Hi Ajshay Verma
1.
A possible approach would be to use the common ODE tools, but, at least in the way I tested them I get the following error messages:
clear all;clc;close all
syms y(t)
[V] = odeToVectorField(diff(y, 2) == 1/t*diff(y) + 2*exp(y)-2*exp(-y))
M = matlabFunction(V,'vars', {'t','Y'})
sol = ode15i(M,[-20 20],[2 3]);
fplot(@(x)deval(sol,x,1), [0 1])
%ode54 ode113 ode23tb return the following
Warning: Failure at t=-1.956270e+01. Unable to meet integration tolerances without
reducing the step size below the smallest value allowed (5.684342e-14) at time t.
> In ode45 (line 308)
Warning: Function behaves unexpectedly on array inputs. To improve performance,
properly vectorize your function to return an output with the same size and shape as
the input arguments.
> In matlab.graphics.function.FunctionLine>getFunction
In matlab.graphics.function.FunctionLine/updateFunction
In matlab.graphics.function.FunctionLine/set.Function_I
In matlab.graphics.function.FunctionLine/set.Function
In matlab.graphics.function.FunctionLine
In fplot>singleFplot (line 226)
In fplot>@(f)singleFplot(cax,{f},limits,extraOpts,args) (line 185)
In fplot>vectorizeFplot (line 185)
In fplot (line 156)
Warning: Error updating FunctionLine.
The following error was reported evaluating the function in FunctionLine update:
Attempting to evaluate the solution outside the interval [-2.000000e+01,
-1.956270e+01] where it is defined.
% In ode23s (line 379)
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate.
RCOND = 6.744364e-24.
Warning: Failure at t=-1.956324e+01. Unable to meet integration tolerances without
reducing the step size below the smallest value allowed (6.950260e-14) at time t.
> In ode23s (line 401)
Warning: Function behaves unexpectedly on array inputs. To improve performance,
properly vectorize your function to return an output with the same size and shape
as the input arguments.
> In matlab.graphics.function.FunctionLine>getFunction
In matlab.graphics.function.FunctionLine/updateFunction
In matlab.graphics.function.FunctionLine/set.Function_I
In matlab.graphics.function.FunctionLine/set.Function
In matlab.graphics.function.FunctionLine
In fplot>singleFplot (line 226)
In fplot>@(f)singleFplot(cax,{f},limits,extraOpts,args) (line 185)
In fplot>vectorizeFplot (line 185)
In fplot (line 156)
Warning: Error updating FunctionLine.
The following error was reported evaluating the function in FunctionLine update:
Attempting to evaluate the solution outside the interval [-2.000000e+01,
-1.956324e+01] where it is defined.
% ode23t returns NaNs
In ode23t>itsolve (line 919)
In ode23t (line 543)
Warning: Matrix is singular, close to singular or badly scaled. Results may be
inaccurate. RCOND = NaN.
> In ode23t>itsolve (line 919)
In ode23t (line 543)
Warning: Matrix is singular, close to singular or badly scaled. Results may be
inaccurate. RCOND = NaN.
> In ode23t>itsolve (line 919)
In ode23t (line 543)
Warning: Matrix is singular, close to singular or badly scaled. Results may be
inaccurate. RCOND = NaN.
Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without
reducing the step size below the smallest value allowed (0.000000e+00) at time t.
> In ode23t (line 554)
.
2.
MuPAD gives a solution but, in the way I have used it, only when prompting without the initial conditions y'(0)=0 y'(1)=4 supplied in the question.
.
3.
I also tried a polynomial approximation of y, but as soon as the amount of polynomial terms were increased or the x resolution improved reducing dx, then MATLAB would spend minutes through calculations
close all
clc
clear all
dx=.1;x=[0:dx:1];
N=3; % amount terms
a=[-5:.1:5]; % accuracy terms
L=combinator(numel(a),N,'c');
err=zeros(length(L),length(x));
for k=1:1:length(L)
y=zeros(1,length(x));
for s=1:1:N
y=y+a(L(k,s))*x;
end
y=a(L(k,1))*ones(1,length(x))+a(L(k,2))*x+a(L(k,3))*x.^2;
d1y=[y(1) diff(y)];
d2y=[diff(d1y) d1y(end)];
err(k,:)=1./x.*d1y+d2y-2*exp(y)+2*exp(-y);
end
figure;hold all;
for k=1:1:length(L)
plot(err(k,:));
end
grid on;
.
even with 10 terms and really long waiting time, there's unbearable error when x surpasses certain value, polynomials only may not be a good idea to solve this differential equation.
4.
also tried a simplified version of the solution returned by MuPAD, seeking the correct values of the 3 constants I have considered convenient may be a good point to carry on.
close all
clc
clear all
dx=.1;x=[0:dx:1];
N=3; % amount terms
a=[-5:.1:5]; % accuracy terms
L=combinator(numel(a),N,'c');
err=zeros(length(L),length(x));
for k=1:1:length(L)
y=zeros(1,length(x));
y=log((a(L(k,1))*x.^2-a(L(k,2))+a(L(k,3))*x)./x.^2);
d1y=[y(1) diff(y)];
d2y=[diff(d1y) d1y(end)];
err(k,:)=d1y+x.*d2y-2*x.*exp(y)+2*x.*exp(-y);
end
figure;hold all;
for k=1:1:length(L)
plot(abs(err(k,:)));
end
grid on;
.
5.
so the function you may want to consider, from MuPAD to MATLAB, is
C1=(2^.5*(C12+2)^.5-2)./x+exp(log(x)*((2*C12+4)^.5-1))./(C13-(2^.5*x.*((2*C12+4).^.5))/(4*(C12+2)^.5))
y=log((x.^2*C1.^2/4-C12/2+x.*C1)/x.^2)
I have not calculated C12 C13, but for instance
C12=1;C13=1;
>>
C1=(2^.5*(C12+2)^.5-2)./x+exp(log(x)*((2*C12+4)^.5-1))./(C13-(2^.5*x.*((2*C12+4).^.5))/(4*(C12+2)^.5))
C1 =
Columns 1 through 8
Inf 4.5323 2.3552 1.7037 1.4549 1.3872 1.4304 1.5595
Columns 9 through 11
1.7679 2.0601 2.4495
y=log((.25*x.^2.*C1.^2-.5*C12+x.*C1)./x.^2)
y =
Columns 1 through 8
NaN -0.7802 -0.4109 -0.1634 0.0407 0.2275 0.4099 0.5964
Columns 9 through 11
0.7930 1.0053 1.2382
.
for a higher resolution x
dx=.01;x=[0:dx:1];
the shape more or less matches with result in Wolfram DE solver
.
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thanks in advance
John BG
