Find y(0.5) for `y'=-2x-y`, `x_0=0, y_0=-1`, with step length 0.1 using Improved Euler / Modified 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`

Improved Euler method / Modified Euler method
`y_(n+1)=y_n+h/2 [f(x_n,y_n) + f(x_n+h,y_n+hf(x_n,y_n))]`

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

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

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

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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,y_0+hk_(1y))`

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

`=f(0.1,-0.9)`

`=0.7`

`y_1=y_0+h/2 [k_(1y) + k_(2y)]`

`=-1+0.1/2 [1 + 0.7]`

`=-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,y_1+hk_(1y))`

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

`=f(0.2,-0.8435)`

`=0.4435`

`y_2=y_1+h/2 [k_(1y) + k_(2y)]`

`=-0.915+0.1/2 [0.715 + 0.4435]`

`=-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,y_2+hk_(1y))`

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

`=f(0.3,-0.8114)`

`=0.2114`

`y_3=y_2+h/2 [k_(1y) + k_(2y)]`

`=-0.8571+0.1/2 [0.4571 + 0.2114]`

`=-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,y_3+hk_(1y))`

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

`=f(0.4,-0.8013)`

`=0.0013`

`y_4=y_3+h/2 [k_(1y) + k_(2y)]`

`=-0.8237+0.1/2 [0.2237 + 0.0013]`

`=-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,y_4+hk_(1y))`

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

`=f(0.5,-0.8112)`

`=-0.1888`

`y_5=y_4+h/2 [k_(1y) + k_(2y)]`

`=-0.8124+0.1/2 [0.0124 + -0.1888]`

`=-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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