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=f(x_0+h,y_0+hk_1)`

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

`k_2=f(x_0+h,y_0+hk_1)=f(0.1,0.9)=-0.9`

`y_1=y_0+h/2(k_1+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=f(x_1,y_1)=f(0.1,0.905)=-0.905`

`k_2=f(x_1+h,y_1+hk_1)=f(0.2,0.8145)=-0.8145`

`y_2=y_1+h/2(k_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	-1	-0.9	0.1	0.905
1	0.1	0.905	-0.905	-0.8145	0.2	0.819

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Method-2 : Using formula `k_2=f(x_0+h/2,y_0+(hk_1)/2)`

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

`k_2=f(x_0+h/2,y_0+(hk_1)/2)=f(0.05,0.95)=-0.95`

`y_1=y_0+hk_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=f(x_1,y_1)=f(0.1,0.905)=-0.905`

`k_2=f(x_1+h/2,y_1+(hk_1)/2)=f(0.15,0.8598)=-0.8598`

`y_2=y_1+hk_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	-1	-0.95	0.1	0.905
1	0.1	0.905	-0.905	-0.8598	0.2	0.819

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