Let y be a differentiable function in $ \Bbb R$ which can be factorised as y = f(x)·sin(x) for some differentiable function f(x) in $ \Bbb R$.
Show that the function y satisfies the differential equation $$ y' = \Biggl[\frac {f'(x)}{f(x)} + cot(x)\Biggr]·y $$
I am not sure how to go about approaching this problem. I started by taking it at face value and getting $$ y' = f'(x)sin(x) + f(x)cos(x) $$ and then plugging it back into the equation. I ended up with $$ \frac{f'(x)f'(x)sin(x)}{f(x)} + 2f'(x)cos(x) + \frac{f(x)cos(x)}{cot(x)} $$ This does not seem to be the way to go about solving the problem.
Guidance on how to solve it would be much appreciated.
$$y' = \Biggl[\frac {f'(x)}{f(x)} + \cot(x)\Biggr]·y$$ $$\dfrac {y'}y -\frac {f'(x)}{f(x)} = \cot(x)$$ $$(\ln y -\ln f(x))' = \cot(x)$$ $$(\ln \dfrac y {f(x)})' = \cot(x)$$ Since $y=f(x)\sin x$: $$(\ln \sin x)' = \cot(x)$$ $$ \dfrac {\cos x} { \sin x} = \cot(x)$$ Which is true.