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=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`

`k_3=hf(x_0+h/2,y_0+k_2/2)=(0.1)f(0.05,0.9525)=(0.1)*(-0.9525)=-0.0952`

`k_4=hf(x_0+h,y_0+k_3)=(0.1)f(0.1,0.9048)=(0.1)*(-0.9048)=-0.0905`

`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`

`y_1=1+1/6[-0.1+2(-0.095)+2(-0.0952)+(-0.0905)]`

`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=hf(x_1,y_1)=(0.1)f(0.1,0.9048)=(0.1)*(-0.9048)=-0.0905`

`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,0.8596)=(0.1)*(-0.8596)=-0.086`

`k_3=hf(x_1+h/2,y_1+k_2/2)=(0.1)f(0.15,0.8619)=(0.1)*(-0.8619)=-0.0862`

`k_4=hf(x_1+h,y_1+k_3)=(0.1)f(0.2,0.8187)=(0.1)*(-0.8187)=-0.0819`

`y_2=y_1+1/6(k_1+2k_2+2k_3+k_4)`

`y_2=0.9048+1/6[-0.0905+2(-0.086)+2(-0.0862)+(-0.0819)]`

`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	-0.1	-0.095	-0.0952	-0.0905	0.1	0.9048
1	0.1	0.9048	-0.0905	-0.086	-0.0862	-0.0819	0.2	0.8187

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