Runge-Kutta method: Difference between revisions

:<math>y' = t *\times \sqrt y</math>

''';Task'''

:Demonstrate the commonly used explicit fourth-order Runge–Kutta method as defined in the [[wp:Runge–Kutta_methods#Common_fourth-order_Runge.E2.80.93Kutta_method|Wikipedia article]] to solve the above differential equation.

:* Solve the given differential equation over the range <math>t = 0 ..\ldots 10</math> with a step value of h=dt<math>\delta t=0.1</math> (101 total points, the first being given)

:* Print the calculated values of <math>y</math> at whole numbered <math>t</math>'s (<math>0.0, 1.0, ...\ldots 10.0</math>) along with error as compared to the exact solution.

''';Method Summary'''summary

:Starting with a given <math>y_n</math> and <math>t_n</math> calculate:

:<math>dy_1\delta y_1 = dt*\delta t\times y'(t_n, y_n)\quad</math>

:<math>dy_2\delta y_2 = dt*\delta t\times y'(t_n + \tfrac{1}{2}dt\delta t , y_n + \tfrac{1}{2}dy_1\delta y_1)</math>

:<math>dy_3\delta y_3 = dt*\delta t\times y'(t_n + \tfrac{1}{2}dt\delta t , y_n + \tfrac{1}{2}dy_2\delta y_2)</math>

:<math>dy_4\delta y_4 = dt*\delta t\times y'(t_n + dt\delta t , y_n + dy_3\delta y_3)\quad</math>

:then:

:<math>y_{n+1} = y_n + \tfrac{1}{6} (dy_1\delta y_1 + 2dy_22\delta y_2 + 2dy_32\delta y_3 + dy_4\delta y_4)</math>

:<math>t_{n+1} = t_n + dt\delta t\quad</math>