Let $y=y(x)$ be the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \sec ^2 x=2 \sec ^2 x+3 \tan x \cdot \sec ^2 x$ such that $y(0)=\frac{5}{4}$. Then $12\left(y\left(\frac{\pi}{4}\right)-\mathrm{e}^{-2}\right)$ is equal to_____________________

<p>$$\begin{aligned} & \frac{d y}{d x}+2 y \sec ^2 x=2 \sec ^2 x+3 \tan x \sec ^2 x \\ & \text { I.F. }=e^{\int 2 \sec ^2 x d x} \end{aligned}$$</p> <p>$$\begin{aligned} &\begin{aligned} & \text { I.F. }=e^{2 \tan x} \\ & y \cdot e^{2 \tan x}=\int e^{2 \tan x}(2+3 \tan x) \sec ^2 x d x \end{aligned}\\ &\text { Put } \tan x=u\\ &\begin{aligned} & \sec ^2 x d x=d u \\ & y \cdot e^{2 u}=\int e^{2 u}(2+3 u) d u \\ & y \cdot e^{2 u} \Rightarrow \frac{2 e^{2 u}}{2}+3 \int e^{2 u} \cdot u d u \\ & y \cdot e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\int \frac{e^{2 u}}{2}\right] \\ & y e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\frac{e^{2 u}}{4}\right]+C \\ & y e^{2 \tan x}=e^{2 \tan x}+3\left[\frac{\tan x e^{2 \tan x}}{2}-\frac{e^{2 \tan x}}{4}\right]+C \end{aligned} \end{aligned}$$</p> <p>$$\begin{aligned} & F(0)=\frac{5}{4} \\ & \frac{5}{4}=1-\frac{3}{4}+C \\ & \frac{5}{4}-\frac{1}{4}=C \\ & 1=C \\ & y=1+3\left(\frac{\tan x}{2}-\frac{1}{4}\right)+1 \cdot e^{-2 \tan x} \\ & y\left(\frac{\pi}{4}\right)=1+3\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{e^2} \\ & y\left(\frac{\pi}{4}\right)=\frac{7}{4}+\frac{1}{e^2} \\ & 12\left(y\left(\frac{x}{4}\right)-\frac{1}{e^2}\right)=12\left(\frac{7}{4}+\frac{1}{e^2}-\frac{1}{e^2}\right)=21 \end{aligned}$$</p>