The vector version of the iterative formula for the Runge–Kutta method isxn+1 = xn + h/6(k1+ 2k2 + 2k3 + k4/), where the vectors k1, k2, k3, and k4 are defined as follows: k1 = f(tn, xn),k2 = f(tn + 1/2 h, xn + 1/2hk1),k3 = f(tn + 1/2 h, xn + 1/2hk2), k4 = f(tn + h, xn + hk3), Explain scalar notation the Runge–Kutta method for the two-dimensionalinitial value problem x' = f (t, x, y), x(t0) = x0;y' = g(t, x, y), y(t0) = y0;
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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
The vector version of the iterative formula for the Runge–Kutta method is
xn+1 = xn + h/6(k1+ 2k2 + 2k3 + k4/),
where the vectors k1, k2, k3, and k4 are defined as follows:
k1 = f(tn, xn),
k2 = f(tn + 1/2 h, xn + 1/2hk1),
k3 = f(tn + 1/2 h, xn + 1/2hk2),
k4 = f(tn + h, xn + hk3),
Explain scalar notation the Runge–Kutta method for the two-dimensional
initial value problem
x' = f (t, x, y), x(t0) = x0;
y' = g(t, x, y), y(t0) = y0;
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