The particular solution of the differential equation $x(1+ y^2)dx-y(1 + x^2)dy = 0,y(0)=1$, is

$y^2 = 2x^2+1$

The correct answer is Option (2) → $y^2 = 2x^2+1$

Given differential equation: $x(1+y^2)dx - y(1+x^2)dy = 0$

Rewrite:

$\frac{dy}{dx} = \frac{x(1+y^2)}{y(1+x^2)} = \frac{x}{y} \cdot \frac{1+y^2}{1+x^2}$

Separate variables:

$\frac{y}{1+y^2} dy = \frac{x}{1+x^2} dx$

Integrate both sides:

$\int \frac{y}{1+y^2} dy = \int \frac{x}{1+x^2} dx$

LHS: $\frac{1}{2} \ln(1+y^2)$, RHS: $\frac{1}{2} \ln(1+x^2)$

So: $\ln(1+y^2) = \ln(1+x^2) + C$ → $1+y^2 = K (1+x^2)$

Apply initial condition $y(0)=1$:

$1 + 1^2 = K (1 + 0^2) \Rightarrow 2 = K \Rightarrow K = 2$

Particular solution:

$1 + y^2 = 2 (1 + x^2) \Rightarrow y^2 = 2(1+x^2) - 1 = 1 + 2x^2$

$y = \sqrt{1 + 2x^2}$