Question

y' = csc x - y cot x

Answer 1

Let us solve the differential equation $y' = \csc x - y \cot x,$ which is equivalent to $y' + y \frac{\cos x}{\sin x}= \frac{1}{\sin x}.$

Let us multiply both part of the equation by $\sin x.$ Then we get the equation $y'\sin x+y\cos x=1,$ which is equivalent to $(y\sin x)'=1.$ It follows that $y\sin x= x+C,$ and hence the general solution is of the form

$y=\frac{1}{\sin x}(x+C)=\csc x(x+C).$