1. **State the problem:** Solve the differential equation $$x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + y = \log x \cdot \sin(\log x).$$ 2. **Rewrite using substitution:** Let $$t = \log x$$, so $$x = e^t$$. Using the chain rule: $$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt} = e^{-t} \frac{dy}{dt}$$ and $$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( e^{-t} \frac{dy}{dt} \right) = \frac{d}{dt} \left( e^{-t} \frac{dy}{dt} \right) \cdot \frac{dt}{dx} = \frac{1}{x} \frac{d}{dt} \left( e^{-t} \frac{dy}{dt} \right) = e^{-t} \frac{d}{dt} \left( e^{-t} \frac{dy}{dt} \right).$$ Expanding the last derivative: $$\frac{d}{dt} \left( e^{-t} \frac{dy}{dt} \right) = -e^{-t} \frac{dy}{dt} + e^{-t} \frac{d^2 y}{dt^2} = e^{-t} \left( \frac{d^2 y}{dt^2} - \frac{dy}{dt} \right).$$ Therefore, $$\frac{d^2 y}{dx^2} = e^{-t} \cdot e^{-t} \left( \frac{d^2 y}{dt^2} - \frac{dy}{dt} \right) = e^{-2t} \left( \frac{d^2 y}{dt^2} - \frac{dy}{dt} \right).$$ 3. **Substitute into original ODE:** $$x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + y = e^{2t} \cdot e^{-2t} \left( \frac{d^2 y}{dt^2} - \frac{dy}{dt} \right) + e^t \cdot e^{-t} \frac{dy}{dt} + y = \frac{d^2 y}{dt^2} - \frac{dy}{dt} + \frac{dy}{dt} + y = \frac{d^2 y}{dt^2} + y.$$ So the equation becomes: $$\frac{d^2 y}{dt^2} + y = t \sin t,$$ where we used $\log x = t$ and $\sin(\log x) = \sin t$. 4. **Solve the transformed ODE:** The homogeneous equation is $$\frac{d^2 y}{dt^2} + y = 0,$$ whose general solution is $$y_h = C_1 \cos t + C_2 \sin t.$$ 5. **Find particular solution using variation of parameters:** Let $y_p = u(t) \cos t + v(t) \sin t$. Then, $$u' \cos t + v' \sin t = 0,$$ $$-u' \sin t + v' \cos t = t \sin t.$$ Solving the system for $u'$ and $v'$: From the system: $$\begin{cases} u' \cos t + v' \sin t = 0 \\ -u' \sin t + v' \cos t = t \sin t \end{cases}$$ The determinant is: $$\cos^2 t + \sin^2 t = 1.$$ Multiply the first equation by $\cos t$ and the second by $\sin t$ and add: $$u' \cos^2 t + v' \sin t \cos t - u' \sin^2 t + v' \sin t \cos t = 0 + t \sin^2 t.$$ Alternatively, solve directly: From first: $$u' \cos t = -v' \sin t \implies u' = -v' \frac{\sin t}{\cos t}.$$ Plug into second: $$-(-v' \frac{\sin t}{\cos t}) \sin t + v' \cos t = v' \frac{\sin^2 t}{\cos t} + v' \cos t = v' \left( \frac{\sin^2 t + \cos^2 t}{\cos t} \right) = \frac{v'}{\cos t} = t \sin t.$$ Therefore, $$v' = t \sin t \cos t,$$ $$u' = -t \sin^2 t.$$ Integrate: $$u = -\int t \sin^2 t dt,$$ $$v = \int t \sin t \cos t dt.$$ 6. **Evaluate integrals:** Use identity $$\sin^2 t = \frac{1 - \cos 2t}{2}$$: $$u = -\int t \frac{1 - \cos 2t}{2} dt = -\frac{1}{2} \int t dt + \frac{1}{2} \int t \cos 2t dt = -\frac{t^2}{4} + \frac{1}{2} \int t \cos 2t dt.$$ By integration by parts for $$\int t \cos 2t dt$$, Let $$I = \int t \cos 2t dt$$: Set $$u = t, dv = \cos 2t dt,$$ $$du = dt, v = \frac{1}{2} \sin 2t,$$ Then, $$I = t \cdot \frac{1}{2} \sin 2t - \int \frac{1}{2} \sin 2t dt = \frac{t}{2} \sin 2t - \left(-\frac{1}{4} \cos 2t \right) + C = \frac{t}{2} \sin 2t + \frac{1}{4} \cos 2t + C.$$ So $$u = -\frac{t^2}{4} + \frac{1}{2} \left( \frac{t}{2} \sin 2t + \frac{1}{4} \cos 2t \right) + C = -\frac{t^2}{4} + \frac{t}{4} \sin 2t + \frac{1}{8} \cos 2t + C.$$ Similarly, for $$v = \int t \sin t \cos t dt,$$ use $$\sin t \cos t = \frac{1}{2} \sin 2t$$: $$v = \frac{1}{2} \int t \sin 2t dt.$$ Let $$J = \int t \sin 2t dt$$: By parts: $$u = t, dv = \sin 2t dt,$$ $$du = dt, v = -\frac{1}{2} \cos 2t,$$ $$J = -\frac{t}{2} \cos 2t + \int \frac{1}{2} \cos 2t dt = -\frac{t}{2} \cos 2t + \frac{1}{4} \sin 2t + C.$$ Hence, $$v = \frac{1}{2} J = -\frac{t}{4} \cos 2t + \frac{1}{8} \sin 2t + C.$$ 7. **Write particular solution:** Ignoring constants of integration since they merge with homogeneous solution, $$y_p = u \cos t + v \sin t = \left(-\frac{t^2}{4} + \frac{t}{4} \sin 2t + \frac{1}{8} \cos 2t \right) \cos t + \left(-\frac{t}{4} \cos 2t + \frac{1}{8} \sin 2t \right) \sin t.$$ 8. **Simplify the particular solution:** Use trigonometric identities: $$\sin 2t \cos t = \frac{1}{2}(\sin 3t + \sin t),$$ $$\cos 2t \cos t = \frac{1}{2}(\cos 3t + \cos t),$$ $$\sin 2t \sin t = \frac{1}{2}(\cos t - \cos 3t),$$ $$\cos 2t \sin t = \frac{1}{2}(\sin 3t - \sin t).$$ Substitute and simplify to express $y_p$ entirely in sums of $t^2$, $t \sin 3t$, $t \sin t$, and constant multiples of $\, \cos 3t$, $\cos t$, $\sin 3t$, and $\sin t$ terms if desired. 9. **Final solution in terms of original variable:** Recall $$t = \log x$$: $$\boxed{y = C_1 \cos(\log x) + C_2 \sin(\log x) + y_p(\log x)}.$$