1. Solve $y'' - y' - 6y = 0$. Step 1: Write the characteristic equation: $$r^2 - r - 6 = 0$$. Step 2: Factor the quadratic: $$(r - 3)(r + 2) = 0$$. Step 3: Roots are $r = 3$ and $r = -2$. Step 4: General solution: $$y = C_1 e^{3x} + C_2 e^{-2x}$$. 2. Solve $y'' + 9y = 0$. Step 1: Characteristic equation: $$r^2 + 9 = 0$$. Step 2: Solve for $r$: $$r^2 = -9 \\ r = \\pm 3i$$. Step 3: Complex roots mean solution in sines and cosines: $$y = C_1 \cos 3x + C_2 \sin 3x$$. 3. Solve $y'' - 8y' + 15y = 30x + 3$. Step 1: Solve homogeneous equation $y'' - 8y' + 15y = 0$. Characteristic equation: $$r^2 - 8r + 15 = 0$$. Step 2: Factor or use quadratic formula: $$(r - 3)(r - 5) = 0 \Rightarrow r = 3, 5$$. Step 3: Homogeneous solution: $$y_h = C_1 e^{3x} + C_2 e^{5x}$$. Step 4: Find particular solution $y_p$ using undetermined coefficients. Since RHS is a polynomial of degree 1, try: $$y_p = Ax + B$$. Step 5: Compute derivatives: $$y_p' = A, \quad y_p'' = 0$$. Step 6: Substitute into original equation: $$0 - 8A + 15(Ax + B) = 30x + 3$$. Simplify: $$15Ax + 15B - 8A = 30x +3$$. Step 7: Equate coefficients: For $x$: $$15A = 30 \Rightarrow A = 2$$. Constant term: $$15B - 8A = 3 \Rightarrow 15B - 16 = 3 \Rightarrow 15B = 19 \Rightarrow B = \frac{19}{15}$$. Step 8: Particular solution: $$y_p = 2x + \frac{19}{15}$$. Step 9: General solution: $$y = y_h + y_p = C_1 e^{3x} + C_2 e^{5x} + 2x + \frac{19}{15}$$.