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)=?`

Forth 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)=-0.09525`

`k_4=hf(x_0+h,y_0+k_3)=(0.1)f(0.1,0.90475)=(0.1)*(-0.90475)=-0.09048`

`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(-0.09525)+(-0.09048)]`

`y_1=0.90484`

`:.y(0.1)=0.90484`

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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(0.1,0.90484)=(0.1)*(-0.90484)=-0.09048`

`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,0.8596)=(0.1)*(-0.8596)=-0.08596`

`k_3=hf(x_1+h/2,y_1+k_2/2)=(0.1)f(0.15,0.86186)=(0.1)*(-0.86186)=-0.08619`

`k_4=hf(x_1+h,y_1+k_3)=(0.1)f(0.2,0.81865)=(0.1)*(-0.81865)=-0.08187`

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

`y_2=0.90484+1/6[-0.09048+2(-0.08596)+2(-0.08619)+(-0.08187)]`

`y_2=0.81873`

`:.y(0.2)=0.81873`

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`:.y(0.2)=0.81873`

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