If $$y(x)$$ is the solution of the differential equation $$\frac{dy}{dx} + \left(\frac{2x+1}{x}\right)y = e^{-2x}$$, $$x \gt 0$$, where $$y(1) = \frac{1}{2}e^{-2}$$, then:

We are given the linear first-order differential equation

$$\frac{dy}{dx}+\left(\frac{2x+1}{x}\right)y=e^{-2x},\qquad x\gt 0,$$

together with the initial condition

$$y(1)=\frac12\,e^{-2}.$$

The standard form for a linear ODE is $$\dfrac{dy}{dx}+P(x)\,y=Q(x).$$ Here we recognise

$$P(x)=\frac{2x+1}{x}=2+\frac1x,\qquad Q(x)=e^{-2x}.$$

For such an equation we first compute the integrating factor (I.F.). The formula for the integrating factor is

$$\text{I.F.}=e^{\displaystyle\int P(x)\,dx}.$$

We therefore integrate $$P(x)$$:

$$\int P(x)\,dx=\int\!\left(2+\frac1x\right)dx=\int 2\,dx+\int \frac1x\,dx=2x+\ln|x|.$$

Since $$x\gt 0$$ we can drop the absolute value and write

$$\text{I.F.}=e^{2x+\ln x}=e^{2x}\,x.$$

Multiplying every term of the differential equation by this integrating factor we get

$$x\,e^{2x}\,\frac{dy}{dx}+\bigl(2+\frac1x\bigr)\,x\,e^{2x}\,y=x\,e^{2x}\,e^{-2x}=x.$$

Notice that the left-hand side is now the derivative of the product of the integrating factor and $$y$$. Indeed, by construction,

$$\frac{d}{dx}\bigl(y\,x\,e^{2x}\bigr)=x.$$

We integrate both sides with respect to $$x$$:

$$\int\frac{d}{dx}\bigl(y\,x\,e^{2x}\bigr)\,dx=\int x\,dx.$$

This yields

$$y\,x\,e^{2x}=\frac{x^{2}}{2}+C,$$

where $$C$$ is the constant of integration. Solving for $$y$$ gives

$$y(x)=\frac{\dfrac{x^{2}}{2}+C}{x\,e^{2x}}=\left(\frac{x}{2}+\frac{C}{x}\right)e^{-2x}.$$

Now we apply the initial condition $$y(1)=\dfrac12\,e^{-2}$$. Substituting $$x=1$$ and $$y(1)$$ we have

$$\left(\frac{1}{2}+\frac{C}{1}\right)e^{-2}=\frac12\,e^{-2}.$$

Cancelling the common factor $$e^{-2}$$ on both sides we obtain

$$\frac12+C=\frac12 \;\Longrightarrow\; C=0.$$

Hence the particular solution is

$$y(x)=\left(\frac{x}{2}\right)e^{-2x}=\frac{x}{2}\,e^{-2x}.$$

We now analyse the four statements one by one.

Statement A: Evaluate $$y(\ln 2)$$.

Substituting $$x=\ln 2$$ we find

$$y(\ln 2)=\frac{\ln 2}{2}\,e^{-2\ln 2}=\frac{\ln 2}{2}\,\bigl(e^{\ln 2}\bigr)^{-2}=\frac{\ln 2}{2}\,\frac{1}{2^{2}}=\frac{\ln 2}{8}.$$

Since $$\dfrac{\ln 2}{8}\neq\ln 4$$, Statement A is false.

Statement B: The same computation above gave $$y(\ln 2)=\dfrac{\ln 2}{8}$$, which is not $$\dfrac{\ln 2}{4}$$. Hence Statement B is also false.

Statement C: To check whether $$y(x)$$ is decreasing on $$\left(\tfrac12,1\right)$$ we look at the derivative.

Differentiate $$y(x)=\dfrac{x}{2}e^{-2x}$$:

We use the product rule $$\dfrac{d}{dx}(uv)=u'v+uv'$$. Here $$u=\dfrac{x}{2}$$ so $$u'=\dfrac12$$, and $$v=e^{-2x}$$ so $$v'=-2e^{-2x}$$. Therefore

$$\frac{dy}{dx}=\frac12\,e^{-2x}+\frac{x}{2}\,(-2)\,e^{-2x}=e^{-2x}\left(\frac12-\!x\right)=\frac{e^{-2x}}{2}\,(1-2x).$$

The exponential $$e^{-2x}$$ is always positive, so the sign of $$\dfrac{dy}{dx}$$ is governed solely by $$1-2x$$. We have

$$1-2x\lt 0\quad\Longleftrightarrow\quad x\gt \frac12.$$

Thus $$y'(x)\lt 0$$ whenever $$x\gt \tfrac12$$. Consequently $$y(x)$$ is indeed decreasing throughout the entire interval $$\left(\tfrac12,1\right)$$. Statement C is true.

Statement D: On the interval $$(0,1)$$ we must split the sign test: for $$0\lt x\lt \tfrac12$$ we have $$1-2x\gt 0$$, so $$y'(x)\gt 0$$; hence $$y$$ is increasing there. Since it is not decreasing on the whole of $$(0,1)$$, Statement D is false.

Only Statement C is correct.

Hence, the correct answer is Option C.