I f a transformation $y = u v$ transforms the given differential equation $f(x) y'' - 4 f'(x) y'+ g(x) y = 0 $ into the equation of the form $v'' + h(x) v = 0 $ then $u$ must be
- $ \frac{1}{f^2} $
- $xf $
- $ \frac{1}{2f} $
- $f^2$
I am stuck on this problem. Can anyone help me please...............