Consider the initial value problem $Consider the initial value problem This question is related to using the predictor-corrector method to estimate the solution y(0.2) using a step size of h = 0.05. For this problem, you will need these values to carry out the computations: Which of the following show a portion of the formula for the corrected value of A) y_{3}+\frac{9 \times 0.05}{24}\left(0.2 \times 2+5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 5\right]\right)\right) B) y_{3}-\frac{9 \times 0.05}{24}\left(0.2 \times 2+5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 5\right]\right)\right) C) y_{3}+\frac{37 \times 0.05}{24}\left(5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 5\right]\right)\right) D) y_{3}-\frac{37 \times 0.05}{24}\left(5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 57\right]\right)\right.$ 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:
$Consider the initial value problem This question is related to using the predictor-corrector method to estimate the solution y(0.2) using a step size of h = 0.05. For this problem, you will need these values to carry out the computations: Which of the following show a portion of the formula for the corrected value of A) y_{3}+\frac{9 \times 0.05}{24}\left(0.2 \times 2+5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 5\right]\right)\right) B) y_{3}-\frac{9 \times 0.05}{24}\left(0.2 \times 2+5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 5\right]\right)\right) C) y_{3}+\frac{37 \times 0.05}{24}\left(5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 5\right]\right)\right) D) y_{3}-\frac{37 \times 0.05}{24}\left(5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 57\right]\right)\right.$
Which of the following show a portion of the formula for the corrected value of $Consider the initial value problem This question is related to using the predictor-corrector method to estimate the solution y(0.2) using a step size of h = 0.05. For this problem, you will need these values to carry out the computations: Which of the following show a portion of the formula for the corrected value of A) y_{3}+\frac{9 \times 0.05}{24}\left(0.2 \times 2+5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 5\right]\right)\right) B) y_{3}-\frac{9 \times 0.05}{24}\left(0.2 \times 2+5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 5\right]\right)\right) C) y_{3}+\frac{37 \times 0.05}{24}\left(5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 5\right]\right)\right) D) y_{3}-\frac{37 \times 0.05}{24}\left(5 \times\left(y_{3}+\frac{0.05}{24}\left[55\left(2 \times 0.15+5 y_{3}\right)-59\left(2 \times 0.10+5 y_{2}\right)+37\left(2 \times 0.05+5 y_{1}\right)-9 \times 57\right]\right)\right.$
A)

y_{3}+\frac{9 \times 0.05}{24}\left(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)\right)

y_{3}-\frac{9 \times 0.05}{24}\left(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)\right)

y_{3}+\frac{37 \times 0.05}{24}\left(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)\right)

y_{3}-\frac{37 \times 0.05}{24}\left(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 57\right]\right)\right.

Consider the Following Initial Value Problem

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