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

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`

Modified Euler method
`y_(m+1)=y_m+hf(x_m+1/2 h,y_m + 1/2 hf(x_m,y_m))`

`f(x_0,y_0)=f(0,-1)=1`

`x_0+1/2 h=0+0.1/2 =0.05`

`y_0 + 1/2 hf(x_0,y_0)=-1+0.1/2 * 1=-0.95`

`f(x_0+1/2 h, y_0 + 1/2 hf(x_0,y_0)=f(0.05,-0.95)=0.85`

`y_1=y_0+hf(x_0+1/2 h, y_0 + 1/2 hf(x_0,y_0))=-1+0.1*0.85=-0.915`

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Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process

`f(x_1,y_1)=f(0.1,-0.915)=0.715`

`x_1+1/2 h=0.1+0.1/2 =0.15`

`y_1 + 1/2 hf(x_1,y_1)=-0.915+0.1/2 * 0.715=-0.8792`

`f(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1)=f(0.15,-0.8792)=0.5792`

`y_2=y_1+hf(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1))=-0.915+0.1*0.5792=-0.8571`

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Again taking `(x_2,y_2)` in place of `(x_1,y_1)` and repeat the process

`f(x_2,y_2)=f(0.2,-0.8571)=0.4571`

`x_2+1/2 h=0.2+0.1/2 =0.25`

`y_2 + 1/2 hf(x_2,y_2)=-0.8571+0.1/2 * 0.4571=-0.8342`

`f(x_2+1/2 h, y_2 + 1/2 hf(x_2,y_2)=f(0.25,-0.8342)=0.3342`

`y_3=y_2+hf(x_2+1/2 h, y_2 + 1/2 hf(x_2,y_2))=-0.8571+0.1*0.3342=-0.8237`

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Again taking `(x_3,y_3)` in place of `(x_2,y_2)` and repeat the process

`f(x_3,y_3)=f(0.3,-0.8237)=0.2237`

`x_3+1/2 h=0.3+0.1/2 =0.35`

`y_3 + 1/2 hf(x_3,y_3)=-0.8237+0.1/2 * 0.2237=-0.8125`

`f(x_3+1/2 h, y_3 + 1/2 hf(x_3,y_3)=f(0.35,-0.8125)=0.1125`

`y_4=y_3+hf(x_3+1/2 h, y_3 + 1/2 hf(x_3,y_3))=-0.8237+0.1*0.1125=-0.8124`

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Again taking `(x_4,y_4)` in place of `(x_3,y_3)` and repeat the process

`f(x_4,y_4)=f(0.4,-0.8124)=0.0124`

`x_4+1/2 h=0.4+0.1/2 =0.45`

`y_4 + 1/2 hf(x_4,y_4)=-0.8124+0.1/2 * 0.0124=-0.8118`

`f(x_4+1/2 h, y_4 + 1/2 hf(x_4,y_4)=f(0.45,-0.8118)=-0.0882`

`y_5=y_4+hf(x_4+1/2 h, y_4 + 1/2 hf(x_4,y_4))=-0.8124+0.1*-0.0882=-0.8212`

`:.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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