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Consider the Initial Value Problem This Question Is Related

Question 3

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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:   To use the corrector formula, you need f<sub>4</sub> . Which of the following is the correct expression for f<sub>4</sub> A)    0.2 \times 2-5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right]   B)    5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right)    C)    0.2 \times 2+5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right]   D)    -5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\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:   To use the corrector formula, you need f<sub>4</sub> . Which of the following is the correct expression for f<sub>4</sub> A)    0.2 \times 2-5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right]   B)    5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right)    C)    0.2 \times 2+5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right]   D)    -5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right]
To use the corrector formula, you need f4 . Which of the following is the correct expression for f4


A) 0.2×25×(y3+0.0524[55(2×0.15+5y3) 59(2×0.10+5y2) +37(2×0.05+5y1) 9×5]] 0.2 \times 2-5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right]
B) 5×(y3+0.0524[55(2×0.15+5y3) 59(2×0.10+5y2) +37(2×0.05+5y1) 9×5]) 5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right)
C) 0.2×2+5×(y3+0.0524[55(2×0.15+5y3) 59(2×0.10+5y2) +37(2×0.05+5y1) 9×5]] 0.2 \times 2+5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right]
D) 5×(y3+0.0524[55(2×0.15+5y3) 59(2×0.10+5y2) +37(2×0.05+5y1) 9×5]] -5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right) -59\left(2 \times 0.10+5 y_{2}\right) +37\left(2 \times 0.05+5 y_{1}\right) -9 \times 5\right]\right]

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