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 }




def calc_rk4(f)
  return ->(t,y,dt){
         ->(dy1   ){
         ->(dy2   ){
         ->(dy3   ){
         ->(dy4   ){ ( dy1 + 2*dy2 + 2*dy3 + dy4 ) / 6 }.call(
           dt * f.call( t + dt  , y + dy3   ))}.call(
           dt * f.call( t + dt/2, y + dy2/2 ))}.call(
           dt * f.call( t + dt/2, y + dy1/2 ))}.call(
           dt * f.call( t       , y         ))}
end
 
TIME_MAXIMUM, WHOLE_TOLERANCE = 10.0, 1.0e-5
T_START, Y_START, DT          =  0.0, 1.0, 0.10
 
def my_diff_eqn(t,y) ; t * Math.sqrt(y)                    ; end
def my_solution(t  ) ; (t**2 + 4)**2 / 16                  ; end
def  find_error(t,y) ; (y - my_solution(t)).abs            ; end
def   is_whole?(t  ) ; (t.round - t).abs < WHOLE_TOLERANCE ; end
 
dy = calc_rk4( ->(t,y){my_diff_eqn(t,y)} )
 
t, y = T_START, Y_START
while t <= TIME_MAXIMUM
  printf("y(%4.1f)\t= %12.6f \t error: %12.6e\n",t,y,find_error(t,y)) if is_whole?(t)
  t, y = t + DT, y + dy.call(t,y,DT)
end