Let `y(x)` be a solution of the differential equation `(1+e^x)y^(prime)+y e^x=1.` If `y(0)=2` , then which of the following statements is (are) true? (a) `( b ) (c) y(( d ) (e)-4( f ))=0( g )` (h) (b) `( i ) (j) y(( k ) (l)-2( m ))=0( n )` (o) (c) `( d ) (e) y(( f ) x (g))( h )` (i) has a critical point in the interval `( j ) (k)(( l ) (m)-1,0( n ))( o )` (p) (q) `( r ) (s) y(( t ) x (u))( v )` (w) has no critical point in the interval `( x ) (y)(( z ) (aa)-1,0( b b ))( c c )` (dd)

Options

A. `y(-4)=0`
B. `y(-2)=0`
C. `y(x)` has a critical point in the interval `(-1,0)`
D. `y(x)` has no critical point in the interval `(-1,0)`

Explanation

To solve the differential equation given by

\[ (1 + e^x) y' + y e^x = 1 \]

with the initial condition \( y(0) = 2 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the equation in a more standard form. We can express it as:

\[ y' + \frac{e^x}{1 + e^x} y = \frac{1}{1 + e^x} \]

### Step 2: Identify the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int \frac{e^x}{1 + e^x} \, dx} \]

To compute this integral, we can simplify it:

\[ \int \frac{e^x}{1 + e^x} \, dx = \int \left( 1 - \frac{1}{1 + e^x} \right) \, dx = x - \ln(1 + e^x) + C \]

Thus, the integrating factor becomes:

\[ \mu(x) = e^{x - \ln(1 + e^x)} = \frac{e^x}{1 + e^x} \]

### Step 3: Multiply the Equation by the Integrating Factor
Now we multiply the entire differential equation by the integrating factor:

\[ \frac{e^x}{1 + e^x} y' + \frac{e^{2x}}{(1 + e^x)^2} y = \frac{e^x}{1 + e^x} \]

This simplifies to:

\[ \frac{d}{dx} \left( \frac{e^x}{1 + e^x} y \right) = \frac{e^x}{1 + e^x} \]

### Step 4: Integrate Both Sides
Integrating both sides gives:

\[ \frac{e^x}{1 + e^x} y = \int \frac{e^x}{1 + e^x} \, dx \]

The right-hand side can be computed as:

\[ \int \frac{e^x}{1 + e^x} \, dx = x - \ln(1 + e^x) + C \]

Thus, we have:

\[ \frac{e^x}{1 + e^x} y = x - \ln(1 + e^x) + C \]

### Step 5: Solve for \( y \)
Now, we can solve for \( y \):

\[ y = \frac{(1 + e^x)(x - \ln(1 + e^x) + C)}{e^x} \]

### Step 6: Apply the Initial Condition
Using the initial condition \( y(0) = 2 \):

\[ y(0) = \frac{(1 + 1)(0 - \ln(2) + C)}{1} = 2 \]

This simplifies to:

\[ 2C - \ln(2) = 2 \implies C = 1 + \frac{\ln(2)}{2} \]

### Step 7: Final Solution
Substituting \( C \) back into the equation for \( y \):

\[ y = \frac{(1 + e^x)(x - \ln(1 + e^x) + 1 + \frac{\ln(2)}{2})}{e^x} \]

### Step 8: Analyze Critical Points
To find critical points, we need to set \( y' = 0 \) and analyze the behavior of the function in the intervals specified.