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

Options

A. `(x^(2)+1)(y^(2)-1)=c`
B. `(x^(2)-1)(y^(2)+1)=c`
C. `(x^(2)+1)(y^(2)+1)=c`
D. None of these

Answer

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

### Step 1: Rewrite the Differential Equation
We start with the equation:
\[ (xy^2 + x)dx + (yx^2 + y)dy = 0 \]
This can be rewritten as:
\[ (xy^2 + x)dx + (yx^2 + y)dy = 0 \]

### Step 2: Identify the Terms
Let's denote:
\[ M = xy^2 + x \quad \text{and} \quad N = yx^2 + y \]
Thus, our equation becomes:
\[ Mdx + Ndy = 0 \]

### Step 3: Check for Exactness
To check if the equation is exact, we need to compute:
\[ \frac{\partial M}{\partial y} \quad \text{and} \quad \frac{\partial N}{\partial x} \]
Calculating these:
\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(xy^2 + x) = 2xy \]
\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(yx^2 + y) = 2yx \]
Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.

### Step 4: Find the Potential Function
To find the potential function \( \psi(x, y) \) such that:
\[ \frac{\partial \psi}{\partial x} = M \quad \text{and} \quad \frac{\partial \psi}{\partial y} = N \]
We integrate \( M \) with respect to \( x \):
\[ \psi(x, y) = \int (xy^2 + x)dx = \frac{1}{2}x^2y^2 + x^2 + g(y) \]
where \( g(y) \) is a function of \( y \).

### Step 5: Differentiate with Respect to \( y \)
Now, we differentiate \( \psi \) with respect to \( y \):
\[ \frac{\partial \psi}{\partial y} = x^2y + g'(y) \]
Setting this equal to \( N \):
\[ x^2y + g'(y) = yx^2 + y \]
This gives:
\[ g'(y) = y \]
Integrating \( g'(y) \):
\[ g(y) = \frac{1}{2}y^2 + C \]

### Step 6: Combine the Results
Now substituting \( g(y) \) back into \( \psi(x, y) \):
\[ \psi(x, y) = \frac{1}{2}x^2y^2 + x^2 + \frac{1}{2}y^2 + C \]

### Step 7: Set the Potential Function Equal to a Constant
Thus, the general solution is given by:
\[ \frac{1}{2}x^2y^2 + x^2 + \frac{1}{2}y^2 = C \]

### Final Answer
The general solution of the differential equation is:
\[ x^2 + y^2 + \frac{1}{2}x^2y^2 = C \]