A bacterial infection in an internal wound grows as `N'(t) = N_(0) exp (t)`, where the time t is in hours.A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as `(dN)/(dt)= -5 N^(2)`. What will be the plot of `N_(0)/N` vs. t after 1 hour?

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Explanation

To solve the problem, we need to analyze the growth of the bacterial population and the effect of the antibiotic over time. Let's break it down step by step.

### Step 1: Determine the bacterial population before the antibiotic reaches the wound

The growth of the bacterial population is given by the equation:
\[ N'(t) = N_0 e^t \]
where \( N_0 \) is the initial population at \( t = 0 \).

To find the population at \( t = 1 \) hour, we can integrate this equation:
\[ N(t) = \int N'(t) dt = \int N_0 e^t dt = N_0 e^t + C \]

At \( t = 0 \), \( N(0) = N_0 \), so:
\[ N(0) = N_0 e^0 + C \]
\[ N_0 = N_0 + C \]
Thus, \( C = 0 \).

Therefore, the bacterial population at \( t = 1 \) hour is:
\[ N(1) = N_0 e^1 = N_0 e \]

### Step 2: Analyze the effect of the antibiotic after it reaches the wound

After 1 hour, the antibiotic starts to take effect, and we have the differential equation:
\[ \frac{dN}{dt} = -5N^2 \]

We can rearrange this equation:
\[ \frac{dN}{N^2} = -5 dt \]

### Step 3: Integrate the equation

Integrating both sides:
\[ \int \frac{dN}{N^2} = \int -5 dt \]
This gives:
\[ -\frac{1}{N} = -5t + C' \]

### Step 4: Determine the constant of integration

At \( t = 1 \), we know \( N(1) = N_0 e \):
\[ -\frac{1}{N_0 e} = -5(1) + C' \]
\[ C' = \frac{1}{N_0 e} + 5 \]

### Step 5: Substitute back to find \( N(t) \)

Now substituting \( C' \) back into the integrated equation:
\[ -\frac{1}{N} = -5t + \left(\frac{1}{N_0 e} + 5\right) \]
Rearranging gives:
\[ \frac{1}{N} = 5t - \frac{1}{N_0 e} - 5 \]

### Step 6: Find \( \frac{N_0}{N} \)

Taking the reciprocal:
\[ \frac{N_0}{N} = \frac{N_0 e}{5N_0 e(5t - 5 - \frac{1}{N_0 e})} \]
This simplifies to:
\[ \frac{N_0}{N} = 5N_0 e(5t - 5 - \frac{1}{N_0 e}) \]

### Step 7: Plot \( \frac{N_0}{N} \) vs. \( t \)

The relationship we derived is linear in terms of \( t \), indicating that the plot of \( \frac{N_0}{N} \) vs. \( t \) will be a straight line.

### Conclusion

The plot of \( \frac{N_0}{N} \) vs. \( t \) after 1 hour will be a straight line with a positive slope.

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