Solved

Consider the Initial Value Problem This Question Is Related

Question 11

Multiple Choice

Consider the initial value problem  Consider the initial value problem   This question is related to using the predictor-corrector method to estimate the solution y(0.2)  using a step size of h = 0.05. For this problem, you will need these values to carry out the computations:   Which of the following show a portion of the formula for the corrected value of   A)    y_{3}+\frac{9 \times 0.05}{24}\left[0.2 \times(-3) +7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)    B)    y_{3}-\frac{9 \times 0.05}{24}\left(0.2 \times(-3) +7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)    C)    y_{3}+\frac{37 \times 0.05}{24}\left(7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)    D)    y_{3}-\frac{37 \times 0.05}{24}\left(7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)  This question is related to using the predictor-corrector method to estimate the solution y(0.2) using a step size of h = 0.05.
For this problem, you will need these values to carry out the computations:
 Consider the initial value problem   This question is related to using the predictor-corrector method to estimate the solution y(0.2)  using a step size of h = 0.05. For this problem, you will need these values to carry out the computations:   Which of the following show a portion of the formula for the corrected value of   A)    y_{3}+\frac{9 \times 0.05}{24}\left[0.2 \times(-3) +7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)    B)    y_{3}-\frac{9 \times 0.05}{24}\left(0.2 \times(-3) +7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)    C)    y_{3}+\frac{37 \times 0.05}{24}\left(7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)    D)    y_{3}-\frac{37 \times 0.05}{24}\left(7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)
Which of the following show a portion of the formula for the corrected value of  Consider the initial value problem   This question is related to using the predictor-corrector method to estimate the solution y(0.2)  using a step size of h = 0.05. For this problem, you will need these values to carry out the computations:   Which of the following show a portion of the formula for the corrected value of   A)    y_{3}+\frac{9 \times 0.05}{24}\left[0.2 \times(-3) +7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)    B)    y_{3}-\frac{9 \times 0.05}{24}\left(0.2 \times(-3) +7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)    C)    y_{3}+\frac{37 \times 0.05}{24}\left(7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)    D)    y_{3}-\frac{37 \times 0.05}{24}\left(7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)


A) y3+9×0.0524[0.2×(3) +7×(y3+0.0524[55(3×0.15+7y3) 59(3×0.10+7y2) +37(3×0.05+7y1) 9×5]) ) y_{3}+\frac{9 \times 0.05}{24}\left[0.2 \times(-3) +7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)
B) y39×0.0524(0.2×(3) +7×(y3+0.0524[55(3×0.15+7y3) 59(3×0.10+7y2) +37(3×0.05+7y1) 9×5]) ) y_{3}-\frac{9 \times 0.05}{24}\left(0.2 \times(-3) +7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)
C) y3+37×0.0524(7×(y3+0.0524[55(3×0.15+7y3) 59(3×0.10+7y2) +37(3×0.05+7y1) 9×5]) ) y_{3}+\frac{37 \times 0.05}{24}\left(7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)
D) y337×0.0524(7×(y3+0.0524[55(3×0.15+7y3) 59(3×0.10+7y2) +37(3×0.05+7y1) 9×5]) ) y_{3}-\frac{37 \times 0.05}{24}\left(7 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(-3 \times 0.15+7 y_{3}\right) -59\left(-3 \times 0.10+7 y_{2}\right) +37\left(-3 \times 0.05+7 y_{1}\right) -9 \times 5\right]\right) \right)

Correct Answer:

verifed

Verified

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

Related Questions