To keep the resonance frequency \(\omega\) unchanged, we know:
\[ \omega' = \omega = \frac{1}{\sqrt{L'C'}} \]
Given:
\[ L'C' = LC \]
Substituting \(C' = 4C\) into the equation:
\[ L' \cdot (4C) = LC \implies L' = \frac{L}{4} \]
Thus, the inductance must be decreased by:
\[ L - L' = L - \frac{L}{4} = \frac{3L}{4} \]
Find output voltage in the given circuit.
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: