Numerical methods to solve first order differential equation - Douis.net
We consider that the gradient of the line is ( approx.) the mean of the gradient at the gradient at . This gives : ( , ) ( , ). 2. The improved Euler's formula. r r.
Numerical methods to solve first order differential equation
In this chapter, we want to solve equations which can be written dy f ( x, y ) and y ( x0 ) y0 dx There are three methods to solve numerically this equation.
Formulae to be used will be stated explicitly in questions.
Knowing P0 ( x0 , y0 ), we work out P1 then P2 then P3 etc.
Euler's formula To work out Pr 1 , we consider that the gradient of the line Pr Pr 1 is (approx.) equal to the gradient at Pr .
This gives: yr 1 yr hf ( xr , yr )
The mid-point formula We consider that the gradient of the line Pr 1 Pr 1 is (approx.) equal to the gradient at Pr : yr 1 yr 1 2hf ( xr , yr )
This gives
The improved Euler's formula We consider that the gradient of the line Pr Pr 1 is ( approx.) the mean of the gradient at Pr and the gradient at Pr 1. h f ( xr , yr ) f ( xr 1 , yr*1 ) 2 with yr*1 yr hf ( xr , yr )
yr 1 yr
This gives :
Or as it is given in the exam question: 1 yr 1 yr k1 k2 2 where k1 hf ( xr , yr ) and k2 hf ( xr h, yr k1 ) Possible layout for your workings out:
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