1. **State the problem:** Solve the differential equation $$y'' - 2y' + y = (x+5)^2$$. 2. **Identify the type of equation:** This is a non-homogeneous linear second-order differential equation with constant coefficients. 3. **Solve the homogeneous equation:** $$y'' - 2y' + y = 0$$. The characteristic equation is $$r^2 - 2r + 1 = 0$$. 4. **Solve the characteristic equation:** $$r^2 - 2r + 1 = (r-1)^2 = 0$$ So, $$r = 1$$ is a repeated root. 5. **Write the general solution of the homogeneous equation:** $$y_h = (C_1 + C_2 x) e^{x}$$ where $$C_1$$ and $$C_2$$ are constants. 6. **Find a particular solution $$y_p$$:** Since the right side is a polynomial $$ (x+5)^2 = x^2 + 10x + 25$$, try a polynomial of degree 2: $$y_p = Ax^2 + Bx + C$$. 7. **Compute derivatives:** $$y_p' = 2Ax + B$$ $$y_p'' = 2A$$. 8. **Substitute into the original equation:** $$y_p'' - 2y_p' + y_p = (x+5)^2$$ Substitute: $$2A - 2(2Ax + B) + (Ax^2 + Bx + C) = x^2 + 10x + 25$$ Simplify: $$2A - 4Ax - 2B + Ax^2 + Bx + C = x^2 + 10x + 25$$ Group terms: $$Ax^2 + (-4A + B)x + (2A - 2B + C) = x^2 + 10x + 25$$ 9. **Equate coefficients:** - For $$x^2$$: $$A = 1$$ - For $$x$$: $$-4A + B = 10$$ - For constant term: $$2A - 2B + C = 25$$ 10. **Solve for $$A, B, C$$:** - From $$A=1$$ - Substitute into $$-4(1) + B = 10 \\ B = 10 + 4 = 14$$ - Substitute $$A$$ and $$B$$ into constant term: $$2(1) - 2(14) + C = 25 \\ 2 - 28 + C = 25 \\ C = 25 + 26 = 51$$ 11. **Write the particular solution:** $$y_p = x^2 + 14x + 51$$ 12. **Write the general solution:** $$y = y_h + y_p = (C_1 + C_2 x) e^{x} + x^2 + 14x + 51$$ This is the complete solution to the differential equation.