Let $$y = y(x)$$ be solution of the following differential equation
$$e^y \frac{dy}{dx} - 2e^y \sin x + \sin x \cos^2 x = 0$$, $$y\left(\frac{\pi}{2}\right) = 0$$.
If $$y(0) = \log_e \alpha + \beta e^{-2}$$, then $$4(\alpha + \beta)$$ is equal to ___.

We are given the differential equation $$e^y \frac{dy}{dx} - 2e^y \sin x + \sin x \cos^2 x = 0$$ with the initial condition $$y\left(\frac{\pi}{2}\right) = 0$$.

Let $$t = e^y$$. Then $$\frac{dt}{dx} = e^y \frac{dy}{dx}$$. Substituting into the equation:

$$\frac{dt}{dx} - 2t\sin x + \sin x \cos^2 x = 0$$

Rearranging: $$\frac{dt}{dx} - 2\sin x \cdot t = -\sin x \cos^2 x$$ $$-(1)$$

This is a first-order linear ODE of the form $$\frac{dt}{dx} + P(x)t = Q(x)$$, where $$P(x) = -2\sin x$$ and $$Q(x) = -\sin x \cos^2 x$$.

The integrating factor is $$\text{IF} = e^{\int P(x)\,dx} = e^{\int -2\sin x\,dx} = e^{2\cos x}$$.

Multiplying both sides of $$(1)$$ by the integrating factor:

$$\frac{d}{dx}\left[t \cdot e^{2\cos x}\right] = -\sin x \cos^2 x \cdot e^{2\cos x}$$

Integrating both sides: $$t \cdot e^{2\cos x} = \int -\sin x \cos^2 x \cdot e^{2\cos x}\,dx + C$$

To evaluate the integral, we substitute $$u = \cos x$$, so $$du = -\sin x\,dx$$:

$$\int -\sin x \cos^2 x \cdot e^{2\cos x}\,dx = \int u^2 e^{2u}\,du$$

We evaluate $$\int u^2 e^{2u}\,du$$ using integration by parts twice.

First application: let $$f = u^2$$, $$dg = e^{2u}\,du$$, so $$df = 2u\,du$$, $$g = \frac{e^{2u}}{2}$$.

$$\int u^2 e^{2u}\,du = \frac{u^2 e^{2u}}{2} - \int u \cdot e^{2u}\,du$$

Second application: let $$f = u$$, $$dg = e^{2u}\,du$$, so $$df = du$$, $$g = \frac{e^{2u}}{2}$$.

$$\int u \cdot e^{2u}\,du = \frac{u \cdot e^{2u}}{2} - \int \frac{e^{2u}}{2}\,du = \frac{u \cdot e^{2u}}{2} - \frac{e^{2u}}{4}$$

Combining: $$\int u^2 e^{2u}\,du = \frac{u^2 e^{2u}}{2} - \frac{u \cdot e^{2u}}{2} + \frac{e^{2u}}{4} = \frac{e^{2u}(2u^2 - 2u + 1)}{4}$$

Substituting back $$u = \cos x$$:

$$t \cdot e^{2\cos x} = \frac{e^{2\cos x}(2\cos^2 x - 2\cos x + 1)}{4} + C$$ $$-(2)$$

Applying the initial condition $$y\left(\frac{\pi}{2}\right) = 0$$: at $$x = \frac{\pi}{2}$$, $$t = e^0 = 1$$ and $$\cos\frac{\pi}{2} = 0$$.

$$1 \cdot e^0 = \frac{e^0(0 - 0 + 1)}{4} + C$$

$$1 = \frac{1}{4} + C \implies C = \frac{3}{4}$$

Now, to find $$y(0)$$: at $$x = 0$$, $$\cos 0 = 1$$. Substituting into $$(2)$$:

$$t \cdot e^{2} = \frac{e^{2}(2 - 2 + 1)}{4} + \frac{3}{4} = \frac{e^{2}}{4} + \frac{3}{4}$$

$$t = \frac{1}{4} + \frac{3}{4}e^{-2}$$

Since $$t = e^y$$, we have $$e^{y(0)} = \frac{1}{4} + \frac{3}{4}e^{-2}$$.

Therefore $$y(0) = \log_e\left(\frac{1}{4} + \frac{3}{4}e^{-2}\right)$$.

Comparing with $$y(0) = \log_e(\alpha + \beta e^{-2})$$, we get $$\alpha = \frac{1}{4}$$ and $$\beta = \frac{3}{4}$$.

So $$4(\alpha + \beta) = 4\left(\frac{1}{4} + \frac{3}{4}\right) = 4 \times 1 = 4$$.