Solution of a second order differential equation

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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

回答(1 个)

Sam Chak
Sam Chak 2024-1-31
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
  2 个评论
Swami
Swami 2024-1-31
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
Sam Chak 2024-1-31
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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