Question 1
Multiple Choice
The fourth order Runge-Kutta method for solving y′′=f(x,y,y′) ,y(x0) =y0,y′(x0) =u0 is
A) yn+1=yn+h(m1−2m2+2m3−m4) /6,un+1=un+h(k1−2k2+2k3−k4) /6
where m1=un,k1=f(xn,yn,un) ,
m2=un+hk1/2,k2=f(xn+h/2,yn+hm1/2,un+hk1/2) m3=un+hk2/2,k3=f(xn+h/2,yn+hm2/2,un+hk2/2) m4=un+hk3,k4=f(xn+h,yn+hm3,un+hk3)
B) yn+1=yn−h(m1+2m2+2m3+m4) /6,un+1=un−h(k1+2k2+2k3+k4) /6
where m1=un,k1=f(xn,yn,un) ,
m2=un+hk1/2,k2=f(xn+h/2,yn+hm1/2,un+hk1/2) m3=un+hk2/2,k3=f(xn+h/2,yn+hm2/2,un+hk2/2) m4=un+hk3,k4=f(xn+h,yn+hm3,un+hk3)
C) yn+1=yn+h(m1+2m2+2m3+m4) /6,un+1=un−h(k1+2k2+2k3+k4) /6
where m1=un,k1=f(xn,yn,un) ,
m2=un+hk1/2,k2=f(xn+h/2,yn+hm1/2,un+hk1/2) m3=un+hk2/2,k3=f(xn+h/2,yn+hm2/2,un+hk2/2) m4=un+hk3,k4=f(xn+h,yn+hm3,un+hk3)
D) yn+1=yn−h(m1+2m2+2m3+m4) /6,un+1=un+h(k1+2k2+2k3+k4) /6
where m1=un,k1=f(xn,yn,un) ,
m2=un+hk1/2,k2=f(xn+h/2,yn+hm1/2,un+hk1/2) m3=un+hk2/2,k3=f(xn+h/2,yn+hm2/2,un+hk2/2) m4=un+hk3,k4=f(xn+h,yn+hm3,un+hk3)
E) yn+1=yn+h(m1+2m2+2m3+m4) /6,un+1=un+h(k1+2k2+2k3+k4) /6
where m1=un,k1=f(xn,yn,un) ,
m2=un+hk1/2,k2=f(xn+h/2,yn+hm1/2,un+hk1/2) m3=un+hk2/2,k3=f(xn+h/2,yn+hm2/2,un+hk2/2) m4=un+hk3,k4=f(xn+h,yn+hm3,un+hk3)
Correct Answer:

Verified