Question

Let \(y = y(x)\) be the solution of the differential equation,
\((x^2 + 1)^2\frac{dy}{dx} + 2x(x^2 + 1)y = 1\)
such that \(y(0) = 0\). If \(\sqrt{a}\,y(1) = \frac{\pi}{32}\), then the value of 'a' is :

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{16}\)
• C. \(1\)
• D. \(\frac{1}{4}\)

Hint:

Divide by \(x^2 + 1\); the left side becomes the derivative of \((x^2 + 1)y\).

Answer 1

The Correct Option is B

Divide by \((x^2 + 1)\): \((x^2 + 1)\frac{dy}{dx} + 2xy = \frac{1}{x^2 + 1}\).
The left side is \(\frac{d}{dx}\left[(x^2 + 1)y\right]\). Integrate: \((x^2 + 1)y = \tan^{-1}x + C\).
\(y(0) = 0 \Rightarrow C = 0\). At \(x = 1\): \(2y(1) = \frac{\pi}{4}\), so \(y(1) = \frac{\pi}{8}\).
\(\sqrt{a}\cdot\frac{\pi}{8} = \frac{\pi}{32} \Rightarrow \sqrt{a} = \frac{1}{4} \Rightarrow a = \frac{1}{16}\).
Answer: (2) \(\frac{1}{16}\)