Find y(0.5) for `y'=-2x-y`, `x_0=0, y_0=-1`, with step length 0.1 using Runge-Kutta 4 method (1st order derivative)

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

Fourth 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`

`k_3=f(x_0+h/2,y_0+(hk_2)/2)=f(0.05,-0.9575)=0.8575`

`k_4=f(x_0+h,y_0+hk_3)=f(0.1,-0.9142)=0.7143`

`y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)`

`y_1=-1+0.1/6[1+2(0.85)+2(0.8575)+(0.7143)]`

`y_1=-0.9145`

`:.y(0.1)=-0.9145`

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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,-0.9145)=0.7145`

`k_2=f(x_1+h/2,y_1+(hk_1)/2)=f(0.15,-0.8788)=0.5788`

`k_3=f(x_1+h/2,y_1+(hk_2)/2)=f(0.15,-0.8856)=0.5856`

`k_4=f(x_1+h,y_1+hk_3)=f(0.2,-0.856)=0.456`

`y_2=y_1+h/6(k_1+2k_2+2k_3+k_4)`

`y_2=-0.9145+0.1/6[0.7145+2(0.5788)+2(0.5856)+(0.456)]`

`y_2=-0.8562`

`:.y(0.2)=-0.8562`

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Again taking `(x_2,y_2)` in place of `(x_0,y_0)` and repeat the process

`k_1=f(x_2,y_2)=f(0.2,-0.8562)=0.4562`

`k_2=f(x_2+h/2,y_2+(hk_1)/2)=f(0.25,-0.8334)=0.3334`

`k_3=f(x_2+h/2,y_2+(hk_2)/2)=f(0.25,-0.8395)=0.3395`

`k_4=f(x_2+h,y_2+hk_3)=f(0.3,-0.8222)=0.2222`

`y_3=y_2+h/6(k_1+2k_2+2k_3+k_4)`

`y_3=-0.8562+0.1/6[0.4562+2(0.3334)+2(0.3395)+(0.2222)]`

`y_3=-0.8225`

`:.y(0.3)=-0.8225`

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Again taking `(x_3,y_3)` in place of `(x_0,y_0)` and repeat the process

`k_1=f(x_3,y_3)=f(0.3,-0.8225)=0.2225`

`k_2=f(x_3+h/2,y_3+(hk_1)/2)=f(0.35,-0.8113)=0.1113`

`k_3=f(x_3+h/2,y_3+(hk_2)/2)=f(0.35,-0.8169)=0.1169`

`k_4=f(x_3+h,y_3+hk_3)=f(0.4,-0.8108)=0.0108`

`y_4=y_3+h/6(k_1+2k_2+2k_3+k_4)`

`y_4=-0.8225+0.1/6[0.2225+2(0.1113)+2(0.1169)+(0.0108)]`

`y_4=-0.811`

`:.y(0.4)=-0.811`

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Again taking `(x_4,y_4)` in place of `(x_0,y_0)` and repeat the process

`k_1=f(x_4,y_4)=f(0.4,-0.811)=0.011`

`k_2=f(x_4+h/2,y_4+(hk_1)/2)=f(0.45,-0.8104)=-0.0896`

`k_3=f(x_4+h/2,y_4+(hk_2)/2)=f(0.45,-0.8154)=-0.0846`

`k_4=f(x_4+h,y_4+hk_3)=f(0.5,-0.8194)=-0.1806`

`y_5=y_4+h/6(k_1+2k_2+2k_3+k_4)`

`y_5=-0.811+0.1/6[0.011+2(-0.0896)+2(-0.0846)+(-0.1806)]`

`y_5=-0.8196`

`:.y(0.5)=-0.8196`

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`:.y(0.5)=-0.8196`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`k_3`	`k_4`	`x_(n+1)`	`y_(n+1)`
0	0	-1	1	0.85	0.8575	0.7143	0.1	-0.9145
1	0.1	-0.9145	0.7145	0.5788	0.5856	0.456	0.2	-0.8562
2	0.2	-0.8562	0.4562	0.3334	0.3395	0.2222	0.3	-0.8225
3	0.3	-0.8225	0.2225	0.1113	0.1169	0.0108	0.4	-0.811
4	0.4	-0.811	0.011	-0.0896	-0.0846	-0.1806	0.5	-0.8196

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