Items tagged with runge-kutta

dy/dx=sqrt(1+(a*x)+(2*y))

for the case a=1, y=1 and x=0 construct a program for the runge-kutta method of order 2 with formulae as follows where f(x,y)=dy/dx.

k_1=h*f(x_n,y_n)

k_2=h*f(x_n+h,y_n+k_1)

y_(n+1)=y_n+1/2(k_1+k_2).

 

After creating a program obtain value of y correct to 4 decimal places when x=1 for h=0.1 and h =0.05.

How do i implement Runge-Kutta of order 6 for a sytem of boundary value problems on maple

Hello

I am trying to slve the second order differential equation with initial conditions  t0=0.dy/dt=0,y0=10000

-(diff(y, t, t))-9.81+0.563e-3*(0.1832e-2*abs(diff(y, t))+0.51702e-1*abs(diff(y, t))^(3/2)+.4*(diff(y, t))^2) = 0

using 4th order runge kutta.do i need to declare a step parameter like (D(y))(t) = u or is a command that can be applied automatically?

Thanks

 rk4.mw

Dear all, I have been trying to use Runge-Kutta method to plot an approximate solution with the following code. However, although I can get the numerical approximation the plot would not show.

h := .1;

x[0] := 0;

y[0] := 1;

xf := 3;

n := floor(xf/h)

f:= (x,y)->1/(3 y-x-2)

x := x[0]

y := y[0]

for i to n do

k1 := f(x, y);

k2 := f(x+(1/2)*h, y+(1/2)*h*k1);

k3 := f(x+(1/2)*h, y+(1/2)*h*k2);

k4 := f(x+h, h*k3+y);

k := (k1+2*k2+2*k3+k4)*(1/6);

y := h*k+y;

x := x+h

end do;

y[n]

data := [seq([x[n], y[n]], n = 0 .. 30)];

p[2] := plot(data, style = point, color = blue);

p[3] := plot(data, style = line, color = blue);
display(seq(p[n], n = 2 .. 3));

Hi, I got a doubtI don’tknow how to make the Runge-Kutta’s and Euler’s method. The Runge-Kutta for thesecond and fourth order. I don’t know how to beginning the algoritms, I needsome help. Thanks.

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