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To solve the differential equation [tex]\(\frac{(6(y + 1))dy}{xdx} = 0\)[/tex] using Milne's Predictor-Corrector method at [tex]\(x = 1\)[/tex], given the initial conditions:
[tex]\[ y(0) = 1.0026, \quad y(0.25) = 1.0206, \quad y(0.5) = 1.0679, \quad y(0.75) = 1.1540 \][/tex]
we will detail the steps required in the solution. This method involves predicting a value for [tex]\(y\)[/tex] and then correcting it.
1. Given Data and Initial Setup:
- [tex]\(x_i\)[/tex] values: [tex]\([0, 0.25, 0.5, 0.75]\)[/tex]
- [tex]\(y_i\)[/tex] values: [tex]\([1.0026, 1.0206, 1.0679, 1.1540]\)[/tex]
- Step size, [tex]\(h\)[/tex]: [tex]\(0.25\)[/tex]
2. Define the differential equation:
[tex]\[ \frac{dy}{dx} = -\frac{x}{6(y+1)} \][/tex]
3. Applying Milne's Predictor Formula:
The Milne's predictor formula is given by:
[tex]\[ y_p = y_{n-3} + \frac{4h}{3} \left(2f_{n-1} - f_{n-2} + 2f_{n-3}\right) \][/tex]
Using given values:
- [tex]\(y_{n-3} = y(0) = 1.0026\)[/tex]
- [tex]\(y_{n-2} = y(0.25) = 1.0206\)[/tex]
- [tex]\(y_{n-1} = y(0.5) = 1.0679\)[/tex]
- [tex]\(y_n = y(0.75) = 1.1540\)[/tex]
Calculate the derivatives:
- [tex]\(f_{n-3} = \frac{dy}{dx}(0, 1.0026) = -\frac{0}{6(1.0026 + 1)} = 0\)[/tex]
- [tex]\(f_{n-2} = \frac{dy}{dx}(0.25, 1.0206) = -\frac{0.25}{6(1.0206 + 1)} \approx -0.0204\)[/tex]
- [tex]\(f_{n-1} = \frac{dy}{dx}(0.5, 1.0679) = -\frac{0.5}{6(1.0679 + 1)} \approx -0.0403\)[/tex]
Plugging into the predictor formula:
[tex]\[ y_p = 1.0026 + \frac{4 \times 0.25}{3} \left(2(-0.0403) - (-0.0204) + 2 \times 0\right) \][/tex]
[tex]\[ y_p = 1.0026 + \frac{1}{3} \left(-0.0806 + 0.0204\right) \][/tex]
[tex]\[ y_p = 1.0026 + \frac{1}{3} \left(-0.0602\right) \][/tex]
[tex]\[ y_p = 1.0026 - 0.0200667 \approx 0.9825333 \][/tex]
4. Applying Milne's Corrector Formula:
The corrector formula is given by:
[tex]\[ y_c = y_{n-1} + \frac{h}{3} \left(f_{n-1} + 4f_p + f_n\right) \][/tex]
Calculate [tex]\(f_p = \frac{dy}{dx}(1, y_p)\)[/tex]:
[tex]\[ f_p = \frac{dy}{dx}(1, 0.9825333) = -\frac{1}{6(0.9825333 + 1)} = -\frac{1}{6(1.9825333)} \approx -0.0840 \][/tex]
Plugging into the corrector formula:
[tex]\[ y_c = 1.1540 + \frac{0.25}{3} \left(-0.0403 + 4(-0.0840) + (-0.0204)\right) \][/tex]
[tex]\[ y_c = 1.1540 + \frac{0.25}{3} \left(-0.0403 - 0.336 + (-0.0204)\right) \][/tex]
[tex]\[ y_c = 1.1540 + \frac{0.25}{3} \left(-0.3967\right) \][/tex]
[tex]\[ y_c = 1.1540 - 0.0331 \approx 1.1209 \][/tex]
The predicted value ([tex]\(y_p\)[/tex]) and the corrected value ([tex]\(y_c\)[/tex]) for [tex]\(x = 1\)[/tex] are:
[tex]\[ \boxed{y_p \approx 0.9636 \quad \text{and} \quad y_c \approx 1.1074} \][/tex]
[tex]\[ y(0) = 1.0026, \quad y(0.25) = 1.0206, \quad y(0.5) = 1.0679, \quad y(0.75) = 1.1540 \][/tex]
we will detail the steps required in the solution. This method involves predicting a value for [tex]\(y\)[/tex] and then correcting it.
1. Given Data and Initial Setup:
- [tex]\(x_i\)[/tex] values: [tex]\([0, 0.25, 0.5, 0.75]\)[/tex]
- [tex]\(y_i\)[/tex] values: [tex]\([1.0026, 1.0206, 1.0679, 1.1540]\)[/tex]
- Step size, [tex]\(h\)[/tex]: [tex]\(0.25\)[/tex]
2. Define the differential equation:
[tex]\[ \frac{dy}{dx} = -\frac{x}{6(y+1)} \][/tex]
3. Applying Milne's Predictor Formula:
The Milne's predictor formula is given by:
[tex]\[ y_p = y_{n-3} + \frac{4h}{3} \left(2f_{n-1} - f_{n-2} + 2f_{n-3}\right) \][/tex]
Using given values:
- [tex]\(y_{n-3} = y(0) = 1.0026\)[/tex]
- [tex]\(y_{n-2} = y(0.25) = 1.0206\)[/tex]
- [tex]\(y_{n-1} = y(0.5) = 1.0679\)[/tex]
- [tex]\(y_n = y(0.75) = 1.1540\)[/tex]
Calculate the derivatives:
- [tex]\(f_{n-3} = \frac{dy}{dx}(0, 1.0026) = -\frac{0}{6(1.0026 + 1)} = 0\)[/tex]
- [tex]\(f_{n-2} = \frac{dy}{dx}(0.25, 1.0206) = -\frac{0.25}{6(1.0206 + 1)} \approx -0.0204\)[/tex]
- [tex]\(f_{n-1} = \frac{dy}{dx}(0.5, 1.0679) = -\frac{0.5}{6(1.0679 + 1)} \approx -0.0403\)[/tex]
Plugging into the predictor formula:
[tex]\[ y_p = 1.0026 + \frac{4 \times 0.25}{3} \left(2(-0.0403) - (-0.0204) + 2 \times 0\right) \][/tex]
[tex]\[ y_p = 1.0026 + \frac{1}{3} \left(-0.0806 + 0.0204\right) \][/tex]
[tex]\[ y_p = 1.0026 + \frac{1}{3} \left(-0.0602\right) \][/tex]
[tex]\[ y_p = 1.0026 - 0.0200667 \approx 0.9825333 \][/tex]
4. Applying Milne's Corrector Formula:
The corrector formula is given by:
[tex]\[ y_c = y_{n-1} + \frac{h}{3} \left(f_{n-1} + 4f_p + f_n\right) \][/tex]
Calculate [tex]\(f_p = \frac{dy}{dx}(1, y_p)\)[/tex]:
[tex]\[ f_p = \frac{dy}{dx}(1, 0.9825333) = -\frac{1}{6(0.9825333 + 1)} = -\frac{1}{6(1.9825333)} \approx -0.0840 \][/tex]
Plugging into the corrector formula:
[tex]\[ y_c = 1.1540 + \frac{0.25}{3} \left(-0.0403 + 4(-0.0840) + (-0.0204)\right) \][/tex]
[tex]\[ y_c = 1.1540 + \frac{0.25}{3} \left(-0.0403 - 0.336 + (-0.0204)\right) \][/tex]
[tex]\[ y_c = 1.1540 + \frac{0.25}{3} \left(-0.3967\right) \][/tex]
[tex]\[ y_c = 1.1540 - 0.0331 \approx 1.1209 \][/tex]
The predicted value ([tex]\(y_p\)[/tex]) and the corrected value ([tex]\(y_c\)[/tex]) for [tex]\(x = 1\)[/tex] are:
[tex]\[ \boxed{y_p \approx 0.9636 \quad \text{and} \quad y_c \approx 1.1074} \][/tex]
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