1. **Problem A1:** Solve the differential equation $$y'' - 5y' + 6y = 0.$$ 2. **Formula and rules:** This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is $$r^2 - 5r + 6 = 0.$$ 3. **Solve characteristic equation:** Factor or use quadratic formula: $$r^2 - 5r + 6 = (r - 2)(r - 3) = 0.$$ So, $$r = 2$$ or $$r = 3.$$ 4. **General solution:** Since roots are real and distinct, the general solution is $$y = C_1 e^{2t} + C_2 e^{3t}.$$ --- 1. **Problem A2:** Find the general solution of $$y'' + 4y' + 13y = 0.$$ 2. **Characteristic equation:** $$r^2 + 4r + 13 = 0.$$ 3. **Solve characteristic equation:** Use quadratic formula: $$r = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 13}}{2} = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i.$$ 4. **General solution:** For complex roots $$\alpha \pm \beta i$$, solution is $$y = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) = e^{-2t}(C_1 \cos 3t + C_2 \sin 3t).$$ --- 1. **Problem A3:** Solve $$y'' - 2y' - 8y = e^{3t}.$$ 2. **Solve homogeneous equation:** Characteristic equation: $$r^2 - 2r - 8 = 0.$$ Factor: $$(r - 4)(r + 2) = 0 \Rightarrow r = 4, -2.$$ Homogeneous solution: $$y_h = C_1 e^{4t} + C_2 e^{-2t}.$$ 3. **Find particular solution:** Since RHS is $$e^{3t}$$, try $$y_p = Ae^{3t}.$$ 4. **Compute derivatives:** $$y_p' = 3Ae^{3t}, \quad y_p'' = 9Ae^{3t}.$$ 5. **Substitute into ODE:** $$9Ae^{3t} - 2(3Ae^{3t}) - 8(Ae^{3t}) = e^{3t}.$$ Simplify: $$9A - 6A - 8A = 1 \Rightarrow (9 - 6 - 8)A = 1 \Rightarrow -5A = 1 \Rightarrow A = -\frac{1}{5}.$$ 6. **Particular solution:** $$y_p = -\frac{1}{5} e^{3t}.$$ 7. **General solution:** $$y = y_h + y_p = C_1 e^{4t} + C_2 e^{-2t} - \frac{1}{5} e^{3t}.$$ --- 1. **Problem A4:** Solve $$y'' + 9y = 7 \sin(3t).$$ Find complementary function and particular integral. 2. **Complementary function (CF):** Characteristic equation: $$r^2 + 9 = 0 \Rightarrow r = \pm 3i.$$ CF: $$y_c = C_1 \cos 3t + C_2 \sin 3t.$$ 3. **Particular integral (PI):** RHS is $$7 \sin 3t$$ which is a solution to the homogeneous equation, so try $$y_p = t(A \cos 3t + B \sin 3t).$$ 4. **Derivatives:** $$y_p' = A \cos 3t + B \sin 3t + t(-3A \sin 3t + 3B \cos 3t),$$ $$y_p'' = -3A \sin 3t + 3B \cos 3t + (-3A \sin 3t + 3B \cos 3t) + t(-9A \cos 3t - 9B \sin 3t).$$ Simplify: $$y_p'' = -6A \sin 3t + 6B \cos 3t + t(-9A \cos 3t - 9B \sin 3t).$$ 5. **Substitute into ODE:** $$y_p'' + 9 y_p = (-6A \sin 3t + 6B \cos 3t + t(-9A \cos 3t - 9B \sin 3t)) + 9 t (A \cos 3t + B \sin 3t) = -6A \sin 3t + 6B \cos 3t.$$ 6. **Set equal to RHS:** $$-6A \sin 3t + 6B \cos 3t = 7 \sin 3t.$$ Equate coefficients: $$-6A = 7 \Rightarrow A = -\frac{7}{6}, \quad 6B = 0 \Rightarrow B = 0.$$ 7. **Particular integral:** $$y_p = t \left(-\frac{7}{6} \cos 3t \right) = -\frac{7}{6} t \cos 3t.$$ 8. **General solution:** $$y = y_c + y_p = C_1 \cos 3t + C_2 \sin 3t - \frac{7}{6} t \cos 3t.$$ --- 1. **Problem A5:** Solve $$y'' - 4y' + 13y = e^{2t} \cos 3t,$$ express answer in simplified real form. 2. **Homogeneous equation:** Characteristic equation: $$r^2 - 4r + 13 = 0.$$ Discriminant: $$16 - 52 = -36.$$ Roots: $$r = \frac{4 \pm 6i}{2} = 2 \pm 3i.$$ 3. **Homogeneous solution:** $$y_h = e^{2t} (C_1 \cos 3t + C_2 \sin 3t).$$ 4. **Particular solution form:** RHS is $$e^{2t} \cos 3t,$$ so try $$y_p = e^{2t} (A t \cos 3t + B t \sin 3t)$$ because $$e^{2t} \cos 3t$$ is part of homogeneous solution, multiply by $$t$$. 5. **Derivatives:** Calculate $$y_p'$$ and $$y_p''$$ carefully using product and chain rules (omitted here for brevity). 6. **Substitute into ODE and equate coefficients:** After substitution and simplification, solve for $$A$$ and $$B$$: $$A = 0, \quad B = \frac{1}{6}.$$ 7. **Particular solution:** $$y_p = e^{2t} \cdot t \cdot \frac{1}{6} \sin 3t = \frac{1}{6} t e^{2t} \sin 3t.$$ 8. **General solution:** $$y = y_h + y_p = e^{2t} (C_1 \cos 3t + C_2 \sin 3t) + \frac{1}{6} t e^{2t} \sin 3t.$$