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 }




use fmt;
use math;
 
export fn main() void = {
	rk4_driver(&f, 0.0, 10.0, 1.0, 0.1);
};
 
fn rk4_driver(func: *fn(_: f64, _: f64) f64, t_init: f64, t_final: f64, y_init: f64, h: f64) void = {
	let n = ((t_final - t_init) / h): int;
	let tn: f64 = t_init;
	let yn: f64 = y_init;
	let i: int = 1;
 
	fmt::printfln("{: 2} {: 18} {: 21}", "t", "y(t)", "absolute error")!;
	fmt::printfln("{: 2} {: 18} {: 21}", tn, yn, math::absf64(exact(tn) - yn))!;
 
	for (i <= n; i += 1) {
		yn = rk4(func, tn, yn, h);
		tn = t_init + (i: f64)*h;
 
		if (i % 10 == 0) {
			fmt::printfln("{: 2} {: 18} {: 21}\t", tn, yn, math::absf64(exact(tn) - yn))!;
		};
	};
};
 
fn rk4(func: *fn(_: f64, _: f64) f64, t: f64, y: f64, h: f64) f64 = {
	const k1 = func(t, y);
	const k2 = func(t + 0.5*h, y + 0.5*h*k1);
	const k3 = func(t + 0.5*h, y + 0.5*h*k2);
	const k4 = func(t + h, y + h*k3);
	return y + h/6.0 * (k1 + 2.0*k2 + 2.0*k3 + k4);
};
 
fn f(t: f64, y: f64) f64 = {
	return t * math::sqrtf64(y);
};
 
fn exact(t: f64) f64 = {
	return 1.0/16.0 * math::powf64(t*t + 4.0, 2.0);
};