Integrating a 2nd order ODE

Question

PetronasAMG 2021-10-16

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1565336-integrating-a-2nd-order-ode

回答： Star Strider 2021-10-17

采纳的回答： Star Strider

在 MATLAB Online 中打开

I am given an equation,

d^2y/dx^2 + q(x) = 0

x ranges from 0 to 1 and y(0) = 1 and y(L) = 1.5

where L = 1

and q(x) = 2*cos((pi*x)/L)

here is what I have

function dydx = yfunc (x,y)
x = linspace(0,1,30);
L = 1;
for i = 1:length(x)
    qx(i)= 2*cos((pi*x(i))/L);
end
dydx= -qx;
end
%main script
[x,y] = ode45(@yfunc,x,[1 1.5]);

I am running into an error stating Dimensions of arrays being concatenated are not consistent. Could you please help me?

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Answer 1

Star Strider 2021-10-17

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1565336-integrating-a-2nd-order-ode#answer_810081

在 MATLAB Online 中打开

This is a boundary value problem. Use bvp4c to solve it.

syms y(x) x L Y

q(x) = 2*cos(pi*x/L);

Dy = diff(y);

D2y = diff(Dy);

ODE = D2y + q

ODE(x) =

[VF,Subs] = odeToVectorField(ODE)

VF =

Subs =

bvpfcn = matlabFunction(VF, 'Vars',{x,Y,L})

bvpfcn = function_handle with value:

@(x,Y,L)[Y(2);cos((x.*pi)./L).*-2.0]

I solved it completely, however I do not want to deprive you of the same feeling of accomplishment, so I leave the rest to you. It is a straightforward solution. Follow the examples in the documentation I linked to.

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