Given the example Differential equation:






y
′

(
t
)
=
t
×


y
(
t
)




{\displaystyle y'(t)=t\times {\sqrt {y(t)}}}


With initial condition:






t

0


=
0


{\displaystyle t_{0}=0}

 and 




y

0


=
y
(

t

0


)
=
y
(
0
)
=
1


{\displaystyle y_{0}=y(t_{0})=y(0)=1}


This equation has an exact solution:





y
(
t
)
=



1
16



(

t

2


+
4

)

2




{\displaystyle y(t)={\tfrac {1}{16}}(t^{2}+4)^{2}}




Task
Demonstrate the commonly used explicit   fourth-order Runge–Kutta method   to solve the above differential equation.

 Solve the given differential equation over the range 



t
=
0
…
10


{\displaystyle t=0\ldots 10}

 with a step value of 



δ
t
=
0.1


{\displaystyle \delta t=0.1}

 (101 total points, the first being given)
 Print the calculated values of 



y


{\displaystyle y}

 at whole numbered 



t


{\displaystyle t}

's (



0.0
,
1.0
,
…
10.0


{\displaystyle 0.0,1.0,\ldots 10.0}

) along with error as compared to the exact solution.


Method summary
Starting with a given 




y

n




{\displaystyle y_{n}}

 and 




t

n




{\displaystyle t_{n}}

 calculate:





δ

y

1


=
δ
t
×

y
′

(

t

n


,

y

n


)



{\displaystyle \delta y_{1}=\delta t\times y'(t_{n},y_{n})\quad }






δ

y

2


=
δ
t
×

y
′

(

t

n


+



1
2



δ
t
,

y

n


+



1
2



δ

y

1


)


{\displaystyle \delta y_{2}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{1})}






δ

y

3


=
δ
t
×

y
′

(

t

n


+



1
2



δ
t
,

y

n


+



1
2



δ

y

2


)


{\displaystyle \delta y_{3}=\delta t\times y'(t_{n}+{\tfrac {1}{2}}\delta t,y_{n}+{\tfrac {1}{2}}\delta y_{2})}






δ

y

4


=
δ
t
×

y
′

(

t

n


+
δ
t
,

y

n


+
δ

y

3


)



{\displaystyle \delta y_{4}=\delta t\times y'(t_{n}+\delta t,y_{n}+\delta y_{3})\quad }


then:






y

n
+
1


=

y

n


+



1
6



(
δ

y

1


+
2
δ

y

2


+
2
δ

y

3


+
δ

y

4


)


{\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{6}}(\delta y_{1}+2\delta y_{2}+2\delta y_{3}+\delta y_{4})}







t

n
+
1


=

t

n


+
δ
t



{\displaystyle t_{n+1}=t_{n}+\delta t\quad }




rk4 <- function(f, x0, y0, x1, n) {
    vx <- double(n + 1)
    vy <- double(n + 1)
    vx[1] <- x <- x0
    vy[1] <- y <- y0
    h <- (x1 - x0)/n
    for(i in 1:n) {
        k1 <- h*f(x, y)
        k2 <- h*f(x + 0.5*h, y + 0.5*k1)
        k3 <- h*f(x + 0.5*h, y + 0.5*k2)
        k4 <- h*f(x + h, y + k3)
        vx[i + 1] <- x <- x0 + i*h
        vy[i + 1] <- y <- y + (k1 + k2 + k2 + k3 + k3 + k4)/6
    }
    cbind(vx, vy)
}
 
sol <- rk4(function(x, y) x*sqrt(y), 0, 1, 10, 100)
cbind(sol, sol[, 2] - (4 + sol[, 1]^2)^2/16)[seq(1, 101, 10), ]
 
      vx         vy              
 [1,]  0   1.000000  0.000000e+00
 [2,]  1   1.562500 -1.457219e-07
 [3,]  2   3.999999 -9.194792e-07
 [4,]  3  10.562497 -2.909562e-06
 [5,]  4  24.999994 -6.234909e-06
 [6,]  5  52.562489 -1.081970e-05
 [7,]  6  99.999983 -1.659460e-05
 [8,]  7 175.562476 -2.351773e-05
 [9,]  8 288.999968 -3.156520e-05
[10,]  9 451.562459 -4.072316e-05
[11,] 10 675.999949 -5.098329e-05