Find y(0.5) for `y'=-2x-y`, `x_0=0, y_0=-1`, with step length 0.1 using Runge-Kutta 3 method (first order differential equation)

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

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)*(0.85)`

`=0.085`

`k_3=hf(x_0+h,y_0+2k_2-k_1)`

`=(0.1)f(0.1,-0.93)`

`=(0.1)*(0.73)`

`=0.073`

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

`=-1+1/6[0.1+4(0.085)+(0.073)]`

`=-0.9145`

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

---

for `n=1,x_1=0.1,y_1=-0.9145`

`k_1=hf(x_1,y_1)`

`=(0.1)f(0.1,-0.9145)`

`=(0.1)*(0.7145)`

`=0.0714`

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

`=(0.1)f(0.15,-0.8788)`

`=(0.1)*(0.5788)`

`=0.0579`

`k_3=hf(x_1+h,y_1+2k_2-k_1)`

`=(0.1)f(0.2,-0.8702)`

`=(0.1)*(0.4702)`

`=0.047`

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

`=-0.9145+1/6[0.0714+4(0.0579)+(0.047)]`

`=-0.8562`

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

---

for `n=2,x_2=0.2,y_2=-0.8562`

`k_1=hf(x_2,y_2)`

`=(0.1)f(0.2,-0.8562)`

`=(0.1)*(0.4562)`

`=0.0456`

`k_2=hf(x_2+h/2,y_2+k_1/2)`

`=(0.1)f(0.25,-0.8334)`

`=(0.1)*(0.3334)`

`=0.0333`

`k_3=hf(x_2+h,y_2+2k_2-k_1)`

`=(0.1)f(0.3,-0.8351)`

`=(0.1)*(0.2351)`

`=0.0235`

`y_3=y_2+1/6(k_1+4k_2+k_3)`

`=-0.8562+1/6[0.0456+4(0.0333)+(0.0235)]`

`=-0.8224`

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

---

for `n=3,x_3=0.3,y_3=-0.8224`

`k_1=hf(x_3,y_3)`

`=(0.1)f(0.3,-0.8224)`

`=(0.1)*(0.2224)`

`=0.0222`

`k_2=hf(x_3+h/2,y_3+k_1/2)`

`=(0.1)f(0.35,-0.8113)`

`=(0.1)*(0.1113)`

`=0.0111`

`k_3=hf(x_3+h,y_3+2k_2-k_1)`

`=(0.1)f(0.4,-0.8224)`

`=(0.1)*(0.0224)`

`=0.0022`

`y_4=y_3+1/6(k_1+4k_2+k_3)`

`=-0.8224+1/6[0.0222+4(0.0111)+(0.0022)]`

`=-0.8109`

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

---

for `n=4,x_4=0.4,y_4=-0.8109`

`k_1=hf(x_4,y_4)`

`=(0.1)f(0.4,-0.8109)`

`=(0.1)*(0.0109)`

`=0.0011`

`k_2=hf(x_4+h/2,y_4+k_1/2)`

`=(0.1)f(0.45,-0.8104)`

`=(0.1)*(-0.0896)`

`=-0.009`

`k_3=hf(x_4+h,y_4+2k_2-k_1)`

`=(0.1)f(0.5,-0.8299)`

`=(0.1)*(-0.1701)`

`=-0.017`

`y_5=y_4+1/6(k_1+4k_2+k_3)`

`=-0.8109+1/6[0.0011+4(-0.009)+(-0.017)]`

`=-0.8196`

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

`:.y(0.5)=-0.8196`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`k_3`	`x_(n+1)`	`y_(n+1)`
0	0	-1	0.1	0.085	0.073	0.1	-0.9145
1	0.1	-0.9145	0.0714	0.0579	0.047	0.2	-0.8562
2	0.2	-0.8562	0.0456	0.0333	0.0235	0.3	-0.8224
3	0.3	-0.8224	0.0222	0.0111	0.0022	0.4	-0.8109
4	0.4	-0.8109	0.0011	-0.009	-0.017	0.5	-0.8196

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