Solution of a second order differential equation

Question

Swami 2024-1-31

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2076731-solution-of-a-second-order-differential-equation

评论： Sam Chak 2024-1-31

I am unable to solve a differential equation using dsove command in Matlab. I get the Warning: Unable to find symbolic solution. Can you please help me with it.

clc
clear
close
syms y(x)
L=1;
A=1;
p=x^2;
s=10*diff(y,x)+100000*(diff(y,x))^3;
DE = diff(s*A,x,1)+p;
cond = [y(0.0)==0.0, y(L)==1.0];
sol = dsolve(DE==0.0,cond)
Warning: Unable to find symbolic solution.
 
sol =
 
[ empty sym ]
 

Also I tried using bvp4c to solve it but was still unsuccessful. I have enclosed the code below. Thanks a lot for your help

!

clc
clear
close
L=1;
A=1;
% Using bvp4c
x=linspace(0,L,12);
yi=bvpinit(x,[1,1])
yi = struct with fields:
    solver: 'bvpinit'
         x: [0 0.0909 0.1818 0.2727 0.3636 0.4545 0.5455 0.6364 0.7273 0.8182 0.9091 1]
         y: [2×12 double]
     yinit: [1 1]
sol=bvp4c(@bvpfcn,@bcfcn,yi)
Not enough input arguments.

Error in solution>bvpfcn (line 26)
        -x^2/(10*A+300000*(y(2))^2)];

Error in bvparguments (line 96)
testODE = ode(x1,y1,odeExtras{:});

Error in bvp4c (line 119)
    bvparguments(solver_name,ode,bc,solinit,options,varargin);
function dydx = bvpfcn(x,y,A)
dydx = zeros(2,1);
dydx = [y(2)
        -x^2/(10*A+300000*(y(2))^2)];
end
function res = bcfcn(ya,yb)
res = [ya(1)-0.0
       yb(1)-1.0];
end

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Answer 1

Sam Chak 2024-1-31

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2076731-solution-of-a-second-order-differential-equation#answer_1400641

在 MATLAB Online 中打开

Hi @Swami

I've got a solution from the bvp4c() solver.

syms y(x)

L = 1;

A = 1;

p = x^2;

s = 10*diff(y,x) + 100000*(diff(y,x))^3

s(x) =

DE = diff(s*A,x,1) + p == 0;

[V, S] = odeToVectorField(DE)

V =

S =

clear

L = 1;

%% Using bvp4c

x = linspace(0, L, 21);

yi = bvpinit(x, [1, 1]);

sol = bvp4c(@bvpfcn, @bcfcn, yi);

x = sol.x;

y = sol.y;

%% Plot results

plot(x, y, '-o', 'linewidth', 1), grid

xlabel('x', 'fontsize', 14)

ylabel('y', 'fontweight', 'bold', 'fontsize', 14)

legend('y_{1}', 'y_{2}', 'location', 'southeast')

%% Differential equations

function dydx = bvpfcn(x,y)

A = 1;

dydx = [y(2);

- (x^2)/(300000*y(2)^2 + 10*A)];

end

%% Boundary condition

function res = bcfcn(ya,yb)

res = [ya(1) - 0;

yb(1) - 1];

end

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Swami 2024-1-31

Hi @Sam Chak,

Thank you very much for the solution. But what changes did you make here? The code looks the same as before.

Best,

Swami

Sam Chak 2024-1-31

在 MATLAB Online 中打开

Hi @Swami

I'm glad it works. What I did, I moved 'A = 1' to the bvpfcn() function so that I don't need to call unnecessary extra parameters. I like to place constants inside the function unless I want to test out some parameters. Generally, your bvp4c code works if you make a change to this line using this syntax to call 'A'.

sol = bvp4c(@(x, y) bvpfcn(x, y, A), @bcfcn, yi)

If you find the solution helpful, please consider clicking 'Accept' ✔ on the answer and voting 👍 for it. Thanks a bunch!

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