Key Concepts and Formulas

• Differential of a Quotient: $d\left(\frac{u}{v}\right) = \frac{v \, du - u \, dv}{v^2}$
• Integration: $\int \frac{dy}{y} = \ln|y| + C$, $\int \sin(x) \, dx = -\cos(x) + C$
• Double Angle Formula: $\cos(2\theta) = 2\cos^2(\theta) - 1$

Step-by-Step Solution

Step 1: Rearranging the Differential Equation

The given differential equation is $y=\left(x-y \frac{\mathrm{~d} x}{\mathrm{~d} y}\right) \sin \left(\frac{x}{y}\right)$. Our aim is to rewrite this in terms of differentials $dx$ and $dy$. Multiplying both sides by $dy$, we get: $y \, dy = \left(x - y \frac{dx}{dy}\right) \sin \left(\frac{x}{y}\right) dy$ $y \, dy = \left(x \, dy - y \, dx\right) \sin \left(\frac{x}{y}\right)$ This removes the fractional derivative.

Step 2: Forming the Exact Differential

We now have $y \, dy = (x \, dy - y \, dx) \sin \left(\frac{x}{y}\right)$. We want to relate $(x \, dy - y \, dx)$ to $d\left(\frac{x}{y}\right)$. Recall that $d\left(\frac{x}{y}\right) = \frac{y \, dx - x \, dy}{y^2}$. Multiplying both sides of the equation by $\frac{1}{y^2}$, we have $\frac{y \, dy}{y^2} = \frac{(x \, dy - y \, dx)}{y^2} \sin \left(\frac{x}{y}\right)$ $\frac{dy}{y} = -\frac{(y \, dx - x \, dy)}{y^2} \sin \left(\frac{x}{y}\right)$ $\frac{dy}{y} = -d\left(\frac{x}{y}\right) \sin \left(\frac{x}{y}\right)$ This isolates the differential of the quotient.

Step 3: Integrating the Differential Equation

The equation is now $\frac{dy}{y} = -\sin \left(\frac{x}{y}\right) d\left(\frac{x}{y}\right)$. Let $u = \frac{x}{y}$. Then $du = d\left(\frac{x}{y}\right)$. Substituting, we have $\frac{dy}{y} = -\sin(u) \, du$ Integrating both sides, we get $\int \frac{dy}{y} = -\int \sin(u) \, du$ $\ln |y| = \cos(u) + C$ Since $y>0$, we have $\ln y = \cos(u) + C$. Substituting back $u = \frac{x}{y}$, we have $\ln y = \cos \left(\frac{x}{y}\right) + C$ This yields the general solution.

Step 4: Applying the Initial Condition

We are given $x(1) = \frac{\pi}{2}$. This means when $y=1$, $x = \frac{\pi}{2}$. Substituting these values into the general solution, we get $\ln(1) = \cos \left(\frac{\pi/2}{1}\right) + C$ $0 = \cos \left(\frac{\pi}{2}\right) + C$ $0 = 0 + C$ $C = 0$ Therefore, the particular solution is $\ln y = \cos \left(\frac{x}{y}\right)$

Step 5: Evaluating $\cos(x(2))$

We need to find $\cos(x(2))$. Substituting $y=2$ into the particular solution, we get $\ln 2 = \cos \left(\frac{x(2)}{2}\right)$ Let $x_2 = x(2)$. Then $\ln 2 = \cos \left(\frac{x_2}{2}\right)$ We want to find $\cos(x_2)$. Using the double angle formula, $\cos(2\theta) = 2\cos^2(\theta) - 1$, with $\theta = \frac{x_2}{2}$, we have $\cos(x_2) = 2\cos^2 \left(\frac{x_2}{2}\right) - 1$ Substituting $\cos \left(\frac{x_2}{2}\right) = \ln 2$, we get $\cos(x_2) = 2(\ln 2)^2 - 1$ Therefore, $\cos(x(2)) = 2(\ln 2)^2 - 1$.

Common Mistakes & Tips

• Remember the sign in the differential of a quotient. It's easy to mix up $d\left(\frac{x}{y}\right)$ and $d\left(\frac{y}{x}\right)$.
• Don't forget the constant of integration, $C$, and remember to apply the initial condition to find its value.
• Be careful when applying trigonometric identities. Choose the correct identity and substitute values carefully.

Summary

We solved the given differential equation by recognizing the exact differential form related to the quotient $\frac{x}{y}$. After integrating and applying the initial condition, we found the particular solution. Finally, we used the double-angle formula to evaluate $\cos(x(2))$, obtaining the result $2(\ln 2)^2 - 1$.

Final Answer

The final answer is $\boxed{2\left(\log _e 2\right)^2-1}$, which corresponds to option (D).