How to solve a differential equation with non-constant coefficients

Hi all

I have equation like this dy/dx = a(x)*y + b

where : a(x) is non constant (a=1/x) and b is a vector (10000 rows)

how can I solve this equation using matlab !!

Someone answer me plz

Answer 1

[Torsten]

Take a look at the example

ODE with Time-Dependent Terms

under

https://de.mathworks.com/help/matlab/ref/ode45.html

Best wishes

Torsten.

[arbia haded]

hi Torsten, Thanks a lot for your help! I checked the link that you sent to me but I still have another problem. In fact my second member b is independent of x. So i don't have this kind of equation : dy/dx = a(x)*y(x)+ b(x) "my second membre is a vector" I succeeded to solve the equation without second member but I could'nt solve it with a second member !!! I will be grateful if can help me

[Torsten]

And what do the 10000 rows of the vector b mean ? This is not an x-dependence ?

Please show the code you are using at the moment.

Best wishes

Torsten.

[arbia haded]

hi Torsten b is not dependent on x here is my code :

 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]

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.

Answer 2

[arbia haded]

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.

Answer 3

[Francisco Rosales]

Hi all

I have equation like this Az'(t) = Bz(t)+ b

how can I solve this equation using matlab !!

Someone answer me plz

Thaks

Grace

[darova]

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