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

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

Method-1 : Using formula `k_2=hf(x_0+h,y_0+k_1)`

Second order R-K method
`k_1=hf(x_0,y_0)=(0.1)f(0,-1)=(0.1)*(1)=0.1`

`k_2=hf(x_0+h,y_0+k_1)=(0.1)f(0.1,-0.9)=(0.1)*(0.7)=0.07`

`y_1=y_0+(k_1+k_2)/2=-1+0.085=-0.915`

`:.y(0.1)=-0.915`

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

`k_1=hf(x_1,y_1)=(0.1)f(0.1,-0.915)=(0.1)*(0.715)=0.0715`

`k_2=hf(x_1+h,y_1+k_1)=(0.1)f(0.2,-0.8435)=(0.1)*(0.4435)=0.0444`

`y_2=y_1+(k_1+k_2)/2=-0.915+0.0579=-0.8571`

`:.y(0.2)=-0.8571`

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

`k_1=hf(x_2,y_2)=(0.1)f(0.2,-0.8571)=(0.1)*(0.4571)=0.0457`

`k_2=hf(x_2+h,y_2+k_1)=(0.1)f(0.3,-0.8114)=(0.1)*(0.2114)=0.0211`

`y_3=y_2+(k_1+k_2)/2=-0.8571+0.0334=-0.8237`

`:.y(0.3)=-0.8237`

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

`k_1=hf(x_3,y_3)=(0.1)f(0.3,-0.8237)=(0.1)*(0.2237)=0.0224`

`k_2=hf(x_3+h,y_3+k_1)=(0.1)f(0.4,-0.8013)=(0.1)*(0.0013)=0.0001`

`y_4=y_3+(k_1+k_2)/2=-0.8237+0.0112=-0.8124`

`:.y(0.4)=-0.8124`

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

`k_1=hf(x_4,y_4)=(0.1)f(0.4,-0.8124)=(0.1)*(0.0124)=0.0012`

`k_2=hf(x_4+h,y_4+k_1)=(0.1)f(0.5,-0.8112)=(0.1)*(-0.1888)=-0.0189`

`y_5=y_4+(k_1+k_2)/2=-0.8124-0.0088=-0.8212`

`:.y(0.5)=-0.8212`

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`:.y(0.5)=-0.8212`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`x_(n+1)`	`y_(n+1)`
0	0	-1	0.1	0.07	0.1	-0.915
1	0.1	-0.915	0.0715	0.0444	0.2	-0.8571
2	0.2	-0.8571	0.0457	0.0211	0.3	-0.8237
3	0.3	-0.8237	0.0224	0.0001	0.4	-0.8124
4	0.4	-0.8124	0.0012	-0.0189	0.5	-0.8212

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Method-2 : Using formula `k_2=hf(x_0+h/2,y_0+k_1/2)`

Second order R-K method
`k_1=hf(x_0,y_0)=(0.1)f(0,-1)=(0.1)*(1)=0.1`

`k_2=hf(x_0+h/2,y_0+k_1/2)=(0.1)f(0.05,-0.95)=(0.1)*(0.85)=0.085`

`y_1=y_0+k_2=-1+0.085=-0.915`

`:.y(0.1)=-0.915`

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

`k_1=hf(x_1,y_1)=(0.1)f(0.1,-0.915)=(0.1)*(0.715)=0.0715`

`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,-0.8792)=(0.1)*(0.5792)=0.0579`

`y_2=y_1+k_2=-0.915+0.0579=-0.8571`

`:.y(0.2)=-0.8571`

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

`k_1=hf(x_2,y_2)=(0.1)f(0.2,-0.8571)=(0.1)*(0.4571)=0.0457`

`k_2=hf(x_2+h/2,y_2+k_1/2)=(0.1)f(0.25,-0.8342)=(0.1)*(0.3342)=0.0334`

`y_3=y_2+k_2=-0.8571+0.0334=-0.8237`

`:.y(0.3)=-0.8237`

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

`k_1=hf(x_3,y_3)=(0.1)f(0.3,-0.8237)=(0.1)*(0.2237)=0.0224`

`k_2=hf(x_3+h/2,y_3+k_1/2)=(0.1)f(0.35,-0.8125)=(0.1)*(0.1125)=0.0112`

`y_4=y_3+k_2=-0.8237+0.0112=-0.8124`

`:.y(0.4)=-0.8124`

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

`k_1=hf(x_4,y_4)=(0.1)f(0.4,-0.8124)=(0.1)*(0.0124)=0.0012`

`k_2=hf(x_4+h/2,y_4+k_1/2)=(0.1)f(0.45,-0.8118)=(0.1)*(-0.0882)=-0.0088`

`y_5=y_4+k_2=-0.8124-0.0088=-0.8212`

`:.y(0.5)=-0.8212`

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`:.y(0.5)=-0.8212`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`x_(n+1)`	`y_(n+1)`
0	0	-1	0.1	0.085	0.1	-0.915
1	0.1	-0.915	0.0715	0.0579	0.2	-0.8571
2	0.2	-0.8571	0.0457	0.0334	0.3	-0.8237
3	0.3	-0.8237	0.0224	0.0112	0.4	-0.8124
4	0.4	-0.8124	0.0012	-0.0088	0.5	-0.8212

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