Find y(0.2) for `y'=-y`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 3 method (1st order derivative)

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

Third 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,y_0+2hk_2-hk_1)=f(0.1,0.91)=-0.91`

`y_1=y_0+h/6(k_1+4k_2+k_3)`

`y_1=1+(0.1)/6[-1+4(-0.95)+(-0.91)]`

`y_1=0.9048`

`:.y(0.1)=0.9048`

---

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,y_1+2hk_2-hk_1)=f(0.2,0.8234)=-0.8234`

`y_2=y_1+h/6(k_1+4k_2+k_3)`

`y_2=0.9048+(0.1)/6[-0.9048+4(-0.8596)+(-0.8234)]`

`y_2=0.8187`

`:.y(0.2)=0.8187`

---

`:.y(0.2)=0.8187`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`k_3`	`x_(n+1)`	`y_(n+1)`
0	0	1	-1	-0.95	-0.91	0.1	0.9048
1	0.1	0.9048	-0.9048	-0.8596	-0.8234	0.2	0.8187

---

---