1. ## 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.

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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.

2. [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.