Obtain the general solution of the equation x2dydx+(xcotx+2)xy= 4sinx0<x<π\begin{align*} x^2\frac{\mathrm{d}y}{\mathrm{d}x} + (x\cot x+2)xy =\,& 4\sin x \qquad 0 < x < \pi\\[2mm] \end{align*}x2dxdy+(xcotx+2)xy=4sinx0<x<π Give your answer in the form y=f(x)y = \mathrm{f}(x)y=f(x) (8)
