The given differential equation is:

\[
x^2 y' + 2xy - y^3 = 0
\]

This is a first-order differential equation, which can be solved by trying a method like separation of variables or using an integrating factor. Let's proceed step by step:

### Step 1: Rewrite the equation
We first rewrite the equation in a more standard form:

\[
x^2 \frac{dy}{dx} + 2xy - y^3 = 0
\]

Now, divide through by \(x^2\) (assuming \(x \neq 0\)) to simplify:

\[
\frac{dy}{dx} + \frac{2}{x} y - \frac{y^3}{x^2} = 0
\]

### Step 2: Recognize a Bernoulli equation
The equation is now in the form of a **Bernoulli differential equation**, which has the general form:

\[
\frac{dy}{dx} + P(x) y = Q(x) y^n
\]

Here, \(P(x) = \frac{2}{x}\), \(Q(x) = -\frac{1}{x^2}\), and \(n = 3\).

### Step 3: Solve the Bernoulli equation
To solve, we use the substitution:

\[
v = y^{1-n} = y^{-2}
\]

Now, differentiate both sides with respect to \(x\):

\[
\frac{dv}{dx} = -2 y^{-3} \frac{dy}{dx}
\]

Using this substitution, we transform the original equation. First, rewrite the equation in terms of \(v\). Multiply the original equation by \(y^{-3}\):

\[
y^{-3} \frac{dy}{dx} + \frac{2}{x} y^{-2} - \frac{1}{x^2} = 0
\]

This becomes:

\[
- \frac{1}{2} \frac{dv}{dx} + \frac{2}{x} v = \frac{1}{x^2}
\]

### Step 4: Solve the linear differential equation for \(v\)
Now we have a linear differential equation in \(v\):

\[
\frac{dv}{dx} - \frac{4}{x} v = - \frac{2}{x^2}
\]

This is a first-order linear differential equation. We solve it using the integrating factor method.

The integrating factor is:

\[
\mu(x) = e^{\int -\frac{4}{x} dx} = e^{-4 \ln |x|} = x^{-4}
\]

Multiply the entire equation by \(x^{-4}\):

\[
x^{-4} \frac{dv}{dx} - \frac{4}{x} x^{-4} v = - \frac{2}{x^6}
\]

This simplifies to:

\[
\frac{d}{dx} \left( x^{-4} v \right) = - \frac{2}{x^6}
\]

Now, integrate both sides with respect to \(x\):

\[
x^{-4} v = \int - \frac{2}{x^6} dx = \frac{1}{2x^5} + C
\]

Thus,

\[
v = x^4 \left( \frac{1}{2x^5} + C \right)
\]

Simplify this:

\[
v = \frac{x^4}{2x^5} + Cx^4 = \frac{1}{2x} + Cx^4
\]

### Step 5: Substitute back \(v = y^{-2}\)
Since \(v = y^{-2}\), we now substitute back to find \(y\):

\[
y^{-2} = \frac{1}{2x} + Cx^4
\]

Take the reciprocal to solve for \(y\):

\[
y = \left( \frac{1}{2x} + Cx^4 \right)^{-1/2}
\]

This is the general solution to the differential equation.

### Final solution:

\[
y(x) = \left( \frac{1}{2x} + Cx^4 \right)^{-1/2}
\]

Let me know if you'd like more details or have questions!

---

### Relative questions:
1. What is a Bernoulli differential equation?
2. How does the integrating factor method work for linear equations?
3. Can you explain the substitution technique used here in other types of differential equations?
4. What is the general solution approach for first-order nonlinear equations?
5. How does the solution change if \(n = 2\) instead of 3 in a Bernoulli equation?

**Tip:** Recognizing the form of an equation (like Bernoulli's) can simplify solving it tremendously.

Learn how to solve the first-order Bernoulli differential equation x^2y' + 2xy - y^3 = 0 using substitution and the integrating factor method. This step-by-step solution covers key concepts such as the Bernoulli equation, substitution techniques, and the integrating factor theorem. Ideal for university-level students studying differential equations.

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