Question

Given that y' = x+y^2, y(0)=1, find y(0.2), using the backward Euler’s method.

Answer 1

Given $y'=x+y^2, y(0)=1,$ find $y(0.2)$ , using the backward Euler's method.

Solution:

Backward difference approximation for first derivative:

$y'$ $\approx \frac{y_n-y_{n-1}}{h}$ , $h=x_n-x_{n-1}$

$x_0=0, x_1=0.2$

$h=x_n-x_{n-1}=x_1-x_0=0.2-0=0.2$

$y_n=y_{n-1}+hy'_n$ , $y'_n=f(y_n,x_n)=x_n+y_n^2$

$y_0=y(x_0)=y(0)=1;$ $y_1=y(x_1)=y(0.2)$

$y_1=y_0+hy'_1=y_0+h(x_1+y_1^2)=1+0.2(0.2+1^2)=1.24$

Answer: $y(0.2)=1.24.$