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

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 Runge-Kutta (RK2) method formula
`k_1=hf(x_n,y_n)`

`k_2=hf(x_n+h,y_n+k_1)`

`y_(n+1)=y_n+(k_1+k_2)/2`

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

`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`

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

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for `n=1,x_1=0.1,y_1=0.905`

`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`

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

`:.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 Runge-Kutta (RK2) method formula
`k_1=hf(x_n,y_n)`

`k_2=hf(x_n+h/2,y_n+k_1/2)`

`y_(n+1)=y_n+k_2`

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

`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`

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

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for `n=1,x_1=0.1,y_1=0.905`

`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`

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

`:.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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