I saw on your other question that you mention y (0) = 0. Let
f(x, y) = x + y ²
Now consider the recurrences in Euler's method,
[tex]\begin{cases}x_0=0\\x_n=x_{n-1}+h&\text{for }n>0\end{cases}[/tex]
[tex]\begin{cases}y_0=y(x_0)=0\\y_n=y_{n-1}+f(x_{n-1},y_{n-1})h&\text{for }n>0\end{cases}[/tex]
where h is the step size.
We then approximate y (0.2) in ...
• ... 2 steps for h = 0.1 :
[tex]\begin{cases}x_0=0\\y_0=0\end{cases}[/tex]
[tex]\begin{cases}x_1=x_0+0.1=0.1\\y_1=y_0+f(x_0,y_0)\times0.1=0\end{cases}[/tex]
[tex]\begin{cases}x_2=x_1+0.1=0.2\\y_2=y_1+f(x_1,y_1)\times0.1=0.01\end{cases}[/tex]
so that y (0.2) ≈ 0.01;
• ... 4 steps for h = 0.05 :
[tex]\begin{cases}x_0=0\\y_0=0\end{cases}[/tex]
[tex]\begin{cases}x_1=x_0+0.05=0.05\\y_1=y_0+f(x_0,y_0)\times0.05=0\end{cases}[/tex]
[tex]\begin{cases}x_2=x_1+0.05=0.1\\y_2=y_1+f(x_1,y_1)\times0.05=0.0025\end{cases}[/tex]
[tex]\begin{cases}x_3=x_2+0.05=0.15\\y_3=y_2+f(x_2,y_2)\times0.05=0.0075003125\end{cases}[/tex]
[tex]\begin{cases}x_4=x_3+0.05=0.2\\y_4=y_3+f(x_3,y_3)\times0.05=0.0150031252343798828125[/tex]
so y (0.2) ≈ 0.015.
