Question - In Pic
My progress is that i found out that $\displaystyle f''(x) = \frac{[f'(x)]^2}{f(x)}$
but i couldn't get to the solution
If you use the identity:
$\displaystyle \cos{x} + i\sin{x} = e^{ix}$,
then
$\displaystyle y = e^{ix} \cdot e^{3ix} \cdot ... \cdot e^{\left(2n-1\right)ix}$
$\displaystyle = e^{ix + 3ix + ... + \left(2n-1\right)ix}$
$\displaystyle = e^{ix\left(1 + 3 + ... + 2n-1\right)$
$\displaystyle = e^{n^2ix}$