Polio drops are delivered to 50K children in a district. The rate at which polio drops are given is directly proportional to the number of children who have not been administered the drops. By the end of `2^("nd")` week half the children have been given the polio drops. How many will have been given the drops by the end of 3rd week can be estimated using the solution to the differential equation `(dy)/(dx)=k(50-y)` where x denotes the number of weeks and y the number of children who have been given the drops.
Which method of solving a differential equation can be used to solve `(dy)/(dx)=k(50-y)`?

Options

A. Variable separable method
B. Solving Homogeneous differential equation
C. Solving Linear differential equation
D. all of the above

Answer

To solve the differential equation \(\frac{dy}{dx} = k(50 - y)\), we can use the method of separation of variables. Here’s a step-by-step solution:

### Step 1: Rewrite the equation
We start with the given differential equation:
\[ \frac{dy}{dx} = k(50 - y) \]

### Step 2: Separate the variables
We can rearrange the equation to separate the variables \(y\) and \(x\):
\[ \frac{dy}{50 - y} = k \, dx \]

### Step 3: Integrate both sides
Now, we integrate both sides. The left side requires the integration of \(\frac{1}{50 - y}\) and the right side is straightforward:
\[ \int \frac{dy}{50 - y} = \int k \, dx \]

The left side integrates to:
\[ -\ln|50 - y| = kx + C \]

### Step 4: Solve for \(y\)
Next, we solve for \(y\) by exponentiating both sides:
\[ |50 - y| = e^{-kx - C} = Ce^{-kx} \quad \text{(where \(C = e^{-C}\))} \]

This gives us:
\[ 50 - y = Ce^{-kx} \quad \text{or} \quad y = 50 - Ce^{-kx} \]

### Step 5: Determine the constant \(C\)
We know that by the end of the 2nd week, half of the children (25,000) have received the drops. Thus, we can substitute \(x = 2\) and \(y = 25000\) into the equation:
\[ 25000 = 50 - Ce^{-2k} \]
This simplifies to:
\[ Ce^{-2k} = 50 - 25000 = -24950 \]

### Step 6: Find the number of children by the end of the 3rd week
Now we want to find \(y\) when \(x = 3\):
\[ y = 50 - Ce^{-3k} \]

Using the value of \(C\) we found earlier, we can substitute:
\[ y = 50 + 24950 e^{-3k} \]

### Step 7: Estimate \(e^{-3k}\)
To find \(e^{-3k}\), we can use the relationship we derived for \(e^{-2k}\):
\[ e^{-3k} = e^{-2k} \cdot e^{-k} \]

Assuming we can estimate \(e^{-k}\) from the previous calculations, we can substitute back into the equation to find \(y\).

### Conclusion
By following these steps, we can estimate how many children will have received the drops by the end of the 3rd week.