Question

PLEASE PLEASE,ONLY ANSWER THIS QUESTION IF YOU COULD GIVE ME THE MATLAB CODE.THANK YOU.

Consider the following Ordinary Differential Equation (ODE): dy = 3.0*** + 1.08 * 210 – 3* y2 dat with initial condition at point xo = 0.375: Yo = 0.0044 Apply Runge-Kutta method of the second order with h = 0.25 and the set of parameters given below to approximate the solution of the ODE at the three points given in the table below. Fill in the blank spaces. Round up your answers to 4 decimals. Ci 0.375 0.0044 0.625 0.1211 0.875 1.125 1.375 Parameters of the Runge-Kutta Method of the second order: ay = 0, a2 = 1, P1 = qu = 1 Approximation formulas: Yi41 = y + (azki + a2k2)h ky = f(xi, vi) k2 = f(x+pih,yi + qukih)

Answer 1

clear;

clc;

f=@(x,y)3*x^4+1.08*x^10-3*y^2; %Write your f(x,y) function, where dy/dx=f(x,y), x(x0)=y0.

x0=0.375; %example x0=0

y0=0.0044; %example y0=0.5

xn=1.375;% where we need to find the value of y example x=2

h=0.25; %example h=0.2

fprintf(' x y \n');

while x0<=xn

fprintf('%.3f %.4f \n',x0,y0); %values of x and y

k1=f(x0,y0);

x1=x0+h;

k2=f(x1,y0+k1*h);

y1=y0+k2*h;

x0=x1;

y0=y1;

end

x y
0.375 0.0044
0.625 0.1210
0.875 0.5931
1.125 2.1421
1.375 10.8901