why do i receive this error

2 Ansichten (letzte 30 Tage)
Mohammad Adeeb
Mohammad Adeeb am 21 Mär. 2021
Kommentiert: Sergey Kasyanov am 22 Mär. 2021
x=0.1;
x= 0:0.1:1;
yic =[1 -2] ' ;
for i=1:(length(x)-1)
K11 = fn (x, yic);
K21 = fn (x , yic);
K12 = fn ((x + h) , (yic + (h*K11)) , (yic + (h*K21)));
K22 = fn ((x + h),(yic + h*K11) , (yic + h*K21));
y1 = ( yic + 0.5*h*(K11 + K12 ));
y2 = ( yic + 0.5*h*(K12 + K22));
end
function f = fn( x , yic )
dy = yic(2);
dy2 = 2*yic(1)-yic(2);
end
when i run the code this error appear :
''
Output argument "f" (and maybe others) not assigned during call to "HW2>fn".
Error in HW2 (line 22)
K11 = fn (x, yic);
''

Antworten (1)

Sergey Kasyanov
Sergey Kasyanov am 21 Mär. 2021
Hello!
You don't define f in fn function.
Are you want to return f = [fy, fy2]? In that case:
function f = fn( x , yic )
dy = yic(2);
dy2 = 2*yic(1)-yic(2);
f = [dy, dy2];
end
Also you have an error in another lines. Maybe you should to correct it in that way:
K12 = fn ((x + h) , [(yic + (h*K11)) , (yic + (h*K21))] );
K22 = fn ((x + h), [(yic + h*K11) , (yic + h*K21)] );
  11 Kommentare
Walter Roberson
Walter Roberson am 21 Mär. 2021
Writing it that way is just a more compact way of writing it.
Sergey Kasyanov
Sergey Kasyanov am 22 Mär. 2021
It is more readable.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Optimization finden Sie in Help Center und File Exchange

Produkte


Version

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by