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

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

Third order Runge-Kutta (RK3) method formula
`k_1=hf(x_n,y_n)`

`k_2=hf(x_n+h/2,y_n+k_1/2)`

`k_3=hf(x_n+h,y_n+2k_2-k_1)`

`y_(n+1)=y_n+1/6(k_1+4k_2+k_3)`

---

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

`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,y_0+2k_2-k_1)`

`=(0.1)f(0.1,0.91)`

`=(0.1)*(-0.91)`

`=-0.091`

`y_1=y_0+1/6(k_1+4k_2+k_3)`

`=1+1/6[-0.1+4(-0.095)+(-0.091)]`

`=0.9048`

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

---

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

`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,y_1+2k_2-k_1)`

`=(0.1)f(0.2,0.8234)`

`=(0.1)*(-0.8234)`

`=-0.0823`

`y_2=y_1+1/6(k_1+4k_2+k_3)`

`=0.9048+1/6[-0.0905+4(-0.086)+(-0.0823)]`

`=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`	`x_(n+1)`	`y_(n+1)`
0	0	1	-0.1	-0.095	-0.091	0.1	0.9048
1	0.1	0.9048	-0.0905	-0.086	-0.0823	0.2	0.8187

---

---