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3. Runge-Kutta 3 method (first order differential equation) example ( Enter your problem )
  1. Formula-1 & Example-1 : `y'=(x-y)/2`
  2. Example-2 : `y'=-2x-y`
  3. Example-3 : `y'=-y`
  4. Formula-2 & Example-1 : `y'=(x-y)/2`
  5. Example-2 : `y'=-2x-y`
  6. Example-3 : `y'=-y`

5. Example-2 : `y'=-2x-y`





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=f(x_n,y_n)`

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

`k_3=f(x_n+h,y_n+2hk_2-hk_1)`

`y_(n+1)=y_n+h/6(k_1+4k_2+k_3)`



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`

`k_3=f(x_0+h,y_0+2hk_2-hk_1)`

`=f(0.1,-0.93)`

`=0.73`

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

`=-1+(0.1)/6[1+4(0.85)+(0.73)]`

`=-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=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,y_1+2hk_2-hk_1)`

`=f(0.2,-0.8702)`

`=0.4702`

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

`=-0.9145+(0.1)/6[0.7145+4(0.5788)+(0.4702)]`

`=-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=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,y_2+2hk_2-hk_1)`

`=f(0.3,-0.8351)`

`=0.2351`

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

`=-0.8562+(0.1)/6[0.4562+4(0.3334)+(0.2351)]`

`=-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=f(x_3,y_3)`

`=f(0.3,-0.8224)`

`=0.2224`

`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,y_3+2hk_2-hk_1)`

`=f(0.4,-0.8224)`

`=0.0224`

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

`=-0.8224+(0.1)/6[0.2224+4(0.1113)+(0.0224)]`

`=-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=f(x_4,y_4)`

`=f(0.4,-0.8109)`

`=0.0109`

`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,y_4+2hk_2-hk_1)`

`=f(0.5,-0.8299)`

`=-0.1701`

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

`=-0.8109+(0.1)/6[0.0109+4(-0.0896)+(-0.1701)]`

`=-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)`
00-110.850.730.1-0.9145
10.1-0.91450.71450.57880.47020.2-0.8562
20.2-0.85620.45620.33340.23510.3-0.8224
30.3-0.82240.22240.11130.02240.4-0.8109
40.4-0.81090.0109-0.0896-0.17010.5-0.8196





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