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

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 Runge-Kutta (RK2) method formula
`k_1=f(x_n,y_n)`

`k_2=f(x_n+h,y_n+hk_1)`

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

---

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

`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`

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

---

for `n=1,x_1=0.1,y_1=-0.915`

`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`

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

---

for `n=2,x_2=0.2,y_2=-0.8571`

`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`

`x_3=x_2+h=0.2+0.1=0.3`

---

for `n=3,x_3=0.3,y_3=-0.8237`

`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`

`x_4=x_3+h=0.3+0.1=0.4`

---

for `n=4,x_4=0.4,y_4=-0.8124`

`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`

`x_5=x_4+h=0.4+0.1=0.5`

`:.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 Runge-Kutta (RK2) method formula
`k_1=f(x_n,y_n)`

`k_2=f(x_n+h/2,y_n+(hk_1)/2)`

`y_(n+1)=y_n+hk_2`

---

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

`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`

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

---

for `n=1,x_1=0.1,y_1=-0.915`

`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`

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

---

for `n=2,x_2=0.2,y_2=-0.8571`

`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`

`x_3=x_2+h=0.2+0.1=0.3`

---

for `n=3,x_3=0.3,y_3=-0.8237`

`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`

`x_4=x_3+h=0.3+0.1=0.4`

---

for `n=4,x_4=0.4,y_4=-0.8124`

`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`

`x_5=x_4+h=0.4+0.1=0.5`

`:.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

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