Question

If \( y = y(x) \) is the solution of \( \frac{dy}{dx} = \frac{x - y \cos x}{1 + \sin x} \), \( y\left(\frac{\pi}{2}\right) = \frac{\pi^2}{8} \), then \( y(\pi) = \)

• A. \( \frac{5\pi^2}{8} \)
• B. \( \frac{7\pi^2}{8} \)
• C. \( \frac{9\pi^2}{8} \)
• D. \( \frac{12\pi^2}{7} \)

Hint:

Recognize the differential equation as a first-order linear ODE. Find the integrating factor. Multiply the equation by the integrating factor and integrate both sides. Use the initial condition to find the constant of integration. Finally, substitute the required value of \( x \) to find \( y \).

Answer 1

The Correct Option is A

The given differential equation is \( \frac{dy}{dx} = \frac{x - y \cos x}{1 + \sin x} \).
Rewrite it as \( \frac{dy}{dx} + \frac{\cos x}{1 + \sin x} y = \frac{x}{1 + \sin x} \).
This is a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x) y = Q(x) \), where \( P(x) = \frac{\cos x}{1 + \sin x} \) and \( Q(x) = \frac{x}{1 + \sin x} \).
The integrating factor is \( IF = e^{\int P(x) dx} = e^{\int \frac{\cos x}{1 + \sin x} dx} \).
Let \( u = 1 + \sin x \), then \( du = \cos x dx \).
\( \int \frac{\cos x}{1 + \sin x} dx = \int \frac{du}{u} = \log |u| = \log |1 + \sin x| \).
Since \( 1 + \sin x \ge 0 \) for all \( x \), \( IF = e^{\log(1 + \sin x)} = 1 + \sin x \).
The solution is given by \( y \cdot IF = \int Q(x) \cdot IF dx + C \).
\( y (1 + \sin x) = \int \frac{x}{1 + \sin x} (1 + \sin x) dx + C \) \( y (1 + \sin x) = \int x dx + C \) \( y (1 + \sin x) = \frac{x^2}{2} + C \) We are given \( y\left(\frac{\pi}{2}\right) = \frac{\pi^2}{8} \).
Substitute \( x = \frac{\pi}{2} \) and \( y = \frac{\pi^2}{8} \) into the solution: \( \frac{\pi^2}{8} \left( 1 + \sin \frac{\pi}{2} \right) = \frac{(\frac{\pi}{2})^2}{2} + C \) \( \frac{\pi^2}{8} (1 + 1) = \frac{\pi^2}{8} + C \) \( \frac{2\pi^2}{8} = \frac{\pi^2}{8} + C \) \( \frac{\pi^2}{4} = \frac{\pi^2}{8} + C \) \( C = \frac{\pi^2}{4} - \frac{\pi^2}{8} = \frac{2\pi^2 - \pi^2}{8} = \frac{\pi^2}{8} \) The particular solution is \( y (1 + \sin x) = \frac{x^2}{2} + \frac{\pi^2}{8} \).
We need to find \( y(\pi) \).
Substitute \( x = \pi \) into the particular solution: \( y (1 + \sin \pi) = \frac{\pi^2}{2} + \frac{\pi^2}{8} \) \( y (1 + 0) = \frac{4\pi^2 + \pi^2}{8} \) \( y = \frac{5\pi^2}{8} \)