1. **Stating the problem:** We are given the nonhomogeneous differential equation: $$\frac{d^2 y}{d x^2} + 4 \frac{d y}{d x} - 12 y = 6 e^{3 x}$$ We need to find: (a) The homogeneous solution $y_h$ (b) The particular solution $y_k$ (c) The general solution (d) The final solution given initial conditions $y(0) = N - 1$ and $y'(0) = N + 3$ 2. **Solving the homogeneous equation:** The homogeneous equation is: $$\frac{d^2 y}{d x^2} + 4 \frac{d y}{d x} - 12 y = 0$$ The characteristic equation is: $$r^2 + 4r - 12 = 0$$ Using the quadratic formula: $$r = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-12)}}{2 \times 1} = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm \sqrt{64}}{2}$$ $$r = \frac{-4 \pm 8}{2}$$ So the roots are: $$r_1 = \frac{-4 + 8}{2} = 2, \quad r_2 = \frac{-4 - 8}{2} = -6$$ Therefore, the homogeneous solution is: $$y_h = C_1 e^{2x} + C_2 e^{-6x}$$ 3. **Finding the particular solution $y_k$:** Since the right side is $6 e^{3x}$, try a particular solution of the form: $$y_k = A e^{3x}$$ Calculate derivatives: $$y_k' = 3 A e^{3x}, \quad y_k'' = 9 A e^{3x}$$ Substitute into the differential equation: $$9 A e^{3x} + 4 (3 A e^{3x}) - 12 (A e^{3x}) = 6 e^{3x}$$ Simplify: $$9 A e^{3x} + 12 A e^{3x} - 12 A e^{3x} = 6 e^{3x}$$ $$9 A e^{3x} = 6 e^{3x}$$ Divide both sides by $e^{3x}$: $$9 A = 6$$ $$A = \frac{6}{9} = \frac{2}{3}$$ So the particular solution is: $$y_k = \frac{2}{3} e^{3x}$$ 4. **General solution:** $$y = y_h + y_k = C_1 e^{2x} + C_2 e^{-6x} + \frac{2}{3} e^{3x}$$ 5. **Applying initial conditions:** Given: $$y(0) = N - 1, \quad y'(0) = N + 3$$ Calculate $y(0)$: $$y(0) = C_1 e^{0} + C_2 e^{0} + \frac{2}{3} e^{0} = C_1 + C_2 + \frac{2}{3} = N - 1$$ Calculate $y'(x)$: $$y' = 2 C_1 e^{2x} - 6 C_2 e^{-6x} + 3 \times \frac{2}{3} e^{3x} = 2 C_1 e^{2x} - 6 C_2 e^{-6x} + 2 e^{3x}$$ Evaluate at $x=0$: $$y'(0) = 2 C_1 - 6 C_2 + 2 = N + 3$$ 6. **Solve the system:** From $y(0)$: $$C_1 + C_2 = N - 1 - \frac{2}{3} = N - \frac{5}{3}$$ From $y'(0)$: $$2 C_1 - 6 C_2 = N + 3 - 2 = N + 1$$ Multiply the first equation by 2: $$2 C_1 + 2 C_2 = 2 N - \frac{10}{3}$$ Subtract the second equation: $$ (2 C_1 + 2 C_2) - (2 C_1 - 6 C_2) = 2 N - \frac{10}{3} - (N + 1)$$ $$2 C_1 + 2 C_2 - 2 C_1 + 6 C_2 = 2 N - \frac{10}{3} - N - 1$$ $$8 C_2 = N - \frac{13}{3}$$ $$C_2 = \frac{N}{8} - \frac{13}{24}$$ Substitute back to find $C_1$: $$C_1 = N - \frac{5}{3} - C_2 = N - \frac{5}{3} - \left( \frac{N}{8} - \frac{13}{24} \right) = N - \frac{5}{3} - \frac{N}{8} + \frac{13}{24}$$ Find common denominators: $$N - \frac{N}{8} = \frac{8N}{8} - \frac{N}{8} = \frac{7N}{8}$$ $$- \frac{5}{3} + \frac{13}{24} = - \frac{40}{24} + \frac{13}{24} = - \frac{27}{24} = - \frac{9}{8}$$ So: $$C_1 = \frac{7N}{8} - \frac{9}{8} = \frac{7N - 9}{8}$$ 7. **Final solution:** $$y = \frac{7N - 9}{8} e^{2x} + \left( \frac{N}{8} - \frac{13}{24} \right) e^{-6x} + \frac{2}{3} e^{3x}$$