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 }




 
function Runge-Kutta (${function:F}, ${function:y}, $y0, $t0, $dt, $tEnd) {
    function RK ($tn,$yn)  {
        $y1 = $dt*(F -t $tn -y $yn)
        $y2 = $dt*(F -t ($tn + (1/2)*$dt) -y ($yn + (1/2)*$y1))
        $y3 = $dt*(F -t ($tn + (1/2)*$dt) -y ($yn + (1/2)*$y2))
        $y4 = $dt*(F -t ($tn + $dt) -y ($yn + $y3))
        $yn + (1/6)*($y1 + 2*$y2 + 2*$y3 + $y4)
    }
    function time ($t0, $dt, $tEnd)  {
        $end = [MATH]::Floor(($tEnd - $t0)/$dt)
        foreach ($_ in 0..$end) { $_*$dt + $t0 }
    }
    $time, $yn, $t = (time $t0 $dt $tEnd), $y0, 0
    foreach ($tn in $time) {
        if($t -eq $tn) {
            [pscustomobject]@{
                t = "$tn"
                y = "$yn"
                error = "$([MATH]::abs($yn - (y $tn)))"
            }
            $t += 1
        }
        $yn = RK $tn $yn
    }
}
function F ($t,$y)  {
    $t * [MATH]::Sqrt($y)
}
function y ($t)  { 
    (1/16) * [MATH]::Pow($t*$t + 4,2)
}
$y0 = 1
$t0 = 0
$dt = 0.1
$tEnd = 10
Runge-Kutta  F y $y0 $t0  $dt  $tEnd
 