MATLAB: Initial conditions on ODE45

numericalnumerical solverode45

Im trying to solve this IVP: e^y +(t*e^y – sin(y))*(dy/dt)=0 with the initial condition y(2)=1.5.

I was just not sure how to do it with the initial condition with Y(2)=1.5, iknow how to do it if it were y(0)=1.5:

f= @(t,y) (exp(y)+(t.*exp(y)-sin(y))); % This is the function.

[t,y]=ode45(f, [0.5,4], 1.5); % trange is from 0.5 to 4

plot(t,y)

can someone please help me out?

Best Answer

This uses the initial value y(0.5)=1.5 ( not y(0)=1.5):

[t, y] = ode45(f, [0.5, 4], 1.5);

So for y(2)=1.5:

[t, y] = ode45(f, [2, 4], 1.5);

Note: The initial value problem starts at the inital point.

[EDITED]: The call to ODE45 is equivalent, if the problem is formulated in backward direction - an "final value problem": tspan is still [ti, tf], but now ti > tf.

Related Solutions

MATLAB: 4th order Runge-Kutta code that can solve for several intial conditions

Here's a version:

% Initial conditions
t0=0;
x0=0:0.2:1;
y0=1:0.2:2;
dt=0.1;
tz=1;
t_range= t0:dt:tz;
X = zeros(numel(x0),numel(t_range));
Y = zeros(numel(y0),numel(t_range));
x_rk=zeros(1,numel(t_range));
y_rk= zeros(1,numel(t_range));
for n = 1:numel(x0)
    x_rk(1)=x0(n);
    y_rk(1)=y0(n);
    for i=1:length(t_range)-1
        [x_rk(i+1), y_rk(i+1)]=rk_method(x_rk(i),y_rk(i),dt);
    end
    X(n,:) = x_rk;
    Y(n,:) = y_rk;
    %% PLots
    subplot(numel(x0)/2,2,n)
    plot (t_range, x_rk,'b-o',t_range, y_rk,'y-o','MarkerSize',5);
    grid
    title([num2str(x0(n)),'   ',num2str(y0(n))])
    legend('x','y')
end
disp('X = '), disp(X)
disp('Y = '), disp(Y)
%% Functions
function [dxdt, dydt]= f(x,y)
dxdt =  5*x-3*y;
dydt = -6*x+2*y;
end 
function [x1, y1]=rk_method(x0,y0,dt)
[k11, k12]=f(x0,y0);     
[k21, k22]=f(x0+0.5*dt*k11,y0+0.5*dt*k12); 
[k31, k32]=f(x0+0.5*dt*k21,y0+0.5*dt*k22); 
[k41, k42]=f(x0+dt*k31, y0+dt*k32); 
x1=x0+dt*((k11/6)+((k21+k31)/3)+(k41/6));
y1=y0+dt*((k12/6)+((k22+k32)/3)+(k42/6));
end

MATLAB: F=@(t,y) 4exp(0.8t)-0.5*y; tspan = [0 4]; y0 = 2; h = 1; [t,y] = ode_euler(f,tspan,y0,h)

tspan = [0 4];
y0 = 2;
[t1,y1] = ode45(@(t,y) 4*exp(0.8*t)-0.5*y,tspan,y0,h);
figure('name','Reda','color','w');
plot(t1,y1, 'r-','LineWidth',2);grid on;
xlabel('Time');ylabel('Solution y(Time)');
title('And you instead of a blanket substitute solver Euler');

Best Answer

Related Solutions

MATLAB: 4th order Runge-Kutta code that can solve for several intial conditions

MATLAB: F=@(t,y) 4*exp(0.8*t)-0.5*y; tspan = [0 4]; y0 = 2; h = 1; [t,y] = ode_euler(f,tspan,y0,h)

Related Question

MATLAB: F=@(t,y) 4exp(0.8t)-0.5*y; tspan = [0 4]; y0 = 2; h = 1; [t,y] = ode_euler(f,tspan,y0,h)