Errors when trying to solve fourth order ODE

Question

Liam Hoyle 2021-11-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1579374-errors-when-trying-to-solve-fourth-order-ode

评论： Liam Hoyle 2021-11-4

This is for a homewok and I've found myself stuck. I wrote the ODE as a function handle, seems right to me but I'm very new to this language...

My code:

close all; clear all; clc
% constants
A = 2;  %coefficient 
%initial conditions
y0 = [1 0]; %initial conditions of y and dy/dt
t0 =0;
tf = 20;
tspan = [t0, tf];
%function handle for system of ODEs
odefun = @(t,y) [y(2);...     %y1'
                  y(3);...    %y2'
                   y(4);...   %y3'
                     -(A*y(3)+y(1))]; %y4'
                      
               
%...Solve using ode45
[t, y] = ode45(odefun, tspan, y0);  %ode45 function call

The actual ODE I need to solve is y'''' + 2y'' +y =0 , I think it's an issue with my function handle.

Here's the errors:

Index exceeds the number of array elements. Index must not exceed 2.

Error in Problem_1_Lab_2>@(t,y)[y(2);y(3);y(4);-(A*y(3)+y(1))] (line 19)

y(3);... %y2'

Error in odearguments (line 90)

f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.

Error in ode45 (line 106)

odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);

Error in Problem_1_Lab_2 (line 25)

[t, y] = ode45(odefun, tspan, y0); %ode45 function call

Apologies if the error messages make my issue blatanly obvious. My train of thought is that if y(1) = x, then y(3) = x'' and y(4)' = x'''', so I solved the ODE for x'''' and have that result as the final ODE in the handle.

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Answer 1

Walter Roberson 2021-11-4

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1579374-errors-when-trying-to-solve-fourth-order-ode#answer_824029

在 MATLAB Online 中打开

y0 = [1 0]; %initial conditions of y and dy/dt

Two initial conditions

%function handle for system of ODEs
odefun = @(t,y) [y(2);...     %y1'
                  y(3);...    %y2'
                   y(4);...   %y3'
                     -(A*y(3)+y(1))]; %y4'
                      

but you use up to y(4) as inputs and you output 4 items. You need 4 input conditions, one for y1, y2, y3, y4.

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Liam Hoyle 2021-11-4

Yes! I just realized this through trial and error. The input conditions for y3 and y4 were both zero, so I set them to that. Now my code spits out no errors and I can graph! Thank you very much for your answer this was my first time on the forum and I very much appreciate any and all assistance in my work (:

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