1. Find y(0.2) for `y'=(x-y)/2`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 2 method (1st order derivative)

Solution:
Given `y'=(x-y)/(2), y(0)=1, h=0.1, y(0.2)=?`

Method-1 : Using formula `k_2=hf(x_0+h,y_0+k_1)`

Second order Runge-Kutta (RK2) method formula
`k_1=hf(x_n,y_n)`

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

`y_(n+1)=y_n+(k_1+k_2)/2`

---

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

`k_1=hf(x_0,y_0)`

`=(0.1)f(0,1)`

`=(0.1)*(-0.5)`

`=-0.05`

`k_2=hf(x_0+h,y_0+k_1)`

`=(0.1)f(0.1,0.95)`

`=(0.1)*(-0.425)`

`=-0.0425`

`y_1=y_0+(k_1+k_2)/2`

`=1-0.0462`

`=0.9538`

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

---

for `n=1,x_1=0.1,y_1=0.9538`

`k_1=hf(x_1,y_1)`

`=(0.1)f(0.1,0.9538)`

`=(0.1)*(-0.4269)`

`=-0.0427`

`k_2=hf(x_1+h,y_1+k_1)`

`=(0.1)f(0.2,0.9111)`

`=(0.1)*(-0.3555)`

`=-0.0356`

`y_2=y_1+(k_1+k_2)/2`

`=0.9538-0.0391`

`=0.9146`

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

`:.y(0.2)=0.9146`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`x_(n+1)`	`y_(n+1)`
0	0	1	-0.05	-0.0425	0.1	0.9538
1	0.1	0.9538	-0.0427	-0.0356	0.2	0.9146

---

Method-2 : Using formula `k_2=hf(x_0+h/2,y_0+k_1/2)`

Second order Runge-Kutta (RK2) method formula
`k_1=hf(x_n,y_n)`

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

`y_(n+1)=y_n+k_2`

---

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

`k_1=hf(x_0,y_0)`

`=(0.1)f(0,1)`

`=(0.1)*(-0.5)`

`=-0.05`

`k_2=hf(x_0+h/2,y_0+k_1/2)`

`=(0.1)f(0.05,0.975)`

`=(0.1)*(-0.4625)`

`=-0.0462`

`y_1=y_0+k_2`

`=1-0.0462`

`=0.9538`

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

---

for `n=1,x_1=0.1,y_1=0.9538`

`k_1=hf(x_1,y_1)`

`=(0.1)f(0.1,0.9538)`

`=(0.1)*(-0.4269)`

`=-0.0427`

`k_2=hf(x_1+h/2,y_1+k_1/2)`

`=(0.1)f(0.15,0.9324)`

`=(0.1)*(-0.3912)`

`=-0.0391`

`y_2=y_1+k_2`

`=0.9538-0.0391`

`=0.9146`

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

`:.y(0.2)=0.9146`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`x_(n+1)`	`y_(n+1)`
0	0	1	-0.05	-0.0462	0.1	0.9538
1	0.1	0.9538	-0.0427	-0.0391	0.2	0.9146