1. Consider the first order initial value problem
y’ = y + 2x – x2, y(0) = 1, (0 ≤ x < ∞) with exact solution y(x) = x2 + ex. For x = 0.1, the percentage diference between the exact solution and the solution obtained using a single iteration of the second-order Runge Kutta method with step size h = 0.1 is
- 0.06%
- 0.07%
- 0.08%
- 0.1%
2. Consider an ordinary differential equation $\frac{\mathrm{d} x}{\mathrm{d} t}=4t+4$. If x = xo at t = 0, the increment in x calculated using Runge Kutta fourth order multistep method with a step size of Δt = 0.2 is
- 0.22
- 0.44
- 0.66
- 0.88