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 by rearranging the equation to isolate \( dx \) and \( dy \):

\[ ye^y \, dx = (y^3 + 2xe^y) \, dy \]

Dividing both sides by \( ye^y \):

\[ dx = \frac{y^3 + 2xe^y}{ye^y} \, dy \]

### Step 2: Simplifying the Equation
Now, we simplify the right-hand side:

\[ dx = \left( \frac{y^3}{ye^y} + \frac{2x e^y}{ye^y} \right) dy \]

This simplifies to:

\[ dx = \left( \frac{y^2}{e^y} + \frac{2x}{y} \right) dy \]

### Step 3: Writing in Standard Form
We can rewrite the equation in the standard linear form:

\[ \frac{dx}{dy} - \frac{2}{y} x = \frac{y^2}{e^y} \]

### Step 4: Finding the Integrating Factor
The integrating factor \( \mu(y) \) is given by:

\[ \mu(y) = e^{\int -\frac{2}{y} \, dy} = e^{-2 \ln |y|} = \frac{1}{y^2} \]

### Step 5: Multiplying by the Integrating Factor
We multiply the entire equation by the integrating factor:

\[ \frac{1}{y^2} \frac{dx}{dy} - \frac{2}{y^3} x = \frac{y^2}{y^2 e^y} \]

This simplifies to:

\[ \frac{1}{y^2} \frac{dx}{dy} - \frac{2}{y^3} x = \frac{1}{e^y} \]

### Step 6: Integrating Both Sides
Now we can integrate both sides:

\[ \frac{d}{dy} \left( \frac{x}{y^2} \right) = \frac{1}{e^y} \]

Integrating the left side gives:

\[ \frac{x}{y^2} = \int \frac{1}{e^y} \, dy = -e^{-y} + C \]

### Step 7: Solving for x
Multiplying through by \( y^2 \):

\[ x = -y^2 e^{-y} + Cy^2 \]

### Step 8: Applying the Initial Condition
Using the initial condition \( y(0) = 1 \):

\[ 1 = -1^2 e^{-1} + C(1^2) \]

This simplifies to:

\[ 1 = -\frac{1}{e} + C \implies C = 1 + \frac{1}{e} \]

### Step 9: Final Equation for x
Substituting \( C \) back into the equation for \( x \):

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

### Step 10: Finding x when y = 0
Now we evaluate \( x \) when \( y = 0 \):

\[ x(0) = -0^2 e^{0} + \left(1 + \frac{1}{e}\right) 0^2 = 0 \]

Thus, the value of \( x \) when \( y = 0 \) is:

\[ \boxed{0} \]