`x(1+y^(2))dx+y(1+x^(2))dy=0` at `(0, 0)`

Options

A. `(1+x^(2))(1+y^(2))=0`
B. `(1+x^(2))(1+y^(2))=1`
C. `x^(2)+y^(2)=1`
D. `(1-x^(2))(1-y^(2))=1`

Answer

To solve the differential equation \( x(1+y^2)dx + y(1+x^2)dy = 0 \) at the point \( (0, 0) \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given differential equation:
\[ x(1+y^2)dx + y(1+x^2)dy = 0 \]
We can rearrange this to express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{x(1+y^2)}{y(1+x^2)} \]

### Step 2: Separate variables
Next, we separate the variables \( x \) and \( y \):
\[ \frac{y}{1+y^2} dy = -\frac{x}{1+x^2} dx \]

### Step 3: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{y}{1+y^2} dy = -\int \frac{x}{1+x^2} dx \]

For the left side, we can use the substitution \( u = 1 + y^2 \), which gives \( du = 2y dy \) or \( dy = \frac{du}{2y} \):
\[ \int \frac{y}{1+y^2} dy = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln |1+y^2| \]

For the right side, we can use the substitution \( v = 1 + x^2 \), which gives \( dv = 2x dx \) or \( dx = \frac{dv}{2x} \):
\[ -\int \frac{x}{1+x^2} dx = -\frac{1}{2} \int \frac{1}{v} dv = -\frac{1}{2} \ln |1+x^2| \]

### Step 4: Combine the results
Combining the results from both integrals, we have:
\[ \frac{1}{2} \ln |1+y^2| = -\frac{1}{2} \ln |1+x^2| + C \]
Multiplying through by 2 gives:
\[ \ln |1+y^2| + \ln |1+x^2| = 2C \]
This can be rewritten as:
\[ \ln |(1+y^2)(1+x^2)| = 2C \]
Exponentiating both sides, we get:
\[ (1+y^2)(1+x^2) = e^{2C} = k \quad \text{(where \( k \) is a constant)} \]

### Step 5: Determine the constant using the initial condition
To find the constant \( k \), we use the initial condition \( (x, y) = (0, 0) \):
\[ (1+0^2)(1+0^2) = k \implies k = 1 \]

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