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Runge–Kutta Method (4th Order, 2nd-Order ODE)

TOP/Numerical Analysis/Ordinary differential equations
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Find the solution y=f(x) of the second-order ordinary differential equation y’’=F(x,y,y’) using the 4th-order Runge–Kutta method. Starting from y0=f(x0), y’0=f’(x0), calculate the solution for x0≦x≦xn.
\(\normalsize y''=F(x,y,y')\hspace{30px} y_0=f(x_0),\ y'_0=f'(x_0) \rightarrow\ y=f(x)\\\)
F(x,y,p(=y'))
x0
Initial conditions
y0
=f(x0) Initial conditions
y'0=p0
=f'(x0) Initial conditions
xn
x0≦x≦xn
[ n]Number of segments
\(\normalsize Runge-Kutta\ method\\ (1)\ y''=F(x,y,y')\hspace{30px} p=y',\ y_0=f(x_0),\ y'_0=f'(x_0) \rightarrow\ y=f(x)\\ (2)\ p_{n+1}=p_n+{\large\frac{1}{6}}(j_1+2j_2+2j_3+j_4)+{\small O}(h^5)\\\\ \hspace{25px} y_{n+1}=y_n+{\large\frac{1}{6}}(k_1+2k_2+2k_3+k_4)+{\small O}(h^5)\\\\ \hspace{25px} j_1=hF(x_n,\ y_n,\ p_n)\\ \hspace{25px} k_1=h \cdot p_n \\ \hspace{25px} j_2=hF(x_n+{\large\frac{h}{2}},\ y_n+{\large\frac{k_1}{2}},\ p_n+{\large\frac{j_1}{2}})\\ \hspace{25px} k_2=h ( p_n + {\large\frac{j_1}{2}} )\\ \hspace{25px} j_3=hF(x_n+{\large\frac{h}{2}},\ y_n+{\large\frac{k_2}{2}},\ p_n+{\large\frac{j_2}{2}})\\ \hspace{25px} k_3=h ( p_n + {\large\frac{j_2}{2}} )\\ \hspace{25px} j_4=hF(x_n+h,\ y_n+k_3,\ p_n+j_3)\\ \hspace{25px} k_4=h ( p_n + j_3)\\ \)

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