I have a good code for the Runge-Kutta method to determine the function for a 2nd order differential equation, but I need to create a code that solves my boundary conditions. My boundaries are y(0)=25 and y(30)=20. I need to determine what y'(0)=? to get y(30)=20 using false position method. When y'(0)=0, y(30)=54.32. When y'(0)=-3, y(30)=-15.3. Basically what I need to do is use the y(30) values [-15.3,54.32] to find what y'(0) would give me y(30)=20 using false position. If anyone has any idea how to code this and can help I would appreciate it! I have put my RK code below. If you need to know more about the problem it is Problem 6 part d in the attached file.
function [KR4] = NewHighway(z1,z2)
C=.053;
f1 = z2 ;
f2 = C*sqrt(1+z2.^2);
KR4 = [f1 ; f2] ;
end
clear all ;
h = 0.01 ;
t = 0:h:30;
a = zeros(1,length(t));
b = zeros(1,length(t));
a(1) = 25 ;
b(1) = -3 ;
for i = 1:(length(t)-1)
k1 = NewHighway(a(i), b(i));
k2 = NewHighway(a(i)+h/2*(k1(1)), b(i) + h/2*(k1(2)));
k3 = NewHighway((a(i)+h/2*k2(1)), b(i) + h/2*(k2(2)));
k4 = NewHighway((a(i)+h*k3(1)), b(i) + h*k3(2));
a(i+1) = a(i) + (h/6)*(k1(1)+ 2*k2(1) + 2*k3(1) + k4(1));
b(i+1) = b(i) + (h/6)*(k1(2)+ 2*k2(2) + 2*k3(2) + k4(2));
end
plot(t,a)
a(3001)
