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 × 10−19,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 − 0.5et as starting values and h = 0.2 to compare the approximations from
(a) by the explicit Adams-Bashforth four-step method.
(b) the implicit Adams-Moulton three-step method.
6.
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
.
[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 Euler’s 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.
