1. **State the problem:** Solve the differential equation $$xy''' + 2y'' = 0$$ for the function $y(x)$. 2. **Rewrite the equation:** The equation is a linear differential equation involving derivatives of $y$. We want to find $y$ such that $$xy''' + 2y'' = 0.$$ 3. **Introduce substitution:** Let $z = y''$. Then $z' = y'''$. Substitute into the equation: $$x z' + 2z = 0.$$ 4. **Solve the first order ODE for $z$:** This is a linear first order ODE in $z$: $$x \frac{dz}{dx} + 2z = 0.$$ Rewrite as: $$\frac{dz}{dx} = -\frac{2}{x} z.$$ 5. **Separate variables:** $$\frac{1}{z} dz = -\frac{2}{x} dx.$$ 6. **Integrate both sides:** $$\int \frac{1}{z} dz = -2 \int \frac{1}{x} dx,$$ which gives $$\ln |z| = -2 \ln |x| + C_1 = \ln |x|^{-2} + C_1.$$ 7. **Exponentiate:** $$|z| = e^{C_1} x^{-2} = C x^{-2},$$ where $C = e^{C_1}$ is an arbitrary constant. 8. **Recall $z = y''$:** $$y'' = C x^{-2}.$$ 9. **Integrate twice to find $y$:** First integration: $$y' = \int y'' dx = \int C x^{-2} dx = C \int x^{-2} dx = C (-x^{-1}) + C_2 = -\frac{C}{x} + C_2.$$ Second integration: $$y = \int y' dx = \int \left(-\frac{C}{x} + C_2\right) dx = -C \int \frac{1}{x} dx + C_2 x + C_3 = -C \ln |x| + C_2 x + C_3.$$ 10. **Final solution:** $$\boxed{y(x) = -C \ln |x| + C_2 x + C_3}$$ where $C, C_2, C_3$ are arbitrary constants determined by initial/boundary conditions. This completes the solution.