Given the initial-value problem
y'=2-2tyt2+1, 0 ≤t≤1, y0=1
With exact solution
yt= 2t+1t2+1
Using MATLAB use Euler’s method with h = 0.1 to approximate the solution of y
i use 'x' in replace of 't'
clear all
clc
f=@(x,y)(2-2.*x.*y)./(x.^2+1); %Write your f(x,y) function, where dy/dx=f(x,y), x(x0)=y0.
x0=input('\n Enter initial value of x i.e. x0: '); %example x0=0
y0=input('\n Enter initial value of y i.e. y0: '); %example y0=0.5
xn=input('\n Enter the final value of x: ');% where we need to find the value of y
%example x=2
h=input('\n Enter the step length h: '); %example h=0.2
%Formula: y1=y0+h/2*[f(x0,y0)+f(x1,y1*)] where y1*=y0+h*f(x0,y0);
fprintf('\n x y ');
while x0<=xn
fprintf('\n%4.3f %4.3f ',x0,y0);%values of x and y
k=y0+h*f(x0,y0);
x1=x0+h;
y1=y0+h/2*(f(x0,y0)+f(x1,k));
x0=x1;
y0=y1;
end
Given the initial-value problem y'=2-2tyt2+1, 0 ≤t≤1, y0=1 With exact solution yt= 2t+1t2+1 Using MATLAB use...
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