If `x^3dy+ xydx= 2ydx +x^2dy` and `y(2) =e` then `y(4)`= ?

Options

A. `((1/2) +sqrte)`
B. `(1/2sqrte)`
C. `sqrte`
D. `3/2sqrte`

Answer

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

### Step 1: Rearranging the Equation
We start with the equation:
\[ x^3 dy + xy dx = 2y dx + x^2 dy \]
Rearranging gives us:
\[ x^3 dy - x^2 dy = 2y dx - xy dx \]
This simplifies to:
\[ (x^3 - x^2) dy = (2y - xy) dx \]
Factoring out common terms:
\[ (x^2(x - 1)) dy = y(2 - x) dx \]

### Step 2: Separating Variables
Now, we can separate the variables:
\[ \frac{dy}{y} = \frac{(2 - x)}{x^2(x - 1)} dx \]

### Step 3: Integrating Both Sides
Next, we integrate both sides:
\[ \int \frac{dy}{y} = \int \frac{(2 - x)}{x^2(x - 1)} dx \]
The left side integrates to:
\[ \ln |y| + C_1 \]
For the right side, we will use partial fraction decomposition:
\[ \frac{(2 - x)}{x^2(x - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1} \]
Multiplying through by the denominator \( x^2(x - 1) \) and solving for \( A \), \( B \), and \( C \) will yield the coefficients.

### Step 4: Finding Coefficients
Setting up the equation:
\[ 2 - x = A x(x - 1) + B(x - 1) + C x^2 \]
We will find \( A \), \( B \), and \( C \) by substituting suitable values for \( x \) or by equating coefficients.

1. Set \( x = 0 \):
\[ 2 = B(-1) \implies B = -2 \]

2. Set \( x = 1 \):
\[ 1 = C(1) \implies C = 1 \]

3. Set \( x = 2 \) to find \( A \):
\[ 0 = A(2)(1) - 2(1) + 2 \implies 0 = 2A - 2 \implies A = 1 \]

Thus, we have \( A = 1 \), \( B = -2 \), \( C = 1 \).

### Step 5: Integrating the Right Side
Now substituting back:
\[ \int \left( \frac{1}{x} - \frac{2}{x^2} + \frac{1}{x - 1} \right) dx \]
This integrates to:
\[ \ln |x| + \frac{2}{x} + \ln |x - 1| + C_2 \]

### Step 6: Combining Results
Combining both integrals, we have:
\[ \ln |y| = \ln |x| + \frac{2}{x} + \ln |x - 1| + C \]
Exponentiating both sides gives:
\[ y = K \cdot x \cdot (x - 1) e^{\frac{2}{x}} \]
where \( K = e^C \).

### Step 7: Using Initial Condition
Using the initial condition \( y(2) = e \):
\[ e = K \cdot 2 \cdot (2 - 1) e^{\frac{2}{2}} \implies e = K \cdot 2 \cdot 1 \cdot e \implies K = \frac{1}{2} \]
Thus, the equation for \( y \) becomes:
\[ y = \frac{1}{2} x (x - 1) e^{\frac{2}{x}} \]

### Step 8: Finding \( y(4) \)
Now we find \( y(4) \):
\[ y(4) = \frac{1}{2} \cdot 4 \cdot (4 - 1) e^{\frac{2}{4}} = \frac{1}{2} \cdot 4 \cdot 3 \cdot e^{\frac{1}{2}} = 6 e^{\frac{1}{2}} = 6 \sqrt{e} \]

Thus, the final answer is:
\[ \boxed{6 \sqrt{e}} \]