1. **State the problem:** We are given the second-order differential equation $$\frac{d^2x}{dt^2} - 9 = 4t$$ with initial conditions $$x=0$$ and $$t=0$$. 2. **Rewrite the equation:** Rearranging, we get $$\frac{d^2x}{dt^2} = 4t + 9$$. 3. **Substitution:** Let $$y = \frac{dx}{dt}$$. Then $$\frac{dy}{dt} = \frac{d^2x}{dt^2}$$. 4. **Write the system:** The system can be written as: $$\frac{dy}{dt} = 4t + 9$$ $$\frac{dx}{dt} = y$$ 5. **Matrix form for part (b)(i):** Given the substitution $$y = \frac{dx}{dt}$$, the system is: $$\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -9 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$ 6. **Find eigenvalues (b)(ii):** The matrix is $$A = \begin{pmatrix} 0 & 1 \\ -9 & 0 \end{pmatrix}$$. The characteristic equation is: $$\det(A - \lambda I) = 0$$ $$\det \begin{pmatrix} -\lambda & 1 \\ -9 & -\lambda \end{pmatrix} = 0$$ $$(-\lambda)(-\lambda) - (1)(-9) = \lambda^2 + 9 = 0$$ So, $$\lambda^2 = -9$$ $$\lambda = \pm 3i$$ 7. **Long-term velocity (b)(iii):** Since eigenvalues are purely imaginary, the system exhibits oscillatory behavior with no exponential growth or decay. Hence, the long-term velocity oscillates and does not settle to a fixed value. 8. **Amended equation:** $$\frac{d^2x}{dt^2} + 4 \frac{dx}{dt} + 6x = 10t + 4$$ 9. **Substitution for amended system (b)(i):** Let $$y = \frac{dx}{dt}$$, then: $$\frac{dy}{dt} = 10t + 4 - 4y - 6x$$ $$\frac{dx}{dt} = y$$ So the system is: $$\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -6 & -4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 0 \\ 10t + 4 \end{pmatrix}$$ 10. **Initial conditions:** At $$t=0$$, the particle is stationary at 0, so: $$x(0) = 0, \quad y(0) = 0$$ 11. **Euler's method (b)(ii):** Step length $$h=0.1$$. Using Euler's method: $$x_{n+1} = x_n + h y_n$$ $$y_{n+1} = y_n + h (10 t_n + 4 - 4 y_n - 6 x_n)$$ At $$t_0=0$$: $$x_0=0, y_0=0$$ Calculate $$x_1, y_1$$ at $$t_1=0.1$$: $$x_1 = 0 + 0.1 \times 0 = 0$$ $$y_1 = 0 + 0.1 (10 \times 0 + 4 - 4 \times 0 - 6 \times 0) = 0.1 \times 4 = 0.4$$ So displacement at $$t=0.1$$ is $$x_1 = 0$$. 12. **Long-term velocity (b)(iii):** The characteristic equation of the homogeneous system is: $$\lambda^2 + 4 \lambda + 6 = 0$$ Calculate discriminant: $$\Delta = 4^2 - 4 \times 6 = 16 - 24 = -8 < 0$$ Eigenvalues: $$\lambda = \frac{-4 \pm \sqrt{-8}}{2} = -2 \pm i \sqrt{2}$$ Since real part is negative, the velocity decays to zero as $$t \to \infty$$. **Final answers:** - Eigenvalues of first matrix: $$\pm 3i$$ - Long-term velocity for first system: oscillatory, no fixed limit - Displacement at $$t=0.1$$ for amended system: $$0$$ - Long-term velocity for amended system: tends to zero