Question

Find the particular solution of the differential equation: \[ \frac{dy}{dx} + y \cot x = 4x \csc x \text{(} x \neq 0 \text{)}. \] Given that \( y = 0 \) \(\text{ when}\) \( x = \frac{\pi}{2} \).

Hint:

For linear differential equations, use the integrating factor to simplify the equation and then integrate both sides.

Answer 1

This is a first-order linear differential equation. The standard form is: \[ \frac{dy}{dx} + P(x) y = Q(x) \] Here, \( P(x) = \cot x \) and \( Q(x) = 4x \csc x \). The integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \cot x \, dx} = e^{\ln \sin x} = \sin x \] Multiply both sides of the differential equation by \( \mu(x) = \sin x \): \[ \sin x \frac{dy}{dx} + y \sin x \cot x = 4x \] This simplifies to: \[ \frac{d}{dx} \left( y \sin x \right) = 4x \] Now, integrate both sides: \[ \int \frac{d}{dx} \left( y \sin x \right) \, dx = \int 4x \, dx \] \[ y \sin x = 2x^2 + C \] Now, use the initial condition \( y = 0 \) when \( x = \frac{\pi}{2} \): \[ 0 \cdot \sin \left( \frac{\pi}{2} \right) = 2 \left( \frac{\pi}{2} \right)^2 + C \] \[ 0 = 2 \times \frac{\pi^2}{4} + C $\Rightarrow$ C = -\frac{\pi^2}{2} \] Thus, the particular solution is: \[ y \sin x = 2x^2 - \frac{\pi^2}{2} \] \[ y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x} \] Final Answer: The particular solution is: \[ y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x} \]