If `ye^(y) dx = (y^(3) + 2xe^(y))dy, y(0) = 1`, then the value of x when y = 0 is

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the differential equation given by \( ye^{y} dx = (y^{3} + 2xe^{y}) dy \) with the initial condition \( y(0) = 1 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ ye^{y} dx = (y^{3} + 2xe^{y}) dy \]
We can rearrange this to express \( \frac{dx}{dy} \):
\[ \frac{dx}{dy} = \frac{y^{3} + 2xe^{y}}{ye^{y}} \]

### Step 2: Simplifying the Right Side
Now, we simplify the right side:
\[ \frac{dx}{dy} = \frac{y^{3}}{ye^{y}} + \frac{2xe^{y}}{ye^{y}} = \frac{y^{2}}{e^{y}} + \frac{2x}{y} \]

### Step 3: Formulating the Linear Differential Equation
We can rewrite this as:
\[ \frac{dx}{dy} - \frac{2x}{y} = \frac{y^{2}}{e^{y}} \]
This is a first-order linear differential equation of the form:
\[ \frac{dx}{dy} + P(x) x = Q(y) \]
where \( P(y) = -\frac{2}{y} \) and \( Q(y) = \frac{y^{2}}{e^{y}} \).

### Step 4: Finding the Integrating Factor
The integrating factor \( \mu(y) \) is given by:
\[ \mu(y) = e^{\int P(y) dy} = e^{\int -\frac{2}{y} dy} = e^{-2 \ln |y|} = \frac{1}{y^{2}} \]

### Step 5: Multiplying by the Integrating Factor
We multiply the entire equation by the integrating factor:
\[ \frac{1}{y^{2}} \frac{dx}{dy} - \frac{2x}{y^{3}} = \frac{y^{2}}{e^{y} y^{2}} = \frac{1}{e^{y}} \]
This simplifies to:
\[ \frac{d}{dy}\left(\frac{x}{y^{2}}\right) = \frac{1}{e^{y}} \]

### Step 6: Integrating Both Sides
Now, we integrate both sides:
\[ \int \frac{d}{dy}\left(\frac{x}{y^{2}}\right) dy = \int \frac{1}{e^{y}} dy \]
The left side gives:
\[ \frac{x}{y^{2}} = -e^{-y} + C \]
where \( C \) is the constant of integration.

### Step 7: Solving for \( x \)
We can express \( x \) as:
\[ x = -y^{2} e^{-y} + Cy^{2} \]

### Step 8: Applying the Initial Condition
Using the initial condition \( y(0) = 1 \), we substitute \( y = 1 \) and \( x = 0 \):
\[ 0 = -1 \cdot e^{-1} + C \cdot 1^{2} \]
This gives:
\[ C = e^{-1} \]

### Step 9: Final Expression for \( x \)
Substituting \( C \) back into the equation for \( x \):
\[ x = -y^{2} e^{-y} + e^{-1} y^{2} \]
or
\[ x = y^{2} (e^{-1} - e^{-y}) \]

### Step 10: Finding \( x \) when \( y = 0 \)
Now we find \( x \) when \( y = 0 \):
\[ x = 0^{2} (e^{-1} - e^{0}) = 0 \cdot (e^{-1} - 1) = 0 \]

Thus, the value of \( x \) when \( y = 0 \) is:
\[ \boxed{0} \]