STEP-BY-STEP ANSWER:
Step 1: Note that the associated homogeneous equation is y'' − y = 0 with general solution y_c = c₁ e^x + c₂ e^(–x).
Step 2: Since the forcing term is 1 + e^x, propose a particular solution of the form y_p = A + B e^x.
Step 3: Differentiate: y_p' = B e^x and y_p'' = B e^x.
Step 4: Substitute into the differential equation: B e^x − (A + B e^x) = 1 + e^x.
Step 5: Simplify to get –A = 1 and (B − B) e^x = e^x, so A = –1 and the coefficient of e^x yields 0 = e^x, which indicates that we must adjust the guess.
Step 6: Since e^x is already a solution of the homogeneous equation, multiply the e^x part by x. Instead, try y_p = A + B x e^x.
Step 7: Compute derivatives: y_p' = B e^x + Bx e^x; y_p'' = 2B e^x + Bx e^x.
Step 8: Substitute into the ODE: (2B e^x + Bx e^x) − [A + B x e^x] = 1 + e^x, yielding 2B e^x − A = 1 + e^x.
Step 9: Equate coefficients: For the constant term, –A = 1 so A = –1; for the e^x term, 2B = 1 so B = 1/2.
Step 10: Thus, the particular solution is y_p = –1 + (1/2) x e^x, and the general solution is y = c₁ e^x + c₂ e^(–x) – 1 + (1/2) x e^x.
Final Answer: