Solve the differential equation ,
`(xy^(2) + x) dx + (x^(2)y +y)dy=0`

Answer

To solve the differential equation \( (xy^2 + x)dx + (x^2y + y)dy = 0 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given differential equation:
\[ (xy^2 + x)dx + (x^2y + y)dy = 0 \]

### Step 2: Group the Terms
We can rearrange the equation to group the terms:
\[ (xy^2 + x)dx + (x^2y + y)dy = 0 \]

### Step 3: Factor Out Common Terms
Notice that we can factor out common terms:
\[ x(y^2 + 1)dx + y(x^2 + 1)dy = 0 \]

### Step 4: Separate Variables
We can rewrite the equation in a separable form:
\[ \frac{dy}{dx} = -\frac{xy^2 + x}{x^2y + y} \]
This can be simplified further:
\[ \frac{dy}{dx} = -\frac{x(y^2 + 1)}{y(x^2 + 1)} \]

### Step 5: Integrate Both Sides
Now we can integrate both sides. We will separate the variables:
\[ \frac{y}{y^2 + 1} dy = -\frac{x}{x^2 + 1} dx \]

Integrating both sides:
\[ \int \frac{y}{y^2 + 1} dy = -\int \frac{x}{x^2 + 1} dx \]

### Step 6: Perform the Integrations
The left side integrates to:
\[ \frac{1}{2} \ln(y^2 + 1) \]
And the right side integrates to:
\[ -\frac{1}{2} \ln(x^2 + 1) \]

### Step 7: Combine the Results
Combining the results gives us:
\[ \frac{1}{2} \ln(y^2 + 1) + \frac{1}{2} \ln(x^2 + 1) = C \]
Where \( C \) is the constant of integration.

### Step 8: Simplify the Expression
We can simplify this to:
\[ \ln((y^2 + 1)(x^2 + 1)) = 2C \]
Letting \( K = e^{2C} \), we have:
\[ (y^2 + 1)(x^2 + 1) = K \]

### Final Solution
Thus, the solution to the differential equation is:
\[ (y^2 + 1)(x^2 + 1) = C \]
where \( C \) is a constant.

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