Find y(0.2) for `y'=-y`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 2 method (1st order derivative)

Solution:
Given `y'=-y, y(0)=1, h=0.1, y(0.2)=?`

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)*(-0.9)=-0.09`

`y_1=y_0+(k_1+k_2)/2=1-0.095=0.905`

`:.y(0.1)=0.905`

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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(0.1,0.905)=(0.1)*(-0.905)=-0.0905`

`k_2=hf(x_1+h,y_1+k_1)=(0.1)f(0.2,0.8145)=(0.1)*(-0.8145)=-0.0815`

`y_2=y_1+(k_1+k_2)/2=0.905-0.086=0.819`

`:.y(0.2)=0.819`

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`:.y(0.2)=0.819`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`x_(n+1)`	`y_(n+1)`
0	0	1	-0.1	-0.09	0.1	0.905
1	0.1	0.905	-0.0905	-0.0815	0.2	0.819

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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)*(-0.95)=-0.095`

`y_1=y_0+k_2=1-0.095=0.905`

`:.y(0.1)=0.905`

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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(0.1,0.905)=(0.1)*(-0.905)=-0.0905`

`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,0.8598)=(0.1)*(-0.8598)=-0.086`

`y_2=y_1+k_2=0.905-0.086=0.819`

`:.y(0.2)=0.819`

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`:.y(0.2)=0.819`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`x_(n+1)`	`y_(n+1)`
0	0	1	-0.1	-0.095	0.1	0.905
1	0.1	0.905	-0.0905	-0.086	0.2	0.819

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