The solution of `(xy^(2)+x)dx +(yx^(2)+y)dy=0` is `(x^(2)+1)(y^(2)+1)=0`

Answer

To solve the differential equation \( (xy^2 + x)dx + (yx^2 + y)dy = 0 \) and verify if its solution is \( (x^2 + 1)(y^2 + 1) = 0 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
The given differential equation can be rewritten as:
\[ (xy^2 + x)dx + (yx^2 + y)dy = 0 \]
This can be rearranged to:
\[ (xy^2 + x)dx = -(yx^2 + y)dy \]

### Step 2: Factor Out Common Terms
We can factor out \( x \) from the first term and \( y \) from the second term:
\[ x(y^2 + 1)dx + y(x^2 + 1)dy = 0 \]

### Step 3: Separate Variables
Now, we can separate the variables:
\[ \frac{(y^2 + 1)}{(x^2 + 1)} = -\frac{dy}{dx} \]
This implies:
\[ \frac{dy}{y^2 + 1} = -\frac{dx}{x^2 + 1} \]

### Step 4: Integrate Both Sides
Now we will integrate both sides:
\[ \int \frac{dy}{y^2 + 1} = -\int \frac{dx}{x^2 + 1} \]
The integrals yield:
\[ \arctan(y) = -\arctan(x) + C \]
where \( C \) is the constant of integration.

### Step 5: Solve for the Relationship
Rearranging gives:
\[ \arctan(y) + \arctan(x) = C \]
Using the tangent addition formula, we can express this as:
\[ \tan(\arctan(y) + \arctan(x)) = \tan(C) \]
This leads us to:
\[ \frac{y + x}{1 - xy} = \tan(C) \]

### Step 6: Express in Terms of Constants
This can be rewritten in a more manageable form, but we need to check if it matches the proposed solution:
\[ (1 + x^2)(1 + y^2) = k \]
for some constant \( k \).

### Step 7: Verify the Proposed Solution
The proposed solution is \( (x^2 + 1)(y^2 + 1) = 0 \). This implies:
\[ x^2 + 1 = 0 \quad \text{or} \quad y^2 + 1 = 0 \]
However, \( x^2 + 1 \) and \( y^2 + 1 \) cannot be zero for real \( x \) and \( y \). Therefore, the proposed solution does not hold.

### Conclusion
The solution we derived does not match the proposed solution \( (x^2 + 1)(y^2 + 1) = 0 \). Thus, the statement is false.

---