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

Given the differential equation: \[ \sec^2 y \frac{dy}{dx} + 2x \sin y \sec y = x^3 \cos y. \]

Rearranging terms: \[ \sec^2 y \frac{dy}{dx} + 2x \tan y = x^3. \] Let \( t = \tan y \). 

Then: \[ \sec^2 y \frac{dy}{dx} = \frac{dt}{dx}. \] 

Substituting into the equation: \[ \frac{dt}{dx} + 2xt = x^3. \] 

This is a linear first-order differential equation. 

To solve it, we find the integrating factor (IF): \[ \text{IF} = e^{\int 2x dx} = e^{x^2}. \] 

Multiplying the entire equation by the integrating factor: \[ e^{x^2} \frac{dt}{dx} + 2x t e^{x^2} = x^3 e^{x^2}. \] 

This simplifies to: \[ \frac{d}{dx} (t e^{x^2}) = x^3 e^{x^2}. \] 

Integrating both sides: \[ t e^{x^2} = \int x^3 e^{x^2} \, dx. \] 

Using the substitution \( z = x^2, \, dz = 2x dx \): \[ \int x^3 e^{x^2} \, dx = \frac{1}{2} \int z e^z \, dz. \] 

Integrating by parts: \[ \int z e^z \, dz = e^z (z - 1), \] 

so: \[ \frac{1}{2} \int z e^z \, dz = \frac{1}{2} e^{x^2} (x^2 - 1). \] 

Thus: \[ t e^{x^2} = \frac{1}{2} e^{x^2} (x^2 - 1) + C. \] 

Dividing by \( e^{x^2} \): \[ t = \frac{1}{2} (x^2 - 1) + C e^{-x^2}. \] 

Recalling that \( t = \tan y \), 

we have: \[ \tan y = \frac{1}{2} (x^2 - 1) + C e^{-x^2}. \] 

Using the initial condition \( y(1) = 0 \): \[ \tan 0 = \frac{1}{2} (1^2 - 1) + C e^{-1^2}, \] \[ 0 = 0 + C e^{-1} \implies C = 0. \] 

Thus: \[ \tan y = \frac{1}{2} (x^2 - 1). \] 

To find \( y(\sqrt{3}) \): \[ \tan y = \frac{1}{2} ((\sqrt{3})^2 - 1) = \frac{1}{2} (3 - 1) = 1. \] 

Therefore: \[ y = \tan^{-1} (1) = \frac{\pi}{4}. \] 

Therefore: \[ \frac{\pi}{4}. \]