Solve the following initial value problems :
`y e^(y)dx=(y^(3)+2x e^(y))dy, y(0)=1`.

Answer

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

### Step 1: Rearranging the Differential Equation
We start by rewriting the given equation in a more manageable form. We can express it as:
\[ \frac{dx}{dy} = \frac{y^{3} + 2x e^{y}}{y e^{y}} \]

### Step 2: Simplifying the Equation
Now, we simplify the right-hand side:
\[ \frac{dx}{dy} = \frac{y^{3}}{y e^{y}} + \frac{2x e^{y}}{y e^{y}} = \frac{y^{2}}{e^{y}} + \frac{2x}{y} \]
Thus, we have:
\[ \frac{dx}{dy} - \frac{2x}{y} = \frac{y^{2}}{e^{y}} \]

### Step 3: Identifying the Linear Form
This is a linear first-order differential equation of the form:
\[ \frac{dx}{dy} + P(y)x = Q(y) \]
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 Through by the Integrating Factor
Multiply the entire equation by the integrating factor:
\[ \frac{1}{y^{2}} \frac{dx}{dy} - \frac{2x}{y^{3}} = \frac{y^{2}}{y^{2} e^{y}} = \frac{1}{e^{y}} \]
This simplifies to:
\[ \frac{d}{dy} \left( \frac{x}{y^{2}} \right) = \frac{1}{e^{y}} \]

### Step 6: Integrating Both Sides
Integrate both sides with respect to \( y \):
\[ \frac{x}{y^{2}} = \int \frac{1}{e^{y}} \, dy + C \]
The integral of \( \frac{1}{e^{y}} \) is \( -e^{-y} \), so:
\[ \frac{x}{y^{2}} = -e^{-y} + C \]

### Step 7: Solving for \( x \)
Now, we can express \( x \) in terms of \( y \):
\[ x = -y^{2} e^{-y} + C y^{2} \]

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

### Step 9: Final Solution
Substituting \( C \) back into the equation for \( x \):
\[ x = -y^{2} e^{-y} + e^{-1} y^{2} \]
Thus, we have:
\[ x = y^{2} \left( e^{-1} - e^{-y} \right) \]