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

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

Here, `x_0=0,y_0=-1,h=0.1,x_n=0.5`

`y'=-2x-y`

`:. f(x,y)=-2x-y`

Midpoint Euler method
`y_(n+1)=y_n+hf(x_n+h/2 ,y_n + h/2 f(x_n,y_n))`

Or
`y_(n+1)=y_n+h k_(2y)`

`k_(2y)=f(x_n+h/2,y_n+h/2 k_(1y))`

`k_(1y)=f(x_n,y_n)`

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for `n=0,x_0=0,y_0=-1`

`k_(1y)=f(x_0,y_0)`

`=f(0,-1)`

`=1`

`k_(2y)=f(x_0+h/2,y_0+h/2 k_(1y))`

`=f(0+0.1/2,-1+0.1/2 *1)`

`=f(0.05,-0.95)`

`=0.85`

`y_(1)=y_0+h k_(2y)`

`=-1+0.1*0.85`

`=-0.915`

`x_1=x_0+h=0+0.1=0.1`

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for `n=1,x_1=0.1,y_1=-0.915`

`k_(1y)=f(x_1,y_1)`

`=f(0.1,-0.915)`

`=0.715`

`k_(2y)=f(x_1+h/2,y_1+h/2 k_(1y))`

`=f(0.1+0.1/2,-0.915+0.1/2 *0.715)`

`=f(0.15,-0.8792)`

`=0.5792`

`y_(2)=y_1+h k_(2y)`

`=-0.915+0.1*0.5792`

`=-0.8571`

`x_2=x_1+h=0.1+0.1=0.2`

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for `n=2,x_2=0.2,y_2=-0.8571`

`k_(1y)=f(x_2,y_2)`

`=f(0.2,-0.8571)`

`=0.4571`

`k_(2y)=f(x_2+h/2,y_2+h/2 k_(1y))`

`=f(0.2+0.1/2,-0.8571+0.1/2 *0.4571)`

`=f(0.25,-0.8342)`

`=0.3342`

`y_(3)=y_2+h k_(2y)`

`=-0.8571+0.1*0.3342`

`=-0.8237`

`x_3=x_2+h=0.2+0.1=0.3`

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for `n=3,x_3=0.3,y_3=-0.8237`

`k_(1y)=f(x_3,y_3)`

`=f(0.3,-0.8237)`

`=0.2237`

`k_(2y)=f(x_3+h/2,y_3+h/2 k_(1y))`

`=f(0.3+0.1/2,-0.8237+0.1/2 *0.2237)`

`=f(0.35,-0.8125)`

`=0.1125`

`y_(4)=y_3+h k_(2y)`

`=-0.8237+0.1*0.1125`

`=-0.8124`

`x_4=x_3+h=0.3+0.1=0.4`

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for `n=4,x_4=0.4,y_4=-0.8124`

`k_(1y)=f(x_4,y_4)`

`=f(0.4,-0.8124)`

`=0.0124`

`k_(2y)=f(x_4+h/2,y_4+h/2 k_(1y))`

`=f(0.4+0.1/2,-0.8124+0.1/2 *0.0124)`

`=f(0.45,-0.8118)`

`=-0.0882`

`y_(5)=y_4+h k_(2y)`

`=-0.8124+0.1*-0.0882`

`=-0.8212`

`x_5=x_4+h=0.4+0.1=0.5`

`:.y(0.5)=-0.8212`

`n`	`x_n`	`y_n`	`x_(n+1)`	`y_(n+1)`
0	0	-1	0.1	-0.915
1	0.1	-0.915	0.2	-0.8571
2	0.2	-0.8571	0.3	-0.8237
3	0.3	-0.8237	0.4	-0.8124
4	0.4	-0.8124	0.5	-0.8212

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