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 }




/*
 * compiled with:
 * g++ (Debian 8.3.0-6) 8.3.0
 * 
 * g++ -std=c++14 -o rk4 %
 *
 */
# include <iostream>
# include <math.h>
 
auto rk4(double f(double, double))
{
  return [f](double t, double y, double dt) -> double {
    double dy1 { dt * f( t     , y       ) },
           dy2 { dt * f( t+dt/2, y+dy1/2 ) },
           dy3 { dt * f( t+dt/2, y+dy2/2 ) },
           dy4 { dt * f( t+dt  , y+dy3   ) };
    return ( dy1 + 2*dy2 + 2*dy3 + dy4 ) / 6;
  };
}
 
int main(void)
{
  constexpr
    double TIME_MAXIMUM    {  10.0 },
           T_START         {   0.0 },
           Y_START         {   1.0 },
           DT              {   0.1 },
           WHOLE_TOLERANCE { 1e-12 };
 
  auto dy = rk4( [](double t, double y) -> double { return t*sqrt(y); } ) ;
 
  for (
      auto y { Y_START }, t { T_START };
      t <= TIME_MAXIMUM;
      y += dy(t,y,DT), t += DT
      )
    if (ceilf(t)-t < WHOLE_TOLERANCE)
      printf("y(%4.1f)\t=%12.6f \t error: %12.6e\n", t, y, std::fabs(y - pow(t*t+4,2)/16));
 
  return 0;
}