Particular solution of the differential equation \(x(1+ y^2)\,dx-y(1+x^2)\,dy = 0\), given \(y = 0\) when \(x = 1\), is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \((1+x^2)=2(1+ y^2)\)
The variables separate:
\(x(1+y^2)\,dx = y(1+x^2)\,dy\)
\( \displaystyle \frac{x}{1+x^2}\,dx = \frac{y}{1+y^2}\,dy \)
Integrating both sides, with the constant written as \(\frac{1}{2}\log k\):
\( \displaystyle \frac{1}{2}\log(1+x^2) = \frac{1}{2}\log(1+y^2) + \frac{1}{2}\log k \)
\(1 + x^2 = k(1 + y^2)\)
Putting x = 1, y = 0:
\(2 = k(1)\), so \(k = 2\)
The particular solution is \((1+x^2) = 2(1+y^2)\).
Shortcut: All four options have the form of the general solution, so the condition y = 0 at x = 1 decides. It gives 2 = 2 in Option (1), but 1 = 4, 2 = 4 and 1 = 8 in Options (2), (3) and (4).