The solution of the differential equation
`(dy)/(dx) + x(2x + y) = x^(3) (2x + y)^(3) - 2` is (C being an arbitrary constant)

Options

A. `1/(2x + xy) = x^(2) + 1 + Ce^(x)`
B. `1/((2x + y)^2) = x^(2) + 1 + Ce^(x^2)`
C. `1/(2x + y) = x^(2) + 1 + Ce^(-x^2)`
D. `1/((2x + y)^2) = x^(2) + 1 + Ce`

Answer

To solve the differential equation

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

we will follow these steps:

### Step 1: Rewrite the equation
We can rewrite the equation as:

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

### Step 2: Substitute \( t = 2x + y \)
Let \( t = 2x + y \). Then, differentiating with respect to \( x \):

\[ \frac{dt}{dx} = 2 + \frac{dy}{dx}. \]

From this, we can express \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{dt}{dx} - 2. \]

### Step 3: Substitute into the equation
Substituting \( t \) and \( \frac{dy}{dx} \) into the original equation gives:

\[ \frac{dt}{dx} - 2 + x t = x^3 t^3 - 2. \]

Rearranging this, we have:

\[ \frac{dt}{dx} + xt = x^3 t^3. \]

### Step 4: Rearranging and isolating \( dt/dx \)
Rearranging gives:

\[ \frac{dt}{dx} = x^3 t^3 - xt. \]

### Step 5: Divide by \( t^3 \)
Now, divide the entire equation by \( t^3 \):

\[ \frac{1}{t^3} \frac{dt}{dx} = x^3 - \frac{x}{t}. \]

### Step 6: Let \( k = \frac{1}{t^2} \)
Let \( k = \frac{1}{t^2} \). Then:

\[ \frac{dt}{dx} = -\frac{2}{t^3} \frac{dk}{dx}. \]

### Step 7: Rewrite the equation
Substituting this back gives:

\[ -\frac{2}{3} \frac{dk}{dx} - 2xk = -2x^3. \]

### Step 8: Linear differential equation
This is a linear differential equation in \( k \):

\[ \frac{dk}{dx} - 2xk = -2x^3. \]

### Step 9: Find the integrating factor
The integrating factor \( e^{\int -2x \, dx} = e^{-x^2} \).

### Step 10: Multiply through by the integrating factor
Multiplying through by the integrating factor gives:

\[ e^{-x^2} \frac{dk}{dx} - 2xe^{-x^2} k = -2x^3 e^{-x^2}. \]

### Step 11: Integrate
Integrating both sides gives:

\[ k e^{-x^2} = \int -2x^3 e^{-x^2} \, dx + C. \]

### Step 12: Solve for \( k \)
Solving for \( k \) gives:

\[ k = -x^2 + C e^{x^2}. \]

### Step 13: Substitute back for \( t \)
Recalling that \( k = \frac{1}{t^2} \):

\[ \frac{1}{t^2} = -x^2 + C e^{x^2}. \]

### Step 14: Substitute back for \( y \)
Substituting back for \( t = 2x + y \):

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

### Final Solution
Thus, the solution of the differential equation is:

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