-term behavior of the solutions.

Learn how to solve the differential equations S(t) + I(t) = N_0 and I(0) = 1 with a detailed step-by-step solution. This problem covers key concepts in differential equations, initial value problems, and exponential growth models. Suitable for advanced undergraduate students.

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