Finding all functions $f(x)$ so that $f'(x)\sin x - f(x)\cos x = \sin^2x$

Question

Find all functions $f(x)$, so that $f'(x)\sin x - f(x)\cos x = \sin^2x$.

Note: I should solve it without using differential equations. I've been trying to write the LHS as a derivative of some function but no success. I would like some hints.

Answer 1

Many such questions can be easily solved if you observe that one of the sides of the equation can be expressed in some standard form. One such form is $f\cdot g' - g\cdot f'$. This can be written as: $$ \dfrac{g\cdot f' - f\cdot g'}{g^2}$$ or $$-\dfrac{f\cdot g' - g\cdot f'}{f^2} $$ Both forms can be useful at various times! In your problem, $f=f(x)$ and $g=\sin x$.

Answer 2

Hint...If you divide by $\sin^2 x$ you have an exact derivative

Answer 3

$$\left(\frac{f(x)}{\sin x}\right)'=1$$