The equilibrium reaction is:
\[ \text{Cr}_2\text{O}_7^{2-} + [\text{H}^+] + \text{OH}^- \rightleftharpoons 2\text{CrO}_4^{2-}. \]
In a basic medium (OH$^-$), the reaction shifts to the right to produce more CrO$_4^{2-}$, as OH$^-$ consumes H$^+$, reducing its concentration. Conversely, in an acidic medium (H$^+$), the equilibrium shifts to the left, favoring Cr$_2$O$_7^{2-}$.
Thus, the equilibrium shifts to the right in a basic medium.
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: