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)*0.1=0\end{cases}`

`\begin{cases}x_2=x_1+0.1=0.2\\y_2=y_1+f(x_1,y_1)*0.1=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)*0.05=0\end{cases}`

`\begin{cases}x_2=x_1+0.05=0.1\\y_2=y_1+f(x_1,y_1)*0.05=0.0025\end{cases}`

`\begin{cases}x_3=x_2+0.05=0.15\\y_3=y_2+f(x_2,y_2)*0.05=0.0075003125\end{cases}`

so y (0.2) ≈ 0.015.