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A Popular Second Order Runge-Kutta Method for the Solution Of y

Question 40

Multiple Choice

A popular second order Runge-Kutta method for the solution of yt=f(x,y) ,y(x0) =y0y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 } is


A) yn+1=yn+h(k1+k2) /2, where k1=f(xn,yn) k2=f(xn+h/2,yn+hk1/2) \begin{array} { l } y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) \\k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } / 2 \right) \end{array}
B) yn+1=yn+h(k1+k2) /2, where k1=f(xn,yn) ,k2=f(xn+h,yn+hk1/2) y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 \text {, where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , k _ { 2 } = f \left( x _ { n } + h , y _ { n } + h k _ { 1 } / 2 \right)
C) yn+1=yn+h(k1+k2) /2, where k1=f(xn,yn) ,k2=f(xn+h/2,yn+hk1) y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 \text {, where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } \right)
D) yn+1=yn+h(k1+k2) /2, where k1=f(xn,yn) ,k2=f(xn+h,yn+hk1) y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 \text {, where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , k _ { 2 } = f \left( x _ { n } + h , y _ { n } + h k _ { 1 } \right)
E) none of the above

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