The equation of the curve passing through the point (1, 0) and satisfying the differential equation $(1+y^{2})dx-xydy=0$ is :

Question

Accepted Answer

To determine why Option 2, $x^2 - y^2 = 1$, is the correct solution to the differential equation $(1 + y^2)dx - xydy = 0$, we need to solve the differential equation and verify that this equation satisfies the given conditions.

### Explanation:

1. **Solving the Differential Equation:**

The given differential equation is:
   $$
   (1 + y^2)dx - xydy = 0
   $$

This can be rearranged as:
   $$
   \frac{dx}{dy} = \frac{xy}{1 + y^2}
   $$

This is a separable differential equation. We can separate the variables as follows:
   $$
   \frac{dx}{x} = \frac{y}{1 + y^2} dy
   $$

Integrating both sides:
   $$
   \int \frac{dx}{x} = \int \frac{y}{1 + y^2} dy
   $$

The left side integrates to:
   $$
   \ln |x|
   $$

For the right side, use the substitution $u = 1 + y^2$, $du = 2y \, dy$, so $\frac{1}{2} du = y \, dy$:
   $$
   \int \frac{y}{1 + y^2} dy = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln |u| = \frac{1}{2} \ln |1 + y^2|
   $$

Thus, the integrated equation becomes:
   $$
   \ln |x| = \frac{1}{2} \ln |1 + y^2| + C
   $$

Exponentiating both sides to eliminate the logarithms:
   $$
   |x| = e^C \sqrt{1 + y^2}
   $$

Let $e^C = k$, a constant, so:
   $$
   x = k \sqrt{1 + y^2}
   $$

2. **Applying the Initial Condition:**

The curve passes through the point $(1, 0)$. Substituting $x = 1$ and $y = 0$ into the equation:
   $$
   1 = k \sqrt{1 + 0^2} \Rightarrow 1 = k
   $$

Therefore, $k = 1$, and the equation simplifies to:
   $$
   x = \sqrt{1 + y^2}
   $$

Squaring both sides gives:
   $$
   x^2 = 1 + y^2 \Rightarrow x^2 - y^2 = 1
   $$

This matches Option 2.

3. **Verification of Other Options:**

- **Option 1: $x^2 + y^2 = 1$**
     - This represents a circle, which does not satisfy the differential equation derived above.

- **Option 3: $2x^2 + y^2 = 2$**
     - This equation does not satisfy the derived relationship $x^2 - y^2 = 1$.

- **Option 4: $x^2 + 2y^2 = 1$**
     - This equation also does not satisfy the derived relationship $x^2 - y^2 = 1$.

Thus, the correct answer is Option 2, $x^2 - y^2 = 1$, as it satisfies both the differential equation and the initial condition.