Question

Solve y′ = x−y^2, y(0) = 1 using the forward Euler method for in [0, 0.6] by taking h = 0.2.

Answer 1

By Euler's formula we know that , $y_{n+1}=y_n+h.f(x_n,y_n)$ .........(1)

Here given that $f(x,y)=x-y^2$ with $y(0)=1.$

$\therefore x_0=0$ and $y_0=1$

Also given that $h=0.2$ . We have to find the value of $y$ in the interval $[0,0.6]$ .

We also know that, $x_{n+1}=x_n+h$

Therefore, $x_1=0.2$ for $n=0$

$x_2=0.4$ for $n=1$

$x_3=0.6$ for $n=2$

Now corresponding to these values of $x_1,x_2,x_3$ we get the values of $y_1,y_2,y_3$ with the help of (1).

For $n=0$ , $y_1=y_0+h.f(x_0,y_0)$

$=1+(0.2)×[0-1^2]$

$=1-0.2=0.8$

For $n=1,$ $y_2=y_1+h.f(x_1,y_1)$

$=0.8+(0.2)×[0.2-(0.8)^2]$

$=(0.8-0.088)=0.712$

For $n=2,$ $y_3=y_2+h.f(x_2,y_2)$

$=0.712+(0.2)×[0.4-(0.712)^2]$

$=0.712-0.0213838$

$=0.6906$

Therefore the required solutions in $[0,0.6]$ are , $y(0)=1$ , $y(0.2)=0.8$ , $y(0.4)=0.712$ and $y(0.6)=0.6906$ .