Solved

The Adams-Bashforth Formula for Finding the Solution Of yt=f(x,y),y(x0)=y0y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }

Question 10

Multiple Choice

The Adams-Bashforth formula for finding 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(55yn′+59yn−1′−37yn−2′−9yn−3′) /24, where yn′=f(xn,yn) yn−1′=f(xn−1,yn−1) ,yn−2′=f(xn−2,yn−2) ,yn−3′=f(xn−3,yn−3) \begin{array} { l } y _ { n + 1 } ^ { * } = y _ { n } + h \left( 55 y _ { n } ^ { \prime } + 59 y _ { n - 1 } ^ { \prime } - 37 y _ { n - 2 } ^ { \prime } - 9 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \\y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right) \end{array}
B) yn+1∗=yn+h(55yn′−59yn−1′+37yn−2′−9yn−3′) /24, where yn′=f(xn,yn) yn−1′=f(xn−1,yn−1) ,yn−2′=f(xn−2,yn−2) ,yn−3′=f(xn−3,yn−3) \begin{array} { l } y _ { n + 1 } ^ { * } = y _ { n } + h \left( 55 y _ { n } ^ { \prime } - 59 y _ { n - 1 } ^ { \prime } + 37 y _ { n - 2 } ^ { \prime } - 9 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \\y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right) \end{array}
C) yn+1∗=yn+h(55yn′−59yn−1′−37yn−2′+65yn−3′) /24, where yn′=f(xn,yn) yn−1′=f(xn−1,yn−1) ,yn−2′=f(xn−2,yn−2) ,yn−3′=f(xn−3,yn−3) \begin{array} { l } y _ { n + 1 } ^ { * } = y _ { n } + h \left( 55 y _ { n } ^ { \prime } - 59 y _ { n - 1 } ^ { \prime } - 37 y _ { n - 2 } ^ { \prime } + 65 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \\y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right) \end{array}
D) yn+1∗=yn+h(59yn′−55yn−1′+37yn−2′−17yn−3′) /24, where yn′=f(xn,yn) yn−1′=f(xn−1,yn−1) ,yn−2′=f(xn−2,yn−2) ,yn−3′=f(xn−3,yn−3) \begin{array} { l } y _ { n + 1 } ^ { * } = y _ { n } + h \left( 59 y _ { n } ^ { \prime } - 55 y _ { n - 1 } ^ { \prime } + 37 y _ { n - 2 } ^ { \prime } - 17 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \\y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right) \end{array}
E) none of the above

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions

Q5: Using the Adams-Bashforth-Moulton method from the

Q6: Using the Adams-Bashforth method from the

Q7: Using the method from the previous

Q8: The standard central difference approximation of

Q9: The improved Euler's method is what type

Q11: Which of the following are second

Q12: In the previous problem, the local

Q13: Using Euler's method on the previous

Q14: When entering the number <span

Q15: The local truncation error for the