Math, asked by jack7218, 11 months ago

how to solve problem of RK order 4?​

Answers

Answered by Anonymous
0

In the last section, Euler's Method gave us one possible approach for solving differential equations numerically.

The problem with Euler's Method is that you have to use a small interval size to get a reasonably accurate result. That is, it's not very efficient.

The Runge-Kutta Method produces a better result in fewer steps.

Mechanics

Springs and dampeners on cars (This spring applet uses RK4.)

Biology

Predator-prey models

Fisheries collapses

Drug delivery

Epidemic prediction

Physics

Climate change models

Ozone protection

Aviation

On-board computers

Aerodynamics

\displaystyle{y}{\left({x}+{h}\right)}y(x+h) \displaystyle={y}{\left({x}\right)}+=y(x)+ \displaystyle\frac{1}{{6}}{\left({F}_{{1}}+{2}{F}_{{2}}+{2}{F}_{{3}}+{F}_{{4}}\right)}

6

1

(F

1

+2F

2

+2F

3

+F

4

)

where

\displaystyle{F}_{{1}}={h} f{{\left({x},{y}\right)}}F

1

=hf(x,y)

\displaystyle{F}_{{2}}={h} f{{\left({x}+\frac{h}{{2}},{y}+\frac{{F}_{{1}}}{{2}}\right)}}F

2

=hf(x+

2

h

,y+

2

F

1

)

\displaystyle{F}_{{3}}={h} f{{\left({x}+\frac{h}{{2}},{y}+\frac{{F}_{{2}}}{{2}}\right)}}F

3

=hf(x+

2

h

,y+

2

F

2

)

\displaystyle{F}_{{4}}={h} f{{\left({x}+{h},{y}+{F}_{{3}}\right)}}F

4

=hf(x+h,y+F

3


jack7218: but how do we come to know that in RK order 4 we have to solve it only once or twice i mean y1 only or till y2?
Similar questions