1. **Problem Statement:** Find the general solution to the differential equation $$y''' - 3y'' + 2y' = \frac{e^x}{1 + e^{-x}}$$ using the method of variation of parameters. 2. **Step 1: Solve the homogeneous equation** $$y''' - 3y'' + 2y' = 0$$ - Characteristic equation: $$r^3 - 3r^2 + 2r = 0$$ - Factor: $$r(r^2 - 3r + 2) = 0$$ - Roots: $$r=0, r=1, r=2$$ - Homogeneous solution: $$y_h = c_1 + c_2 e^x + c_3 e^{2x}$$ 3. **Step 2: Set up variation of parameters** - Let $$y = u_1(x) \cdot 1 + u_2(x) e^x + u_3(x) e^{2x}$$ - Compute derivatives and apply the method to find $$u_1', u_2', u_3'$$ using the Wronskian and the right side $$\frac{e^x}{1 + e^{-x}}$$. 4. **Step 3: Solve for particular solution** - After applying variation of parameters and integrating, the particular solution is found to be: $$y_p = \frac{1}{2}(e^x + 1) - \frac{1}{2} \ln(e^x + 1) - e^x \ln(e^x + 1) - \frac{1}{2} e^{2x} \ln(1 + e^{-x})$$ 5. **Step 4: Write the general solution** - Combine homogeneous and particular solutions: $$y = c_1 + c_2 e^x + c_3 e^{2x} + \frac{1}{2}(e^x + 1) - \frac{1}{2} \ln(e^x + 1) - e^x \ln(e^x + 1) - \frac{1}{2} e^{2x} \ln(1 + e^{-x})$$ 6. **Step 5: Match with given options** - This matches option (d). **Final answer:** $$y = c_1 + c_2 e^x + c_3 e^{2x} + \frac{1}{2}(e^x + 1) - \frac{1}{2} \ln(e^x + 1) - e^x \ln(e^x + 1) - \frac{1}{2} e^{2x} \ln(1 + e^{-x})$$