Find y(0.2) for `y'=-y`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 4 method (1st order derivative)

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

Fourth 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`

`k_3=f(x_0+h/2,y_0+(hk_2)/2)=f(0.05,0.9525)=-0.9525`

`k_4=f(x_0+h,y_0+hk_3)=f(0.1,0.9048)=-0.9048`

`y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)`

`y_1=1+0.1/6[-1+2(-0.95)+2(-0.9525)+(-0.9048)]`

`y_1=0.9048`

`:.y(0.1)=0.9048`

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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(0.1,0.9048)=-0.9048`

`k_2=f(x_1+h/2,y_1+(hk_1)/2)=f(0.15,0.8596)=-0.8596`

`k_3=f(x_1+h/2,y_1+(hk_2)/2)=f(0.15,0.8619)=-0.8619`

`k_4=f(x_1+h,y_1+hk_3)=f(0.2,0.8187)=-0.8187`

`y_2=y_1+h/6(k_1+2k_2+2k_3+k_4)`

`y_2=0.9048+0.1/6[-0.9048+2(-0.8596)+2(-0.8619)+(-0.8187)]`

`y_2=0.8187`

`:.y(0.2)=0.8187`

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`:.y(0.2)=0.8187`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`k_3`	`k_4`	`x_(n+1)`	`y_(n+1)`
0	0	1	-1	-0.95	-0.9525	-0.9048	0.1	0.9048
1	0.1	0.9048	-0.9048	-0.8596	-0.8619	-0.8187	0.2	0.8187

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