How to define the order of approximation in ODE?
0
$begingroup$
How to define the order of the left part of this equation?
$$
frac{y_{i+1}-y_i}{Delta x}
=frac12left(f(x_i,y_i,y_i')+f(x_{i+1},y_{i+1},y_{i+1}')right)
$$
ordinary-differential-equations approximation
edited Dec 21 '18 at 12:28
LutzL
58.7k42055
asked Dec 21 '18 at 11:14
Полина ГамаюноваПолина Гамаюнова
12
$endgroup$
add a comment |
0
$begingroup$
How to define the order of the left part of this equation?
$$
frac{y_{i+1}-y_i}{Delta x}
=frac12left(f(x_i,y_i,y_i')+f(x_{i+1},y_{i+1},y_{i+1}')right)
$$
ordinary-differential-equations approximation
edited Dec 21 '18 at 12:28
LutzL
58.7k42055
asked Dec 21 '18 at 11:14
Полина ГамаюноваПолина Гамаюнова
12
$endgroup$
$begingroup$
What exactly is $y_i'$ in this situation? Do you solve $y'=f(x,y,y')$? Or $y''=f(x,y,y')$?
$endgroup$
– LutzL
Dec 21 '18 at 12:29
add a comment |
0
0
0
$begingroup$
How to define the order of the left part of this equation?
$$
frac{y_{i+1}-y_i}{Delta x}
=frac12left(f(x_i,y_i,y_i')+f(x_{i+1},y_{i+1},y_{i+1}')right)
$$
ordinary-differential-equations approximation
edited Dec 21 '18 at 12:28
LutzL
58.7k42055
asked Dec 21 '18 at 11:14
Полина ГамаюноваПолина Гамаюнова
12
$endgroup$
How to define the order of the left part of this equation?
$$
frac{y_{i+1}-y_i}{Delta x}
=frac12left(f(x_i,y_i,y_i')+f(x_{i+1},y_{i+1},y_{i+1}')right)
$$
ordinary-differential-equations approximation
ordinary-differential-equations approximation
edited Dec 21 '18 at 12:28
LutzL
58.7k42055
asked Dec 21 '18 at 11:14
Полина ГамаюноваПолина Гамаюнова
12
edited Dec 21 '18 at 12:28
LutzL
58.7k42055
asked Dec 21 '18 at 11:14
Полина ГамаюноваПолина Гамаюнова
12
edited Dec 21 '18 at 12:28
LutzL
58.7k42055
edited Dec 21 '18 at 12:28
LutzL
58.7k42055
edited Dec 21 '18 at 12:28
LutzL
58.7k42055
58.7k42055
asked Dec 21 '18 at 11:14
Полина ГамаюноваПолина Гамаюнова
12
asked Dec 21 '18 at 11:14
Полина ГамаюноваПолина Гамаюнова
12
asked Dec 21 '18 at 11:14
Полина ГамаюноваПолина Гамаюнова
12
12
$begingroup$
What exactly is $y_i'$ in this situation? Do you solve $y'=f(x,y,y')$? Or $y''=f(x,y,y')$?
$endgroup$
– LutzL
Dec 21 '18 at 12:29
add a comment |
$begingroup$
What exactly is $y_i'$ in this situation? Do you solve $y'=f(x,y,y')$? Or $y''=f(x,y,y')$?
$endgroup$
– LutzL
Dec 21 '18 at 12:29
$begingroup$
What exactly is $y_i'$ in this situation? Do you solve $y'=f(x,y,y')$? Or $y''=f(x,y,y')$?
$endgroup$
– LutzL
Dec 21 '18 at 12:29
$begingroup$
What exactly is $y_i'$ in this situation? Do you solve $y'=f(x,y,y')$? Or $y''=f(x,y,y')$?
$endgroup$
– LutzL
Dec 21 '18 at 12:29
add a comment |
0
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$begingroup$
What exactly is $y_i'$ in this situation? Do you solve $y'=f(x,y,y')$? Or $y''=f(x,y,y')$?
$endgroup$
– LutzL
Dec 21 '18 at 12:29