Problem Statement
Consider the first-order linear differential equation $\frac{dy}{dx} + y \cot x = 2x \csc x$ for $0 < x < \pi$, with boundary condition $y\left(\frac{\pi}{2}\right) = 0$.
**(a)** [3 marks] Show that an integrating factor for the differential equation is $I(x) = \sin x$.
**(b)** [4 marks] Hence, find the particular solution $y$ in terms of $x$.
Verified Solution & Marking Scheme
Part (a) [M1]
Identify $P(x) = \cot x$, so $I(x) = e^{\int \cot x \, dx}$
Part (a) [A1]
$\int \cot x \, dx = \ln|\sin x|$
Part (a) [A1]
$I(x) = e^{\ln(\sin x)} = \sin x$ (for $0 < x < \pi$, $\sin x > 0$) (AG)
Part (b) [M1]
Multiply by integrating factor: $\frac{d}{dx}[y \sin x] = 2x \csc x \sin x = 2x$
Part (b) [A1]
Integrate both sides: $y \sin x = \int 2x \, dx = x^2 + C$
Part (b) [M1]
Apply boundary condition $x = \frac{\pi}{2}, y = 0$: $0 \cdot \sin(\frac{\pi}{2}) = (\frac{\pi}{2})^2 + C \implies C = -\frac{\pi^2}{4}$
Part (b) [A1]
$y = \frac{x^2 - \frac{\pi^2}{4}}{\sin x} = (x^2 - \frac{\pi^2}{4})\csc x$