Find y(0.5) for `y'=-2x-y`, `x_0=0, y_0=-1`, with step length 0.1 using Runge-Kutta 2 method (1st order derivative)

Solution:
Given `y'=-2x-y, y(0)=-1, h=0.1, y(0.5)=?`

Method-1 : Using formula `k_2=f(x_0+h,y_0+hk_1)`

Second order R-K method
`k_1=f(x_0,y_0)=f(0,-1)=1`

`k_2=f(x_0+h,y_0+hk_1)=f(0.1,-0.9)=0.7`

`y_1=y_0+h/2(k_1+k_2)=-1+0.085=-0.915`

`:.y(0.1)=-0.915`

---

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,-0.915)=0.715`

`k_2=f(x_1+h,y_1+hk_1)=f(0.2,-0.8435)=0.4435`

`y_2=y_1+h/2(k_1+k_2)=-0.915+0.0579=-0.8571`

`:.y(0.2)=-0.8571`

---

Again taking `(x_2,y_2)` in place of `(x_1,y_1)` and repeat the process

`k_1=f(x_2,y_2)=f(0.2,-0.8571)=0.4571`

`k_2=f(x_2+h,y_2+hk_1)=f(0.3,-0.8114)=0.2114`

`y_3=y_2+h/2(k_1+k_2)=-0.8571+0.0334=-0.8237`

`:.y(0.3)=-0.8237`

---

Again taking `(x_3,y_3)` in place of `(x_2,y_2)` and repeat the process

`k_1=f(x_3,y_3)=f(0.3,-0.8237)=0.2237`

`k_2=f(x_3+h,y_3+hk_1)=f(0.4,-0.8013)=0.0013`

`y_4=y_3+h/2(k_1+k_2)=-0.8237+0.0112=-0.8124`

`:.y(0.4)=-0.8124`

---

Again taking `(x_4,y_4)` in place of `(x_3,y_3)` and repeat the process

`k_1=f(x_4,y_4)=f(0.4,-0.8124)=0.0124`

`k_2=f(x_4+h,y_4+hk_1)=f(0.5,-0.8112)=-0.1888`

`y_5=y_4+h/2(k_1+k_2)=-0.8124-0.0088=-0.8212`

`:.y(0.5)=-0.8212`

---

`:.y(0.5)=-0.8212`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`x_(n+1)`	`y_(n+1)`
0	0	-1	1	0.7	0.1	-0.915
1	0.1	-0.915	0.715	0.4435	0.2	-0.8571
2	0.2	-0.8571	0.4571	0.2114	0.3	-0.8237
3	0.3	-0.8237	0.2237	0.0013	0.4	-0.8124
4	0.4	-0.8124	0.0124	-0.1888	0.5	-0.8212

---

Method-2 : Using formula `k_2=f(x_0+h/2,y_0+(hk_1)/2)`

Second 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)=0.85`

`y_1=y_0+hk_2=-1+0.085=-0.915`

`:.y(0.1)=-0.915`

---

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,-0.915)=0.715`

`k_2=f(x_1+h/2,y_1+(hk_1)/2)=f(0.15,-0.8792)=0.5792`

`y_2=y_1+hk_2=-0.915+0.0579=-0.8571`

`:.y(0.2)=-0.8571`

---

Again taking `(x_2,y_2)` in place of `(x_1,y_1)` and repeat the process

`k_1=f(x_2,y_2)=f(0.2,-0.8571)=0.4571`

`k_2=f(x_2+h/2,y_2+(hk_1)/2)=f(0.25,-0.8342)=0.3342`

`y_3=y_2+hk_2=-0.8571+0.0334=-0.8237`

`:.y(0.3)=-0.8237`

---

Again taking `(x_3,y_3)` in place of `(x_2,y_2)` and repeat the process

`k_1=f(x_3,y_3)=f(0.3,-0.8237)=0.2237`

`k_2=f(x_3+h/2,y_3+(hk_1)/2)=f(0.35,-0.8125)=0.1125`

`y_4=y_3+hk_2=-0.8237+0.0112=-0.8124`

`:.y(0.4)=-0.8124`

---

Again taking `(x_4,y_4)` in place of `(x_3,y_3)` and repeat the process

`k_1=f(x_4,y_4)=f(0.4,-0.8124)=0.0124`

`k_2=f(x_4+h/2,y_4+(hk_1)/2)=f(0.45,-0.8118)=-0.0882`

`y_5=y_4+hk_2=-0.8124-0.0088=-0.8212`

`:.y(0.5)=-0.8212`

---

`:.y(0.5)=-0.8212`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`x_(n+1)`	`y_(n+1)`
0	0	-1	1	0.85	0.1	-0.915
1	0.1	-0.915	0.715	0.5792	0.2	-0.8571
2	0.2	-0.8571	0.4571	0.3342	0.3	-0.8237
3	0.3	-0.8237	0.2237	0.1125	0.4	-0.8124
4	0.4	-0.8124	0.0124	-0.0882	0.5	-0.8212

---

---