why do i receive this error

2021 3 21
1 个回答
更新时间：2021 3 22
1 次查看（30 天）

0 个投票

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);
''

请先登录，再回答此问题。

回答（1 个）

Sergey Kasyanov
Sergey Kasyanov 2021-3-21

1 个投票

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 个评论
显示 9更早的评论 隐藏 9更早的评论

Mohammad Adeeb
Mohammad Adeeb 2021-3-21
i want the code to take those values of [dy , dy2] and then get thme in the for loop
Sergey Kasyanov
Sergey Kasyanov 2021-3-21
What are you need to do in global?
Mohammad Adeeb
Mohammad Adeeb 2021-3-21
im trying to write a code to solve 2nd order ode by using reduction of order and runge-kutta 4th order method
Mohammad Adeeb
Mohammad Adeeb 2021-3-21
the code run , but the value of y is only matrix 2*2 , and its should be 11*2
Sergey Kasyanov
Sergey Kasyanov 2021-3-21
Can you write the equation?
Mohammad Adeeb
Mohammad Adeeb 2021-3-21
编辑：Mohammad Adeeb 2021-3-21
d2y/dx2 +dy/dx-2y=0
but i should use my own function not a built in one
Mohammad Adeeb
Mohammad Adeeb 2021-3-21
i did the following changes to my code
clear all;
close all;
clc;
h=0.1; %step size
x=0:h:1; %the X domain range
yic = [1 -2]'; %intial condition
t=exp(-2*x);
for i=1:(length(x)-1)
K11(i) = fn (x(i), yic(i));
K21(i) = fn (x(i) , yic(i));
K12(i) = fn ((x(i) + h) , [(yic(i) + (h*K11(i))) , (yic(i) + (h*K21(i)))] );
K22(i) = fn ((x(i) + h), [(yic(i) + h*K11(i)) , (yic(i) + h*K21(i))] );
y1(i+1) = ( yic(i) + 0.5*h*(K11(i) + K12(i) ));
y2(i+1) = ( yic(i) + 0.5*h*(K12(i) + K22(i)));
end
plot(x,y1,'k*')
hold on;
plot(x,y_exact,'r.-')
xlabel('X axis')
ylabel('Y axis')
legend('Y_e_x_a_c_t','Y_a_p_p')
function f = fn( x , yic )
dy = yic(2);
dy2 = 2*yic(1)-yic(2);
end
.
but i recive the following error :
Index exceeds the number of array elements (1).
Error in HW2>fn (line 41)
dy = yic(2);
Error in HW2 (line 25)
K11(i) = fn (x(i), yic(i));
Sergey Kasyanov
Sergey Kasyanov 2021-3-21
h = 0.1;
x = 0:h:1;
y = [[1;-2],zeros(2, length(x)-1)];
for i = 1:length(x)-1
K1 = f(x(i), y(:, i));
K2 = f(x(i) + h/2, y(:, i) + h*K1/2);
K3 = f(x(i) + h/2, y(:, i) + h*K2/2);
K4 = f(x(i) + h, y(:, i) + h*K3);
y(:, i+1) = y(:, i) + h/6*(K1 + 2*K2 + 2*K3 + K4);
end
function dy = f(x, y)
dy = [0, 1
2, -1] * y;
end
plot(x, y(1,:), 'k*');
Mohammad Adeeb
Mohammad Adeeb 2021-3-21
thank u so much ; but i have qustion why do you use dy as matrix and multiply it by y
Walter Roberson
Walter Roberson 2021-3-21
Writing it that way is just a more compact way of writing it.
Sergey Kasyanov
Sergey Kasyanov 2021-3-22
It is more readable.

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Communications Toolbox 的更多信息

产品

版本

R2020b

标签

Community Treasure Hunt

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

Start Hunting!

Translated by