If \(y = y(x)\) is the solution of the differential equation \(\displaystyle\frac{dy}{dx} = x(1 + y^2)\) satisfying \(y(0) = 1\), then the value of y at \(x = \sqrt{\displaystyle\frac{\pi}{6}}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sqrt{3}\)
The equation is variables separable:
\( \displaystyle \frac{dy}{1 + y^2} = x\,dx \)
Integrating both sides:
\( \displaystyle \tan^{-1} y = \frac{x^2}{2} + C \)
Putting x = 0, y = 1:
\( \displaystyle C = \tan^{-1} 1 = \frac{\pi}{4} \)
At \(x = \sqrt{\frac{\pi}{6}}\), \(x^2 = \frac{\pi}{6}\):
\( \displaystyle \tan^{-1} y = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi}{3} \)
\( \displaystyle y = \tan\frac{\pi}{3} = \sqrt{3} \)