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Math Help - Numerical Method

  1. #1
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    Numerical Method

    Q: Use the formula (\frac{\mathrm{d^2}y}{\mathrm{d}x^2})_0 \approx \frac{y_1 - 2y_0 + y_{-1}}{h^2} and (\frac{\mathrm{d^2}y}{\mathrm{d}x})_0 \approx \frac{y_1 - y_{-1}}{2h} where necessary.

    \frac{\mathrm{d^2}y}{\mathrm{d}x^2} = 1 + \frac{1}{2} \frac{\mathrm{d}y}{\mathrm{d}x},  y = 2 at x = 0 and  y = 2.1 at x = 0.1. Use step length of 0.1 (=h) to estimate the value at x=0.2 to 3 decimal places.


    _______________________________________________

    I differentiated to find \frac{\mathrm{d}y}{\mathrm{d}x} and substituted values into the equation at the point x_1 and used to find x_2. However, I got the answer to be 2.22262625 = 2.223 to 3 decimal places but the answer is 2.222. Can someone try this question and see if they get correct answer. If possible, can you post the method so I can compare with mine and spot me mistake. Thanks in advance.
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  2. #2
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    [Heinemann FP3 Chapter 5A Question 17]

    Air you made a mistake in posting the question.

    It should be \frac{\mathrm{d^2}y}{\mathrm{d}x^2} = 1 + \frac{1}{2}  y \frac{\mathrm{d}y}{\mathrm{d}x}


    Okay i did the question once, got the correct answer but made a mistake in my working.

    Initially i made the mistake of doing
     \left ( \frac{d^{2}y}{dx^{2}} \right )_1 \approx 1 + \frac{1}{2} y_2   \left ( \frac{dy}{dx} \right )_1

    \frac{y_2 - 2y_1 + y_{0}}{0.1^2} \approx 1 + \frac{1}{2} y_2 \left( \frac{y_2 - y_{0}}{0.2} \right)

    then subsituting the values into the equation. solved the quad and got 2.222.
    did you notice the error though ?

    Here is my solution, not guarantee that is is correct.

    \frac{y_2 - 2y_1 + y_{0}}{0.1^2} \approx 1 + \frac{1}{2} y_1 \left( \frac{y_2 - y_{0}}{0.2} \right)

    then sub the values

    \frac{y_2 - 2 \cdot 2.1  +2}{0.1^2} \approx 1 + \frac{1}{2} 2.1  \left( \frac{y_2 - 2}{0.2} \right)

    giving y_2 = 2.127337

    which disagrees with everybody.
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