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

The given differential equation is a first-order linear form
$$\frac{dy}{dx}+P(x)\,y = Q(x)$$
with $$P(x)=2\sec^{2}x$$ and $$Q(x)=2\sec^{2}x+3\tan x\,\sec^{2}x$$.

Step 1: Find the integrating factor (IF).
$$\text{IF}=e^{\int P(x)\,dx}=e^{\int 2\sec^{2}x\,dx}=e^{2\tan x}$$

Step 2: Multiply every term of the differential equation by the IF.
$$e^{2\tan x}\frac{dy}{dx}+2\sec^{2}x\,e^{2\tan x}\,y=(2\sec^{2}x+3\tan x\,\sec^{2}x)\,e^{2\tan x}$$

The left side is now the derivative of $$y\,e^{2\tan x}$$ because
$$\frac{d}{dx}\bigl(y\,e^{2\tan x}\bigr)=e^{2\tan x}\frac{dy}{dx}+2\sec^{2}x\,e^{2\tan x}\,y.$$

Therefore,
$$\frac{d}{dx}\bigl(y\,e^{2\tan x}\bigr)=\sec^{2}x\,e^{2\tan x}\,(2+3\tan x).$$

Step 3: Integrate both sides with respect to $$x$$.
$$\int\frac{d}{dx}\bigl(y\,e^{2\tan x}\bigr)\,dx=\int\sec^{2}x\,e^{2\tan x}\,(2+3\tan x)\,dx.$$

Put $$u=\tan x \;\Rightarrow\; du=\sec^{2}x\,dx$$ to transform the right integral:
$$\int e^{2u}\,(2+3u)\,du.$$

Break this into two parts:
$$2\int e^{2u}\,du+3\int u\,e^{2u}\,du.$$

First integral: $$2\int e^{2u}\,du=2\left(\frac{e^{2u}}{2}\right)=e^{2u}.$$

Second integral: use integration by parts, $$A=u,\; dB=e^{2u}du$$.
Then $$B=\frac{e^{2u}}{2}$$ and
$$\int u\,e^{2u}\,du=u\frac{e^{2u}}{2}-\int\frac{e^{2u}}{2}\,du =\frac{u\,e^{2u}}{2}-\frac{e^{2u}}{4}.$$
Multiply by the coefficient $$3$$:
$$3\int u\,e^{2u}\,du=\frac{3u\,e^{2u}}{2}-\frac{3e^{2u}}{4}.$$

Add the two parts:
$$e^{2u}+\left(\frac{3u\,e^{2u}}{2}-\frac{3e^{2u}}{4}\right) =\frac{3u\,e^{2u}}{2}+\frac{e^{2u}}{4}.$$

Thus
$$y\,e^{2\tan x}=\frac{3\tan x\,e^{2\tan x}}{2}+\frac{e^{2\tan x}}{4}+C,$$
where $$C$$ is the constant of integration.

Step 4: Divide by $$e^{2\tan x}$$ to isolate $$y$$:
$$y=\frac{3}{2}\tan x+\frac{1}{4}+C\,e^{-2\tan x}.$$

Step 5: Use the initial condition $$y(0)=\frac{5}{4}$$.
At $$x=0$$, $$\tan 0=0$$ and $$e^{-2\tan 0}=1$$, so
$$\frac{5}{4}=0+\frac{1}{4}+C\;(1)\;\Longrightarrow\;C=1.$$

Hence the particular solution is
$$y=\frac{3}{2}\tan x+\frac{1}{4}+e^{-2\tan x}.$$

Step 6: Evaluate $$y$$ at $$x=\frac{\pi}{4}$$.
$$\tan\!\left(\frac{\pi}{4}\right)=1,\quad e^{-2\tan(\pi/4)}=e^{-2}.$$
Therefore
$$y\!\left(\frac{\pi}{4}\right)=\frac{3}{2}(1)+\frac{1}{4}+e^{-2} =\frac{7}{4}+e^{-2}.$$

Step 7: Compute the required expression:
$$12\left(y\!\left(\frac{\pi}{4}\right)-e^{-2}\right) =12\left(\frac{7}{4}+e^{-2}-e^{-2}\right) =12\cdot\frac{7}{4}=21.$$

Final answer: $$21$$.