1. Given differential equation

$e^y\frac{dy}{dx}-2e^y\sin x+\sin x\cos^2 x=0$ with condition $y\left(\frac{\pi}{2}\right)=0.$

We need to find $y(0)$ in the form $y(0)=\ln(\alpha+\beta e^{-2}),$ and then compute $4(\alpha+\beta)$.

2. Rewrite the differential equation

Notice that $\frac{d}{dx}(e^y)=e^y\frac{dy}{dx}.$ So the equation becomes $\frac{d}{dx}(e^y)-2e^y\sin x+\sin x\cos^2 x=0.$ Let $u=e^y.$ Then the equation is $\frac{du}{dx}-2u\sin x=-\sin x\cos^2 x.$

This is a linear differential equation in $u$.

3. Solve the linear differential equation

We have $\frac{du}{dx}+Pu=Q,$ where $P=-2\sin x, \qquad Q=-\sin x\cos^2 x.$

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

Multiplying throughout by the integrating factor: $e^{2\cos x}\frac{du}{dx}-2\sin x\,e^{2\cos x}u=-\sin x\cos^2 x\,e^{2\cos x}.$

So, $\frac{d}{dx}\left(ue^{2\cos x}\right)=-\sin x\cos^2 x\,e^{2\cos x}.$

4. Integrate

$ue^{2\cos x}=\int -\sin x\cos^2 x\,e^{2\cos x}\,dx+C.$

Put $t=\cos x \implies dt=-\sin x\,dx.$ Then $\int -\sin x\cos^2 x\,e^{2\cos x}\,dx=\int t^2 e^{2t}\,dt.$

Now compute $I=\int t^2 e^{2t}\,dt.$ Using standard integration, $\int t^2 e^{2t}dt=e^{2t}\left(\frac{t^2}{2}-\frac{t}{2}+\frac14\right)+C.$

Hence, $ue^{2\cos x}=e^{2\cos x}\left(\frac{\cos^2 x}{2}-\frac{\cos x}{2}+\frac14\right)+C.$

Therefore, $u=\frac{\cos^2 x}{2}-\frac{\cos x}{2}+\frac14+Ce^{-2\cos x}.$

Since $u=e^y$, $e^y=\frac{\cos^2 x}{2}-\frac{\cos x}{2}+\frac14+Ce^{-2\cos x}.$

5. Use the initial condition

Given $y\left(\frac{\pi}{2}\right)=0,$ so $e^{y(\pi/2)}=1.$ At $x=\frac{\pi}{2}$, we have $\cos\frac{\pi}{2}=0$. Thus $1=\frac{0}{2}-\frac{0}{2}+\frac14+C e^0=\frac14+C.$ So, $C=\frac34.$

Hence, $e^y=\frac{\cos^2 x}{2}-\frac{\cos x}{2}+\frac14+\frac34 e^{-2\cos x}.$

6. Find $y(0)$

At $x=0$, $\cos 0=1$. So $e^{y(0)}=\frac{1}{2}-\frac{1}{2}+\frac14+\frac34 e^{-2}.$ Thus $e^{y(0)}=\frac14+\frac34 e^{-2}.$ Therefore, $y(0)=\ln\left(\frac14+\frac34 e^{-2}\right).$

Comparing with $y(0)=\ln(\alpha+\beta e^{-2}),$ we get $\alpha=\frac14, \qquad \beta=\frac34.$ So $\alpha+\beta=1.$

Hence, $4(\alpha+\beta)=4.$

7. Final answer

$\boxed{4}$

The derived answer matches the stored correct answer.