Let `y =f (x)` satisfies the differential equation `xy (1+y) dx =dy . If f (0)=1 and f (2) =(e ^(2))/(k-e ^(2)),` then find the value of k.

Answer

To solve the differential equation \( xy(1+y) \, dx = dy \) with the initial conditions \( f(0) = 1 \) and \( f(2) = \frac{e^2}{k - e^2} \), we will proceed step by step.

### Step 1: Rearranging the Differential Equation
We start with the given differential equation:
\[ xy(1+y) \, dx = dy \]
We can rearrange this to separate variables:
\[ \frac{dy}{y(1+y)} = x \, dx \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. The left side requires partial fraction decomposition:
\[ \frac{1}{y(1+y)} = \frac{A}{y} + \frac{B}{1+y} \]
Multiplying through by \( y(1+y) \) gives:
\[ 1 = A(1+y) + By \]
Setting \( y = 0 \):
\[ 1 = A \implies A = 1 \]
Setting \( y = -1 \):
\[ 1 = -B \implies B = -1 \]
Thus, we have:
\[ \frac{1}{y(1+y)} = \frac{1}{y} - \frac{1}{1+y} \]
Now, we can integrate:
\[ \int \left( \frac{1}{y} - \frac{1}{1+y} \right) dy = \int x \, dx \]
This results in:
\[ \ln |y| - \ln |1+y| = \frac{x^2}{2} + C \]

### Step 3: Simplifying the Equation
Using the properties of logarithms, we can rewrite the left side:
\[ \ln \left| \frac{y}{1+y} \right| = \frac{x^2}{2} + C \]
Exponentiating both sides gives:
\[ \frac{y}{1+y} = e^{\frac{x^2}{2} + C} = Ae^{\frac{x^2}{2}} \quad \text{(where \( A = e^C \))} \]

### Step 4: Solving for \( y \)
Rearranging gives:
\[ y = \frac{Ae^{\frac{x^2}{2}}}{1 - Ae^{\frac{x^2}{2}}} \]

### Step 5: Applying Initial Condition \( f(0) = 1 \)
Substituting \( x = 0 \) and \( y = 1 \):
\[ 1 = \frac{A}{1 - A} \]
This implies:
\[ 1 - A = A \implies 1 = 2A \implies A = \frac{1}{2} \]
Thus, the function becomes:
\[ y = \frac{\frac{1}{2} e^{\frac{x^2}{2}}}{1 - \frac{1}{2} e^{\frac{x^2}{2}}} \]

### Step 6: Applying Second Condition \( f(2) = \frac{e^2}{k - e^2} \)
Now, substituting \( x = 2 \):
\[ f(2) = \frac{\frac{1}{2} e^{2}}{1 - \frac{1}{2} e^{2}} = \frac{e^2/2}{1 - e^2/2} \]
Setting this equal to \( \frac{e^2}{k - e^2} \):
\[ \frac{e^2/2}{1 - e^2/2} = \frac{e^2}{k - e^2} \]
Cross-multiplying gives:
\[ e^2(k - e^2) = 2e^2(1 - e^2/2) \]
Simplifying:
\[ k e^2 - e^4 = 2e^2 - e^4 \]
Thus:
\[ k e^2 = 2e^2 \implies k = 2 \]

### Final Answer
The value of \( k \) is:
\[ \boxed{2} \]