Step 1: Solve the homogeneous part first:
Set the right side to zero: $\dfrac{dy}{dx}+y\cot x=0$.
Separate variables: $\dfrac{dy}{y}=-\cot x\,dx$.
Integrate both sides: $\ln y=-\ln(\sin x)+\ln k$, so $y=\dfrac{k}{\sin x}$ for a constant k.
Step 2: Let the constant become a function of x:
Assume the full solution has the form $y=\dfrac{v(x)}{\sin x}$, where v is now a function to be found.
Differentiate using the quotient rule:
\[ \frac{dy}{dx}=\frac{v'\sin x-v\cos x}{\sin^2x}=\frac{v'}{\sin x}-\frac{v\cos x}{\sin^2x} \]
Step 3: Substitute into the original equation:
Put y and dy/dx into $\dfrac{dy}{dx}+y\cot x=4x\csc x$.
\[ \frac{v'}{\sin x}-\frac{v\cos x}{\sin^2x}+\frac{v}{\sin x}\cdot\frac{\cos x}{\sin x}=4x\csc x \]
The last two terms cancel exactly, leaving:
\[ \frac{v'}{\sin x}=\frac{4x}{\sin x}\implies v'=4x \]
Integrate: $v=2x^2+C$.
Step 4: Write y and apply the given condition:
\[ y=\frac{v}{\sin x}=\frac{2x^2+C}{\sin x} \]
At $x=\dfrac{\pi}{2}$, $y=0$: $0=\dfrac{2(\pi/2)^2+C}{1}$, so $C=-\dfrac{\pi^2}{2}$.
Final Answer:
The variation of parameters method gives the same particular solution.
\[ \boxed{y=\frac{4x^2-\pi^2}{2\sin x}} \]