1. **State the problem:** Solve the differential equation $$y'' + 4y = 5t^2 e^t$$ for the general and particular solutions. 2. **Identify the type of equation:** This is a nonhomogeneous linear second-order differential equation with constant coefficients. 3. **Solve the homogeneous equation:** $$y'' + 4y = 0$$ Characteristic equation: $$r^2 + 4 = 0$$ Solve for $$r$$: $$r^2 = -4$$ $$r = \pm 2i$$ 4. **Write the homogeneous solution:** $$y_h = C_1 \cos(2t) + C_2 \sin(2t)$$ where $$C_1$$ and $$C_2$$ are arbitrary constants. 5. **Find a particular solution $$y_p$$:** Since the right side is $$5t^2 e^t$$, try a particular solution of the form: $$y_p = e^t (A t^2 + B t + C)$$ where $$A$$, $$B$$, and $$C$$ are constants to be determined. 6. **Compute derivatives:** $$y_p = e^t (A t^2 + B t + C)$$ $$y_p' = e^t (A t^2 + B t + C) + e^t (2 A t + B) = e^t (A t^2 + (B + 2 A) t + (C + B))$$ $$y_p'' = e^t (A t^2 + (B + 2 A) t + (C + B)) + e^t (2 A t + (B + 2 A)) = e^t (A t^2 + (B + 4 A) t + (C + 2 B + 2 A))$$ 7. **Substitute into the differential equation:** $$y_p'' + 4 y_p = e^t (A t^2 + (B + 4 A) t + (C + 2 B + 2 A)) + 4 e^t (A t^2 + B t + C) = 5 t^2 e^t$$ 8. **Combine like terms:** $$e^t \left[ (A + 4 A) t^2 + ((B + 4 A) + 4 B) t + (C + 2 B + 2 A + 4 C) \right] = 5 t^2 e^t$$ Simplify coefficients: $$e^t \left[ 5 A t^2 + (B + 4 A + 4 B) t + (C + 2 B + 2 A + 4 C) \right] = 5 t^2 e^t$$ $$e^t \left[ 5 A t^2 + (5 B + 4 A) t + (5 C + 2 B + 2 A) \right] = 5 t^2 e^t$$ 9. **Match coefficients with the right side:** For $$t^2$$ term: $$5 A = 5 \implies A = 1$$ For $$t$$ term: $$5 B + 4 A = 0 \implies 5 B + 4 = 0 \implies B = -\frac{4}{5}$$ For constant term: $$5 C + 2 B + 2 A = 0 \implies 5 C + 2 \left(-\frac{4}{5}\right) + 2 = 0$$ $$5 C - \frac{8}{5} + 2 = 0$$ $$5 C + \frac{2}{5} = 0 \implies 5 C = -\frac{2}{5} \implies C = -\frac{2}{25}$$ 10. **Write the particular solution:** $$y_p = e^t \left(t^2 - \frac{4}{5} t - \frac{2}{25} \right)$$ 11. **Write the general solution:** $$y = y_h + y_p = C_1 \cos(2t) + C_2 \sin(2t) + e^t \left(t^2 - \frac{4}{5} t - \frac{2}{25} \right)$$