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

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

Euler method

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

`y_1=y_0+hf(x_0,y_0)`

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

`=-1+(0.1)*(1)`

`=-1+(0.1)`

`=-0.9`

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

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for `n=1,x_1=0.1,y_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`

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

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for `n=2,x_2=0.2,y_2=-0.83`

`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`

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

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for `n=3,x_3=0.3,y_3=-0.787`

`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`

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

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for `n=4,x_4=0.4,y_4=-0.7683`

`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`

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

`:.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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