1. **State the problem:** Solve the differential equation $$x^2 y'' + 4x y' + 2x = 4 \ln x$$ using the same method (likely variation of parameters or reduction of order). 2. **Rewrite the equation:** Divide through by $$x^2$$ to get a standard form: $$y'' + \frac{4}{x} y' + \frac{2}{x} = \frac{4 \ln x}{x^2}$$ 3. **Identify the homogeneous equation:** $$y'' + \frac{4}{x} y' = 0$$ 4. **Solve the homogeneous equation:** Let $$y' = u$$, then $$u' + \frac{4}{x} u = 0$$ This is a first order linear ODE for $$u$$. 5. **Solve for $$u$$:** The integrating factor is $$\mu = e^{\int \frac{4}{x} dx} = e^{4 \ln x} = x^4$$. Multiply both sides by $$x^4$$: $$x^4 u' + 4 x^3 u = (x^4 u)' = 0$$ Integrate: $$(x^4 u) = C_1 \implies u = y' = \frac{C_1}{x^4}$$ 6. **Integrate to find $$y$$:** $$y = \int y' dx = \int \frac{C_1}{x^4} dx = C_1 \int x^{-4} dx = C_1 \left(-\frac{1}{3 x^3}\right) + C_2 = -\frac{C_1}{3 x^3} + C_2$$ 7. **General solution to homogeneous equation:** $$y_h = C_2 - \frac{C_1}{3 x^3}$$ 8. **Find a particular solution $$y_p$$:** Rewrite original equation as: $$y'' + \frac{4}{x} y' = \frac{4 \ln x}{x^2} - \frac{2}{x}$$ 9. **Use variation of parameters:** The homogeneous solutions are constants and $$x^{-3}$$ (from step 6). Let: $$y_1 = 1, \quad y_2 = x^{-3}$$ 10. **Wronskian:** $$W = y_1 y_2' - y_2 y_1' = 1 \cdot (-3 x^{-4}) - x^{-3} \cdot 0 = -3 x^{-4}$$ 11. **Formulas for variation of parameters:** $$u_1' = - \frac{y_2 g(x)}{W} = - \frac{x^{-3} \left(\frac{4 \ln x}{x^2} - \frac{2}{x}\right)}{-3 x^{-4}} = \frac{x^{-3} \left(\frac{4 \ln x}{x^2} - \frac{2}{x}\right)}{3 x^{-4}}$$ Simplify powers of $$x$$: $$u_1' = \frac{4 \ln x}{3 x} - \frac{2}{3}$$ Similarly, $$u_2' = \frac{y_1 g(x)}{W} = \frac{1 \cdot \left(\frac{4 \ln x}{x^2} - \frac{2}{x}\right)}{-3 x^{-4}} = -\frac{4 \ln x}{3 x^2} + \frac{2}{3 x}$$ 12. **Integrate $$u_1'$$:** $$u_1 = \int \left(\frac{4 \ln x}{3 x} - \frac{2}{3}\right) dx = \frac{4}{3} \int \frac{\ln x}{x} dx - \frac{2}{3} \int dx$$ Recall: $$\int \frac{\ln x}{x} dx = \frac{(\ln x)^2}{2}$$ So, $$u_1 = \frac{4}{3} \cdot \frac{(\ln x)^2}{2} - \frac{2}{3} x + C = \frac{2}{3} (\ln x)^2 - \frac{2}{3} x + C$$ 13. **Integrate $$u_2'$$:** $$u_2 = \int \left(-\frac{4 \ln x}{3 x^2} + \frac{2}{3 x}\right) dx$$ Split integrals: $$u_2 = -\frac{4}{3} \int \frac{\ln x}{x^2} dx + \frac{2}{3} \int \frac{1}{x} dx$$ The second integral: $$\int \frac{1}{x} dx = \ln x$$ For the first integral, use integration by parts: Let $$I = \int \frac{\ln x}{x^2} dx$$ Set $$u = \ln x$$, $$dv = x^{-2} dx$$ Then $$du = \frac{1}{x} dx$$, $$v = -x^{-1}$$ So, $$I = u v - \int v du = -\frac{\ln x}{x} + \int \frac{1}{x} \cdot \frac{1}{x} dx = -\frac{\ln x}{x} + \int x^{-2} dx = -\frac{\ln x}{x} - \frac{1}{x} + C$$ 14. **Substitute back:** $$u_2 = -\frac{4}{3} \left(-\frac{\ln x}{x} - \frac{1}{x}\right) + \frac{2}{3} \ln x + C = \frac{4}{3} \frac{\ln x}{x} + \frac{4}{3 x} + \frac{2}{3} \ln x + C$$ 15. **Write particular solution:** $$y_p = u_1 y_1 + u_2 y_2 = u_1 \cdot 1 + u_2 \cdot x^{-3}$$ Substitute: $$y_p = \left(\frac{2}{3} (\ln x)^2 - \frac{2}{3} x\right) + \left(\frac{4}{3} \frac{\ln x}{x} + \frac{4}{3 x} + \frac{2}{3} \ln x\right) x^{-3}$$ Simplify powers: $$y_p = \frac{2}{3} (\ln x)^2 - \frac{2}{3} x + \frac{4}{3} x^{-4} \ln x + \frac{4}{3} x^{-4} + \frac{2}{3} x^{-3} \ln x$$ 16. **General solution:** $$y = y_h + y_p = C_2 - \frac{C_1}{3 x^3} + \frac{2}{3} (\ln x)^2 - \frac{2}{3} x + \frac{4}{3} x^{-4} \ln x + \frac{4}{3} x^{-4} + \frac{2}{3} x^{-3} \ln x$$ This is the complete solution to the given differential equation.