Let x = x(y) be the solution of the differential equation $$y = \left(x-y\frac{dx}{dy}\right)\sin \left(\frac{x}{y}\right),y > 0$$ and $$x(1)=\frac{\pi}{2}$$. Then $$\cos (x(2))$$ is equals to :

We need to solve the differential equation $$y = (x - y\frac{dx}{dy})\sin(\frac{x}{y})$$ with $$x(1) = \frac{\pi}{2}$$.

Rearrange

$$y = x\sin(\frac{x}{y}) - y\frac{dx}{dy}\sin(\frac{x}{y})$$

$$y\frac{dx}{dy}\sin(\frac{x}{y}) = x\sin(\frac{x}{y}) - y$$

Substitute $$v = \frac{x}{y}$$, so $$x = vy$$

$$\frac{dx}{dy} = v + y\frac{dv}{dy}$$

Substituting:

$$y(v + y\frac{dv}{dy})\sin v = vy\sin v - y$$

$$vy\sin v + y^2\frac{dv}{dy}\sin v = vy\sin v - y$$

$$y^2\frac{dv}{dy}\sin v = -y$$

$$y\frac{dv}{dy}\sin v = -1$$

$$\sin v \, dv = -\frac{dy}{y}$$

Integrate both sides

$$\int \sin v \, dv = -\int \frac{dy}{y}$$

$$-\cos v = -\ln y + C$$

$$\cos v = \ln y - C$$

$$\cos(\frac{x}{y}) = \ln y + C$$

Apply initial condition $$x(1) = \frac{\pi}{2}$$

$$\cos(\frac{\pi/2}{1}) = \ln 1 + C$$

$$0 = 0 + C$$

$$C = 0$$

So $$\cos(\frac{x}{y}) = \ln y$$.

Find $$\cos(x(2))$$

At y = 2: $$\cos(\frac{x}{2}) = \ln 2$$

$$\frac{x}{2} = \cos^{-1}(\ln 2)$$

We need $$\cos(x(2)) = \cos(2 \cdot \cos^{-1}(\ln 2))$$.

Using the double angle formula: $$\cos(2\theta) = 2\cos^2\theta - 1$$

Let $$\theta = \cos^{-1}(\ln 2)$$, so $$\cos\theta = \ln 2$$.

$$\cos(x(2)) = 2(\ln 2)^2 - 1$$

The correct answer is Option 4: $$2(\log_e 2)^2 - 1$$.