Use Euler's method with step size 0.2 to estimate y(1.4), where y(x) is the solution of the initial-value problem y'=x-xy, y(1)=0

Question

Use Euler's method  with step size 0.2 to estimate y(1.4), where  y(x) is the solution of the initial-value problem y'=x-xy, y(1)=0

anonymous · Answer

$ y'=f(x,y)=x-xy $      and the formula is   

$ y_{n+1}=y_{n}+h f(x_{n} , y_{n}) $

anonymous · Answer

oh ok. Thanks

anonymous · Answer

h is your step size =0.2   $ x_{0}=1  \ \ y_{0}=0 $

anonymous · Answer

What's the big picture? What am I trying to do here?

anonymous · Answer

well thats a numerical method for solving differential equations and coming up from taylor series http://en.wikipedia.org/wiki/Numerical_methods_for_ordinary_differential_equations

anonymous · Answer

$$y_1=y_0+0.2f(1,0)$$

anonymous · Answer

what's my $$y_0$$?

anonymous · Answer

y0=0

anonymous · Answer

$$y_1=0+0.2f(1,0)$$
$$y_2=0+0.4f(1,0)$$
$$y_3=0+0.6f(1,0)$$
$$y_4=0+0.8f(1,0)$$
$$y_5=0+0.10f(1,0)$$
$$y_6=0+0.12f(1,0)$$
$$y_7=0+0.14f(1,0)$$
Is this the idea?

anonymous · Answer

Ooops....It's supposed to + y_2 and y_3 and so on and so forth  instead of 0's

anonymous · Answer

y_7=y_6+0.14f(1,0)

anonymous · Answer

oh no
actually   $ x_{n+1}=x_{n}+h $

anonymous · Answer

I wrote the question wrong on here...it's supposed to be 
y'=y+xy

anonymous · Answer

no problem with that
u have to find y(1.4)

1.4=1.2+0.2
     =(1+0.2)+0.2

and note that actually  $ y_{n}=y(x_{n}) $