If `y=f(x)` is the solution of equation `ydx+dy=-e^(x)y^(2)`dy, f(0)=1 and area bounded by the curve `y=f(x), y=e^(x)` and x=1 is A, then

Options

A. curve y=f(x) is passing through `(-2,e)`.
B. Curve `y=f(x)` is passing through `(1,1//e)`
C. curve `y=f(x)` is passing through `(1,1//3)`
D. `A=e+2/sqrt(e )-3`

Explanation

To solve the given differential equation and find the area bounded by the curve \( y = f(x) \), \( y = e^x \), and \( x = 1 \), we will follow these steps:

### Step 1: Rewrite the given differential equation
The given equation is:
\[ y \, dx + dy = -e^x y^2 \, dy \]
Rearranging this, we can write:
\[ y \, dx + dy + e^x y^2 \, dy = 0 \]
This can be expressed as:
\[ y \, dx + (1 + e^x y^2) \, dy = 0 \]

### Step 2: Separate variables
We can separate the variables by dividing through by \( y^2 \):
\[ \frac{dx}{y^2} + \frac{dy}{1 + e^x y^2} = 0 \]
This leads to:
\[ \frac{dx}{y^2} = -\frac{dy}{1 + e^x y^2} \]

### Step 3: Integrate both sides
Integrating both sides, we can express this as:
\[ \int \frac{dx}{y^2} = -\int \frac{dy}{1 + e^x y^2} \]
This requires some manipulation, but we can also use a different approach based on the structure of the equation.

### Step 4: Recognize the form of the equation
We can rewrite the original equation as:
\[ \frac{dy}{dx} = -\frac{y}{e^x y^2} = -\frac{1}{e^x y} \]
This suggests that we can express it in terms of a derivative:
\[ \frac{d}{dx} \left( e^{-x} y \right) = -y \]

### Step 5: Solve the differential equation
We can solve the equation:
\[ \frac{d}{dx} \left( e^{-x} y \right) = -y \]
This leads to:
\[ e^{-x} y = -\int y \, dx \]
Integrating both sides gives us:
\[ e^{-x} y = -y + C \]
Rearranging gives:
\[ e^{-x} y + y = C \]
Factoring out \( y \):
\[ y (e^{-x} + 1) = C \]
Thus, we find:
\[ y = \frac{C}{e^{-x} + 1} \]

### Step 6: Apply the initial condition
Using the condition \( f(0) = 1 \):
\[ 1 = \frac{C}{e^0 + 1} = \frac{C}{2} \implies C = 2 \]
Thus, we have:
\[ y = \frac{2}{e^{-x} + 1} = \frac{2 e^x}{1 + e^x} \]

### Step 7: Find the area A
To find the area \( A \) bounded by \( y = f(x) \), \( y = e^x \), and \( x = 1 \):
\[ A = \int_0^1 \left( e^x - \frac{2 e^x}{1 + e^x} \right) dx \]
This simplifies to:
\[ A = \int_0^1 e^x \left( 1 - \frac{2}{1 + e^x} \right) dx \]
Calculating this integral:
\[ A = \int_0^1 \left( \frac{e^x + 2 - 2}{1 + e^x} \right) dx = \int_0^1 \frac{e^x}{1 + e^x} dx \]

### Step 8: Evaluate the integral
Using substitution or integration techniques, we can evaluate this integral to find the area \( A \).

### Final Result
After evaluating the integral, we find the area \( A \) as:
\[ A = e - 1 \]