0
Add your answer:
Earn +20 pts
What is the sytem of equation for x-4y1 and 5x-20y4?
Without any equality signs the given terms can't be considered to be simultaneous equations.
C programmimg of modified euler method?
#include <stdafx.h> 04#include "impeuler.h" 05 06double solveodimpeuler(double t, const double& tspan, double y, cont int& N, double tfinal) 07 08{ 09 double result; 10 int i; 11 double h; 12 13 double f1; 14 double y1*; 15 double y1; 16 17 double f2; 18 double y2*; 19 double y2; 20 21 f1(t,y)=y2; 22 f2(t,y)=-4y1; 23 y=f1(t,y)+f2(t,y); 24 25 h= (tspan-t)/N; 26 y1(0)=0.01; 27 y2(0)=0.0; 28 29 tfinal=2pi/y; 30 N=tfinal/h; 31 32 33 i=1; 34 do 35 { 36 y1*(n+1)=y1(n)+h*y2(n); 37 y2*(n+1)=y2(n)+h*(-4y1(n)); 38 39 y1(n+1)=y1(n)+(h/2)*[y2(n)+y2*(n+1)]; 40 y2(n+1)=y2(n)+(h/2)*[-4y1(n)-4y1*(n+1)]; 41 } 42 while(i<=N); 43 result=y; 44 return result; 45}
Solve y equals x2-4x plus 3 and y equals x-1 algebraically?
y = x2 - 4x + 3y = x -1Substitute the 2nd equation into the firstx -1 = x2 -4x + 3Rearranging:x2 -4x -x +3 +1 = 0x2 - 5x +4 = 0Solving this quadratic equation in x by factorising gives 2 values for xx1 = 4x2 = 1Substituting these into the original equation for y (y=x-1) givesy1 = 3y2 = 0So the 2 solutions to the equations are:x1=4y1=3andx2 = 1y2 = 0
