MATLAB: How can i solve y’=t/t^2+1 by programming a runge kutta 4’th order method

runge-kutta 4

i need a code

Best Answer

As written, this can't be done, because there's a singularity at t = 0
Assuming that you actually mean y' = t/(t^2 + 1)... why would you need RK4? The ODE can be solved by simple integration: y(t) = log(sqrt(t^2 + 1))
If you really really want to do it numerically:

% Define ODE rate equation
ratefun = @(t,y) t/(t^2+1);
% Solve for t in [0,1] with initial condition y(0) = 0
[t,y] = ode45(ratefun,[0,1],0);
% View solution
plot(t,y,t,log(sqrt(t.^2+1)))
legend('RK45','exact','Location','NW')

But... ode45 is a variable-step RK45 integrator, not the classical fixed-step RK4 method. If you really want the fixed-step RK4 method for an equation with a simple analytic solution... well, then I have to assume that this is a homework problem.

If this is a homework problem, please do not just post a request for the solution. Show what you have done, and ask a specific question for guidance on a particular part of the problem.

Related Solutions

MATLAB: Runge-Kutta method for 2×2 IVP

I have made some modifications to the shapes of the arrays in your code and changed the representation of the system of ODEs as a cell array of function handles to an array of function handles. The code below also demonstrates how "ode45" can be used to solve the problem. Both methods give the same result!

function testODE
f = @(t,y) [y(1)-4.*y(2);-y(1)+y(2)];
y0 = [1;0];
t0 = 0;
T = 1;
N = 11;
[Y,t] = RK42d(f, t0, T, y0, N);
figure;
hold on;
plot(t,Y(:,1),'r');
plot(t,Y(:,2),'b');
sol = ode45(f,[0,1],y0);
plot(sol.x,sol.y(1,:),'go');
plot(sol.x,sol.y(2,:),'ko');
legend('RK42d - y1','RK42d - y2','ode45 - y1','ode45 - y2');
end
function [Y, t] = RK42d(f, t0, T, y0, N)
%the input of f and y0 must both be vectors
%composed of two fuctions and their respective
%intial conditions
%the vector Y will return a vector as well
h = (T - t0)/(N - 1); %Calculate and store the step size
Y = zeros(N,2); %Initialize the X and Y vector
t = linspace(t0,T,N); % A vector to store the time values
Y(1,:) = y0; % Start Y vector at the intial values.
for i = 1:(N-1)
    y = Y(i,:)';
    k1 = f(t(i),y);
    k2 = f(t(i) +0.5*h, y +0.5*h*k1);
    k3 = f(t(i) +0.5*h, y +0.5*h*k2);
    k4 = f(t(i) +h    , y +h*k3);
    Y(i+1,:) = y + (h/6)*(k1+ 2.*k2 + 2*k3 + k4);
    %Update approximation
end
end

For more information on "ode45", please see

http://www.mathworks.com/help/matlab/ref/ode45.html

MATLAB: Solving system of ODE using ode45 in matlab

clc

clear

n = 30;

t = [0 0.52 1.04 1.75 2.37 3.25 3.83]';

x = [153 185 208 249 261 271 273]';

tt = linspace(t(1),t(end),n);

xx = spline(t,x,tt);

v = calc_derv(tt,xx);

a = calc_derv(tt,v);

subplot(2,1,1),plot(tt,v)

title('Jet Fighters Velocity')

xlabel('Time (s)')

ylabel('Velocity (m/s)')

subplot(2,1,2),plot(tt,a)

title('Jet Fighters Acceleration')

xlabel('Time (s)')

ylabel('Acceleration (m/s^2)')

% calculate 1st order derivative using finite differences

function dydx = calc_derv(x,y);

n = length(x);

h = x(2)-x(1);

dydx(1) = (-y(3) + 4*y(2) - 3*y(1))/(2*h);

dydx(n) = (3*y(n) - 4*y(n-1) + y(n-2))/(2*h);

for i = 2:n-1

dydx(i) = (y(i+1) - y(i-1))/(2*h);

end

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Related Solutions

MATLAB: Runge-Kutta method for 2×2 IVP

MATLAB: Solving system of ODE using ode45 in matlab

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