Let $$y=y(x)$$ satisfy $$\frac{1}{y}\frac{dy}{dx}-\frac{2e^x\cos x}{y}=1$$ and $$y(0)=0$$. Then $$y\left(\frac{\pi}{2}\right)=$$

The given differential equation can be rearranged into a standard linear form.

$$\frac{1}{y} \frac{dy}{dx} - \frac{2e^x \cos x}{y} = 1$$

Multiply the entire equation by $$y$$ to remove the denominator.

$$\frac{dy}{dx} - 2e^x \cos x = y$$

Bring the $$y$$ term to the left side to form a first-order linear differential equation.

$$\frac{dy}{dx} - y = 2e^x \cos x$$

This equation is of the standard form $$\frac{dy}{dx} + P(x)y = Q(x)$$, where $$P(x) = -1$$ and $$Q(x) = 2e^x \cos x$$.

Next, calculate the integrating factor (IF).

$$\text{IF} = e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$$

Multiply both sides of the differential equation by the integrating factor.

The left side simplifies to the derivative of the product of $$y$$ and the IF.

$$e^{-x} \frac{dy}{dx} - e^{-x} y = 2e^x e^{-x} \cos x$$

$$\frac{d}{dx} (y e^{-x}) = 2 \cos x$$

Integrate both sides with respect to $$x$$.

$$\int \frac{d}{dx} (y e^{-x}) dx = \int 2 \cos x dx$$

$$y e^{-x} = 2 \sin x + C$$

Use the given initial condition $$y(0) = 0$$ to find the constant of integration $$C$$.

Substitute $$x = 0$$ and $$y = 0$$ into the equation.

$$0 \cdot e^{-0} = 2 \sin(0) + C$$

$$0 = 0 + C$$

$$C = 0$$

Substitute $$C = 0$$ back into the general solution to obtain the particular solution.

$$y e^{-x} = 2 \sin x$$

$$y = 2e^x \sin x$$

Finally, evaluate the function at $$x = \frac{\pi}{2}$$.

$$y\left(\frac{\pi}{2}\right) = 2e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right)$$

Since $$\sin(\frac{\pi}{2}) = 1$$, the expression simplifies to the final answer.

$$y\left(\frac{\pi}{2}\right) = 2e^{\frac{\pi}{2}}$$