1. Consider the first order initial value problem

y’ = y + 2x – x^{2}, y(0) = 1, (0 ≤ x < ∞) with exact solution y(x) = x^{2} + e^{x}. 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 . If x = x_{o} 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