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[⎕]∇
    ∇
[0]   Z←R(Y¯ RK4)Y;T;YN;TN;∆T;∆Y1;∆Y2;∆Y3;∆Y4
[1]   (T R ∆T)←R
[2]  LOOP:→(R≤TN←¯1↑T)/EXIT
[3]   ∆Y1←∆T×TN Y¯ YN←¯1↑Y
[4]   ∆Y2←∆T×(TN+∆T÷2)Y¯ YN+∆Y1÷2
[5]   ∆Y3←∆T×(TN+∆T÷2)Y¯ YN+∆Y2÷2
[6]   ∆Y4←∆T×(TN+∆T)Y¯ YN+∆Y3
[7]   Y←Y,YN+(∆Y1+(2×∆Y2)+(2×∆Y3)+∆Y4)÷6
[8]   T←T,TN+∆T
[9]   →LOOP
[10] EXIT:Z←T,[⎕IO+.5]Y
    ∇
 
      ∇PRINT[⎕]∇
    ∇
[0]   PRINT;TABLE
[1]   TABLE←0 10 .1({⍺×⍵*.5}RK4)1
[2]   ⎕←'T' 'RK4 Y' 'ERROR'⍪TABLE,TABLE[;2]-{((4+⍵*2)*2)÷16}TABLE[;1]
    ∇
 