Question

Let \(y = y(x)\) be the solution curve of the differential equation \[ \sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y, \] \(y(1) = 0\). Then \(y\left(\sqrt{3}\right)\) is equal to:

• A. \(\frac{\pi}{3}\)
• B. \(\frac{\pi}{6}\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{\pi}{12}\)

Answer 1

The Correct Option is C

\( \sec^2 y \dfrac{dy}{dx} + 2x \sin y \sec y = x^3 \cos y \sec y \)

\( \sec^2 y \dfrac{dy}{dx} + 2x \tan y = x^3 \)

Let \( \tan y = t \). Then \( \sec^2 y \dfrac{dy}{dx} = \dfrac{dt}{dx} \).

The equation becomes \( \dfrac{dt}{dx} + 2xt = x^3 \), given \( e^{2x} dx = e^{x^2} \).

\( te^{x^2} = \int x^3 e^{x^2} dx + c \)

Let \( x^2 = Z \). Then \( t \cdot e^Z = \dfrac{1}{2} \int e^Z \cdot Z dZ = \dfrac{1}{2} \left[ e^Z \cdot Z - e^Z \right] + c \).

\( 2 \tan y = (x^2 - 1) + 2c e^{-x^2} \)

Given \( y(1) = 0 \), we find \( c = 0 \). Thus, \( y(\sqrt{3}) = \dfrac{\pi}{4} \).