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 }




#include <stdio.h>
#include <stdlib.h>
#include <math.h>
 
double rk4(double(*f)(double, double), double dx, double x, double y)
{
	double	k1 = dx * f(x, y),
		k2 = dx * f(x + dx / 2, y + k1 / 2),
		k3 = dx * f(x + dx / 2, y + k2 / 2),
		k4 = dx * f(x + dx, y + k3);
	return y + (k1 + 2 * k2 + 2 * k3 + k4) / 6;
}
 
double rate(double x, double y)
{
	return x * sqrt(y);
}
 
int main(void)
{
	double *y, x, y2;
	double x0 = 0, x1 = 10, dx = .1;
	int i, n = 1 + (x1 - x0)/dx;
	y = (double *)malloc(sizeof(double) * n);
 
	for (y[0] = 1, i = 1; i < n; i++)
		y[i] = rk4(rate, dx, x0 + dx * (i - 1), y[i-1]);
 
	printf("x\ty\trel. err.\n------------\n");
	for (i = 0; i < n; i += 10) {
		x = x0 + dx * i;
		y2 = pow(x * x / 4 + 1, 2);
		printf("%g\t%g\t%g\n", x, y[i], y[i]/y2 - 1);
	}
 
	return 0;
}