## Solving the Differential Equation: (y^2 + xy^2)y' = 1

This article aims to provide a comprehensive solution to the given differential equation:

**(y^2 + xy^2)y' = 1**

We will utilize various techniques to arrive at the general solution.

### 1. Simplifying the Equation

First, let's simplify the equation by factoring out y² on the left-hand side:

**y²(1 + x)y' = 1**

### 2. Separating Variables

To solve this differential equation, we need to separate the variables 'x' and 'y'. Divide both sides of the equation by y²(1+x):

**y' = 1/(y²(1+x))**

Now, we can rewrite y' as dy/dx:

**dy/dx = 1/(y²(1+x))**

Multiply both sides by dx and y²:

**y² dy = dx/(1+x)**

### 3. Integrating Both Sides

Now we have successfully separated the variables. Let's integrate both sides of the equation:

∫ **y² dy** = ∫ **dx/(1+x)**

The left side integrates to (y³/3) and the right side integrates to ln|1+x| + C, where C is the constant of integration:

**(y³/3) = ln|1+x| + C**

### 4. Solving for y

To isolate y, we can perform the following steps:

- Multiply both sides by 3:
**y³ = 3ln|1+x| + 3C** - Take the cube root of both sides:
**y = (3ln|1+x| + 3C)^(1/3)**

### 5. Simplifying the Constant

Since 3C is also a constant, we can rewrite the solution as:

**y = (3ln|1+x| + C)^(1/3)**

where C represents a general constant of integration.

### Conclusion

The general solution to the differential equation (y² + xy²)y' = 1 is:

**y = (3ln|1+x| + C)^(1/3)**

This solution represents a family of curves. The specific curve that satisfies an initial condition can be obtained by plugging in the initial condition and solving for the constant C.