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

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

Fourth order Runge-Kutta (RK4) method formula
`k_1=f(x_n,y_n)`

`k_2=f(x_n+h/2,y_n+(hk_1)/2)`

`k_3=f(x_n+h/2,y_n+(hk_2)/2)`

`k_4=f(x_n+h,y_n+hk_3)`

`y_(n+1)=y_n+h/6(k_1+2k_2+2k_3+k_4)`

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

`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)`

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

`=0.9048`

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

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for `n=1,x_1=0.1,y_1=0.9048`

`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)`

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

`=0.8187`

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

`:.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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