The particular solution of the differential equation \(x(1+y^2)\,dx - y(1+x^2)\,dy = 0,\ y(0) = 1\), is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(y^2 = 2x^2 + 1\)
The variables separate. Dividing by \((1+x^2)(1+y^2)\):
\( \displaystyle \frac{y}{1+y^2}\,dy = \frac{x}{1+x^2}\,dx \)
Multiplying by 2, each side has the form \(f'/f\), so integrating gives logarithms:
\( \log(1+y^2) = \log(1+x^2) + \log K \)
\( 1 + y^2 = K(1+x^2) \)
Putting \(x = 0,\ y = 1\):
\( 1 + 1 = K(1 + 0),\quad K = 2 \)
So the particular solution is
\( 1 + y^2 = 2 + 2x^2 \)
\( y^2 = 2x^2 + 1 \)