Question

‏In Problems 1 and 2 use Euler's method to obtain a four decimal approximation of the indicated value . Carry out the recursion of ( 3 ) by hand , first using h = 0.1 and then using h = 0.05 .
y’ = x + y^2 , y(0); y(0.2)

Answer 1

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,

$\begin{cases}x_0=0\\x_n=x_{n-1}+h&\text{for }n>0\end{cases}$

$\begin{cases}y_0=y(x_0)=0\\y_n=y_{n-1}+f(x_{n-1},y_{n-1})h&\text{for }n>0\end{cases}$

where h is the step size.

We then approximate y (0.2) in ...

• ... 2 steps for h = 0.1 :

$\begin{cases}x_0=0\\y_0=0\end{cases}$

$\begin{cases}x_1=x_0+0.1=0.1\\y_1=y_0+f(x_0,y_0)\times0.1=0\end{cases}$

$\begin{cases}x_2=x_1+0.1=0.2\\y_2=y_1+f(x_1,y_1)\times0.1=0.01\end{cases}$

so that y (0.2) ≈ 0.01;

• ... 4 steps for h = 0.05 :

$\begin{cases}x_0=0\\y_0=0\end{cases}$

$\begin{cases}x_1=x_0+0.05=0.05\\y_1=y_0+f(x_0,y_0)\times0.05=0\end{cases}$

$\begin{cases}x_2=x_1+0.05=0.1\\y_2=y_1+f(x_1,y_1)\times0.05=0.0025\end{cases}$

$\begin{cases}x_3=x_2+0.05=0.15\\y_3=y_2+f(x_2,y_2)\times0.05=0.0075003125\end{cases}$

$\begin{cases}x_4=x_3+0.05=0.2\\y_4=y_3+f(x_3,y_3)\times0.05=0.0150031252343798828125$

so y (0.2) ≈ 0.015.