The general solution of the differential equaiton `(1+y^(2))dx+(1+x^(2))dy=0`, is

Options

A. `x-y=C(1-xy)`
B. `x-y=C(1+xy)`
C. `x+y=C(1-xy)`
D. `x+y=C(1+xy)`

Answer

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

### Step 1: Rearranging the Equation
We start by rearranging the given differential equation:

\[ (1+y^2)dx + (1+x^2)dy = 0 \]

This can be rewritten as:

\[ (1+y^2)dx = -(1+x^2)dy \]

Dividing both sides by \((1+y^2)(1+x^2)\):

\[ \frac{dx}{1+x^2} = -\frac{dy}{1+y^2} \]

### Step 2: Integrating Both Sides
Now, we will integrate both sides:

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

The integral of \(\frac{1}{1+x^2}\) is \(\tan^{-1}(x)\) and the integral of \(\frac{1}{1+y^2}\) is \(\tan^{-1}(y)\). Therefore, we have:

\[ \tan^{-1}(x) = -\tan^{-1}(y) + C \]

where \(C\) is the constant of integration.

### Step 3: Rearranging the Equation
Rearranging the equation gives us:

\[ \tan^{-1}(x) + \tan^{-1}(y) = C \]

### Step 4: Using the Formula for Inverse Tangents
Using the formula for the sum of inverse tangents:

\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]

Thus, we can write:

\[ \tan^{-1}\left(\frac{x+y}{1-xy}\right) = C \]

### Step 5: Final Form of the General Solution
Exponentiating both sides, we can express the general solution as:

\[ \frac{x+y}{1-xy} = k \]

where \(k = \tan(C)\). Rearranging gives us:

\[ x + y = k(1 - xy) \]

This leads us to the final general solution:

\[ x + y = C(1 - xy) \]

where \(C\) is a constant.

### Conclusion
The general solution of the differential equation \((1+y^2)dx + (1+x^2)dy = 0\) is:

\[ x + y = C(1 - xy) \]