Find y(0.5) for `y'=-2x-y`, `x_0=0, y_0=-1`, with step length 0.1 using Euler method (1st order derivative)

Solution:
Given `y'=-2x-y, y(0)=-1, h=0.1, y(0.5)=?`

Euler method
`y_1=y_0+hf(x_0,y_0)=-1+(0.1)f(0,-1)=-1+(0.1)*(1)=-1+(0.1)=-0.9`

`y_2=y_1+hf(x_1,y_1)=-0.9+(0.1)f(0.1,-0.9)=-0.9+(0.1)*(0.7)=-0.9+(0.07)=-0.83`

`y_3=y_2+hf(x_2,y_2)=-0.83+(0.1)f(0.2,-0.83)=-0.83+(0.1)*(0.43)=-0.83+(0.043)=-0.787`

`y_4=y_3+hf(x_3,y_3)=-0.787+(0.1)f(0.3,-0.787)=-0.787+(0.1)*(0.187)=-0.787+(0.0187)=-0.7683`

`y_5=y_4+hf(x_4,y_4)=-0.7683+(0.1)f(0.4,-0.7683)=-0.7683+(0.1)*(-0.0317)=-0.7683+(-0.0032)=-0.7715`

`:.y(0.5)=-0.7715`

`n`	`x_n`	`y_n`	`x_(n+1)`	`y_(n+1)`
0	0	-1	0.1	-0.9
1	0.1	-0.9	0.2	-0.83
2	0.2	-0.83	0.3	-0.787
3	0.3	-0.787	0.4	-0.7683
4	0.4	-0.7683	0.5	-0.7715

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