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

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

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I differentiated to find $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ and substituted values into the equation at the point $\displaystyle x_1$ and used to find $\displaystyle x_2$. However, I got the answer to be $\displaystyle 2.22262625 = 2.223 $ to $\displaystyle 3$ decimal places but the answer is $\displaystyle 2.222$. 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.