[Solved] MA 322 Lab 7

1. The irreversible chemical reaction in which two molecules of solid potassium dichromate (K^[1]Cr₂O₇), two molecules of water (H₂O), and three atoms of solid sulfur (S) combine to yield three molecules of the gas sulfur dioxide (SO₂), four molecules of solid potassium hydroxide (KOH), and two molecules of solid chromic oxide (Cr₂O₃) can be represented symbolically by the stoichiometric equation:

2K₂Cr₂O₇ + 2H₂O + 3S 4KOH + 2Cr₂O₃ + 3SO₂.

If n₁ molecules of K₂Cr₂O₇, n₂ molecules of H₂O, and n₃ molecules of S are originally available, the following differential equation describes the amount x(t) of KOH after time t:

,

where k is the velocity constant of the reaction. If k = 6.22 10¹⁹,n₁ = n₂ = 2 10^[2][3], and n₃ = 3 10³, use the Runge-Kutta method of order four to determine how many units of potassium hydroxide will have been formed after 0.2s?

5. Consider the initial-value problem

.

Use the exact values given from y(t) = (t + 1)² 0.5e^t as starting values and h = 0.2 to compare the approximations from

• by the explicit Adams-Bashforth four-step method.
• the implicit Adams-Moulton three-step method.

• Use each of the Adams-Bashforth methods to approximate the solutions to the followinginitial-value problems. In each case use starting values obtained from the Runge-Kutta method of order four. Compare the results to the actual values.

, with h = 0.1; actual solution.

, with h = 0.1; actual solution.

, with h = 0.2; actual solution.

7. Apply the Adams fourth-order predictor-corrector method with h = 0.2 and starting values from the Runge-Kutta fourth order method to the initial-value problem

.

END

[1] . Show that the Runge-Kutta method of order second and the Modified Euler method give the same approximations to the initial-value problem

.

for any choice of h. Why is this true?

[2] . Use the Modified Euler method to approximate the solutions to each of the following initial-value problems, and compare the results to the actual values.

, with h = 0.5; actual solution y(t) = t² + 6 + 2t , with h = 0.25; actual solution y(t) =

[3] . Consider the problem

.

Use Eulers method with h = 0.025, the Runge-Kutta second-order method with h = 0.05, and the Runge-Kutta fourth-order method with h = 0.1 and compared at the common mesh points of these methods 0.1,0.2,0.3,0.4, and 0.5.