How to solve a differential equation with non-constant coefficients

 a = @(x) 1/x;
xdomain = [1 100];
%b = rand(10000,1);% aléatoire
load ('Bx');
load ('By');
load ('Bz');
b= sqrt(Bx.^2+By.^2+Bz.^2);
y = ones(10000,1);
[x,y] = ode45(@(x,y,a,b) a(x)*y - b ,rdomain,y,[],a,b);
plot(x,y)

Torsten 2017-6-13

在 MATLAB Online 中打开

So you want to solve the ODE

y'=y/x-sqrt(Bx^2+By^2+Bz^2)

for all combinations (Bx,By,Bz) from your loaded arrays ?

Then use a loop over the length of Bx:

a = @(x) 1/x;
xdomain = linspace(1,100,10);
y0=1;
%b = rand(10000,1);% aléatoire
load ('Bx');
load ('By');
load ('Bz');
for i=1:numel(Bx)
  f=@(x,y) a(x)*y-sqrt(Bx(i)^2+By(i)^2+Bz(i)^2);
  [x,y]=ode45(f,xdomain,y0);
  yvec(i,:)=y(:);
end

Best wishes

Torsten.

请先登录，再进行评论。

Answer 2

arbia haded 2017-6-13

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/343851-how-to-solve-a-differential-equation-with-non-constant-coefficients#answer_270533

在 MATLAB Online 中打开

hi Torsten Thank u very much for your help :) Yesterday I tried to simplify the problem, so I started with a very simple sinusoidal signal of the following form: b = A sin (2 pi f t), I calculated the solution of this equation analytically , I found this expression : y(x,t) = -A x pi f cos (2 pi f t), It is clear that the solution has a sinusoidal shape. But when I switched to the solution with matlab I didn't get the right solution. here is my code :

if true
A =1.25; % amplitude de la sinusoide
f =50; % frequence de la sinusoide
Fe =5000; % fréquence d'échantillonnage
t=0:1/ Fe :0.3; % base temps discretisée
B=A*sin (2* pi*f*t); % sinusoide à temps discret
b=diff(B)./diff(t); % dérivé 
figure (1)
subplot(2,2,1),plot (B,'r') % affichage de la courbe 
title ('signal sinusoidal')
grid on
subplot(2,2,2),plot(b,'m')
title ('la dérivée')
grid on
a = @(x) 1/x;
xdomain = linspace(1,100,10);
y0=1;
for i=1:numel(b)
f=@(x,y) -a(x)*y-b(i);
[x,y]=ode45(f,xdomain,y0);
yvec(i,:)=y(:);
end
subplot(2,2,3), plot(x,y)
end

Here I took the derivative of b as a second member (In reality I need the signal derivative). As an attachment you will find the result of simulation.