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 }




import static java.lang.Math.*;
import java.util.function.BiFunction;
 
public class RungeKutta {
 
    static void runge(BiFunction<Double, Double, Double> yp_func, double[] t,
            double[] y, double dt) {
 
        for (int n = 0; n < t.length - 1; n++) {
            double dy1 = dt * yp_func.apply(t[n], y[n]);
            double dy2 = dt * yp_func.apply(t[n] + dt / 2.0, y[n] + dy1 / 2.0);
            double dy3 = dt * yp_func.apply(t[n] + dt / 2.0, y[n] + dy2 / 2.0);
            double dy4 = dt * yp_func.apply(t[n] + dt, y[n] + dy3);
            t[n + 1] = t[n] + dt;
            y[n + 1] = y[n] + (dy1 + 2.0 * (dy2 + dy3) + dy4) / 6.0;
        }
    }
 
    static double calc_err(double t, double calc) {
        double actual = pow(pow(t, 2.0) + 4.0, 2) / 16.0;
        return abs(actual - calc);
    }
 
    public static void main(String[] args) {
        double dt = 0.10;
        double[] t_arr = new double[101];
        double[] y_arr = new double[101];
        y_arr[0] = 1.0;
 
        runge((t, y) -> t * sqrt(y), t_arr, y_arr, dt);
 
        for (int i = 0; i < t_arr.length; i++)
            if (i % 10 == 0)
                System.out.printf("y(%.1f) = %.8f Error: %.6f%n",
                        t_arr[i], y_arr[i],
                        calc_err(t_arr[i], y_arr[i]));
    }
}