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 }




fun step y' (tn,yn) dt =
    let
        val dy1 = dt * y'(tn,yn)
        val dy2 = dt * y'(tn + 0.5 * dt, yn + 0.5 * dy1)
        val dy3 = dt * y'(tn + 0.5 * dt, yn + 0.5 * dy2)
        val dy4 = dt * y'(tn + dt, yn + dy3)
    in
        (tn + dt, yn + (1.0 / 6.0) * (dy1 + 2.0*dy2 + 2.0*dy3 + dy4))
    end
 
(* Suggested test case *)
fun testy' (t,y) =
    t * Math.sqrt y
 
fun testy t =
    (1.0 / 16.0) * Math.pow(Math.pow(t,2.0) + 4.0, 2.0)
 
(* Test-runner that iterates the step function and prints the results. *)
fun test t0 y0 dt steps print_freq y y' =
    let
        fun loop i (tn,yn) =
            if i = steps then ()
            else
                let
                    val (t1,y1) = step y' (tn,yn) dt
                    val y1' = y tn
                    val () = if i mod print_freq = 0 then
                                 (print ("Time: " ^ Real.toString tn ^ "\n");
                                  print ("Exact: " ^ Real.toString y1' ^ "\n");
                                  print ("Approx: " ^ Real.toString yn ^ "\n");
                                  print ("Error: " ^ Real.toString (y1' - yn) ^ "\n\n"))
                             else ()
                 in
                     loop (i+1) (t1,y1)
                end
    in
        loop 0 (t0,y0)
    end
 
(* Run the suggested test case *)
val () = test 0.0 1.0 0.1 101 10 testy testy'