Below is my code for a 4th order Runge-Kutta but I keep getting an error and I don't see what I am doing wrong. I was hoping somebody could help point me in the right direction to fix my code.

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James Tursa
James Tursa 2016 年 4 月 20 日
編集済み: James Tursa 2016 年 4 月 20 日

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You never set the z2 elements inside your loop, so on the second iteration when z2(i-1) is used (with i=3) z2(2) does not exist. I am guessing you need to add a line like this after setting z1(i):
z2(i) = z2(i-1)+((1/6)*(k1z2+2*k2z2+2*k3z2+k4z2));
Also, it appears you have a typo in your z1(i) assignment. That last * should be a + instead. E.g.,
z1(i) = z1(i-1)+((1/6)*(k1z1+2*k2z1+2*k3z1*k4z1));
should be
z1(i) = z1(i-1)+((1/6)*(k1z1+2*k2z1+2*k3z1+k4z1));
Finally, as a readability advice I would suggest you put some spaces in your code. E.g., I think this is much more readable (and easier to spot bugs):
z1(i) = z1(i-1) + (1/6) * (k1z1 + 2*k2z1 + 2*k3z1 + k4z1);
z2(i) = z2(i-1) + (1/6) * (k1z2 + 2*k2z2 + 2*k3z2 + k4z2);

1 件のコメント

WhatIsMatlab-
WhatIsMatlab- 2016 年 4 月 20 日
Thank you!! That did the trick. I took your advice on adding the spacing. You are 100% right it makes it a lot easier to see what is happening in my code. Thank you again.

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その他の回答 (1 件)

Torsten
Torsten 2016 年 4 月 20 日

0 投票

dz2dx = @(z1,z2) 2*z2 5*z1;
There is no connector between z2 and 5 in the expression above.
Best wishes
Torsten.

1 件のコメント

WhatIsMatlab-
WhatIsMatlab- 2016 年 4 月 20 日
Sorry there should be a plus sign there. But that doesn't fix the problem. That was an error from copying the code over.

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