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 }




package require Tcl 8.5
 
# Hack to bring argument function into expression
proc tcl::mathfunc::dy {t y} {upvar 1 dyFn dyFn; $dyFn $t $y}
 
proc rk4step {dyFn y* t* dt} {
    upvar 1 ${y*} y ${t*} t
    set dy1 [expr {$dt * dy($t,       $y)}]
    set dy2 [expr {$dt * dy($t+$dt/2, $y+$dy1/2)}]
    set dy3 [expr {$dt * dy($t+$dt/2, $y+$dy2/2)}]
    set dy4 [expr {$dt * dy($t+$dt,   $y+$dy3)}]
    set y [expr {$y + ($dy1 + 2*$dy2 + 2*$dy3 + $dy4)/6.0}]
    set t [expr {$t + $dt}]
}
 
proc y {t} {expr {($t**2 + 4)**2 / 16}}
proc δy {t y} {expr {$t * sqrt($y)}}
 
proc printvals {t y} {
    set err [expr {abs($y - [y $t])}]
    puts [format "y(%.1f) = %.8f\tError: %.8e" $t $y $err]
}
 
set t 0.0
set y 1.0
set dt 0.1
printvals $t $y
for {set i 1} {$i <= 101} {incr i} {
    rk4step  δy  y t  $dt
    if {$i%10 == 0} {
	printvals $t $y
    }
}