The solution of the differential equation `ydx+ (x +x^2 y) dy =0` is

Explanation

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

### Step 1: Rewrite the Equation
We start by rewriting the given differential equation:
\[ y \, dx + (x + x^2 y) \, dy = 0 \]
This can be rearranged as:
\[ y \, dx + x \, dy + x^2 y \, dy = 0 \]

### Step 2: Group Terms
Next, we group the terms:
\[ y \, dx + (x + x^2 y) \, dy = 0 \]
This can be expressed as:
\[ y \, dx + x \, dy + x^2 y \, dy = 0 \]

### Step 3: Recognize the Total Differential
We can recognize that:
\[ d(xy) = x \, dy + y \, dx \]
Thus, we can rewrite our equation as:
\[ d(xy) + x^2 y \, dy = 0 \]

### Step 4: Divide by \( dy \)
Now, we divide the entire equation by \( dy \):
\[ \frac{d(xy)}{dy} + x^2 y = 0 \]

### Step 5: Integrate Both Sides
We can integrate both sides:
\[ \int \frac{d(xy)}{xy^2} + \int \frac{dy}{y} = 0 \]

### Step 6: Change of Variables
Let \( t = xy \). Then, \( d(xy) = dt \):
\[ \int \frac{dt}{t^2} + \int \frac{dy}{y} = 0 \]

### Step 7: Solve the Integrals
Now we can solve the integrals:
\[ -\frac{1}{t} + \log |y| = C \]

### Step 8: Substitute Back
Substituting back \( t = xy \):
\[ -\frac{1}{xy} + \log |y| = C \]

### Final Solution
Thus, the solution of the given differential equation is:
\[ -\frac{1}{xy} + \log |y| = C \]