Use Euler's method to numerically integrate $y' = -2x^3 + 12x^2 - 20x + 8.5$ from $x = 0$ to $x = 0.5$ with a step size of $h = 0.5$. The initial condition at $x = 0$ is $y = 1$. The exact solution at $x = 0.5$ is $3.21875$.

Use Heun's method to integrate $y' = -2x^3 + 12x^2 - 20x + 8.5$ from $x = 0$ to $x = 0.5$ using $h = 0.5$. Initial condition: $y(0) = 1$. True value is $3.21875$.

Use the classic fourth-order Runge-Kutta method to integrate $y' = 4e^{0.8x} - 0.5y$ from $x = 0$ to $x = 0.5$ using $h = 0.5$. Initial condition: $y(0) = 2$. The true value is $3.751521$.

Assume you have solved an ODE for the first few steps using RK4 and have the points $(x_0, y_0), (x_1, y_1), (x_2, y_2)$. Formulate the predictor equation for $y_3$ using the open multistep method (Adams-Bashforth) combined with Heun's approach.

Convert the second-order unforced damped spring-mass equation $m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0$ into a system of first-order ODEs.

Consider the equation $\frac{dy}{dt} = -1000y + 3000 - 2000e^{-t}$. Analyze why this is a stiff equation and the implications for using an explicit method like Euler's method.

A $10 \text{ m}$ long heated rod is described by $\frac{d^2T}{dx^2} + h'(T_a - T) = 0$. Given $h' = 0.01 \text{ m}^{-2}$, ambient temperature $T_a = 20^\circ\text{C}$, $T(0) = 40^\circ\text{C}$, and $T(10) = 200^\circ\text{C}$. Use the finite difference method with a step size $\Delta x = 2 \text{ m}$ to set up the linear equations for the interior nodes.

Find the eigenvalues of the matrix: