1. **Problem Statement:** Find the general solution to the differential equation $$y'' - 2y' + y = \frac{e^t}{t^2 + 1}$$ using the method of variation of parameters. 2. **Step 1: Solve the homogeneous equation** $$y'' - 2y' + y = 0$$. The characteristic equation is $$r^2 - 2r + 1 = 0$$. 3. **Step 2: Find roots of the characteristic equation:** $$r = 1$$ is a repeated root. 4. **Step 3: Write the homogeneous solution:** $$y_h = c_1 e^t + c_2 t e^t$$. 5. **Step 4: Set up variation of parameters:** Let $$y_p = u_1(t) e^t + u_2(t) t e^t$$ where $$u_1$$ and $$u_2$$ are functions to be determined. 6. **Step 5: Compute Wronskian:** $$W = \begin{vmatrix} e^t & t e^t \\ e^t & e^t + t e^t \end{vmatrix} = e^t (e^t + t e^t) - t e^t e^t = e^{2t}$$. 7. **Step 6: Formulas for $$u_1'$$ and $$u_2'$$:** $$u_1' = -\frac{y_2 g(t)}{W} = -\frac{t e^t \cdot \frac{e^t}{t^2 + 1}}{e^{2t}} = -\frac{t}{t^2 + 1}$$ $$u_2' = \frac{y_1 g(t)}{W} = \frac{e^t \cdot \frac{e^t}{t^2 + 1}}{e^{2t}} = \frac{1}{t^2 + 1}$$ 8. **Step 7: Integrate to find $$u_1$$ and $$u_2$$:** $$u_1 = -\int \frac{t}{t^2 + 1} dt = -\frac{1}{2} \ln(1 + t^2) + C$$ $$u_2 = \int \frac{1}{t^2 + 1} dt = \tan^{-1}(t) + C$$ Ignore constants of integration as they are absorbed in homogeneous solution. 9. **Step 8: Write particular solution:** $$y_p = u_1 e^t + u_2 t e^t = -\frac{1}{2} e^t \ln(1 + t^2) + t e^t \tan^{-1}(t)$$ 10. **Step 9: Write general solution:** $$y = y_h + y_p = c_1 e^t + c_2 t e^t - \frac{1}{2} e^t \ln(1 + t^2) + t e^t \tan^{-1}(t)$$ **Answer:** The correct general solution matches the first option: $$y = c_1 e^t + c_2 t e^t - \frac{1}{2} e^t \ln(1 + t^2) + t e^t \tan^{-1}(t)$$ This solution uses variation of parameters and the properties of logarithmic and inverse tangent integrals.