If `x(y)=x` and `y dx/dy=2x+y^3(y+1)e^y` and `x(1)=0` then `x(e)` is equal to

Options

A. `(e^e-1)`
B. `e^3(e^e-1)`
C. `e^2+1`
D. `(e^e-3)`

Answer

To solve the problem, we start with the given differential equation and initial condition. The equation is:

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

with the initial condition \( x(1) = 0 \).

### Step 1: Rearranging the Equation
First, we can rearrange the equation to isolate \( \frac{dx}{dy} \):

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

### Step 2: 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 \( Q(y) = y^2(y + 1)e^y \).

### Step 3: Finding the Integrating Factor
To solve this, we need to find the integrating factor \( \mu(y) \):

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

### Step 4: Multiplying by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor:

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

This simplifies to:

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

### Step 5: Recognizing the Left Side as a Derivative
The left-hand side can be recognized as the derivative of \( x \cdot \frac{1}{y^2} \):

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

### Step 6: Integrating Both Sides
Integrate both sides with respect to \( y \):

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

The left side simplifies to:

\[ x \cdot \frac{1}{y^2} = \int (y + 1)e^y \, dy \]

### Step 7: Solving the Integral
To solve the integral on the right, we can use integration by parts:

Let \( u = y + 1 \) and \( dv = e^y dy \), then \( du = dy \) and \( v = e^y \).

Thus:

\[ \int (y + 1)e^y \, dy = e^y(y + 1) - \int e^y \, dy = e^y(y + 1) - e^y + C = e^y y + C \]

### Step 8: Plugging Back into the Equation
Substituting back, we have:

\[ x \cdot \frac{1}{y^2} = e^y y + C \]

Multiplying through by \( y^2 \):

\[ x = y^2(e^y y + C) \]

### Step 9: Applying Initial Condition
Now we apply the initial condition \( x(1) = 0 \):

\[ 0 = 1^2(e^1 \cdot 1 + C) \implies 0 = e + C \implies C = -e \]

### Step 10: Final Solution
Substituting \( C \) back into the equation for \( x \):

\[ x = y^2(e^y y - e) \]

### Step 11: Finding \( x(e) \)
Now we need to find \( x(e) \):

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

### Final Answer
Thus, the value of \( x(e) \) is:

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