Question

Use Runge-Kutta’s method of order two to determine approximate solutions for y(0;1) and y(0;2)

y′ = f(t,y) = ycos(t) y(0) = 2

Answer 1

$$y'=f(t, y)=y\cos t$$

$$t_0=0, y_0=2$$

$h=0.1$

$$y_{n+1}=y_n+k_2+O(h^3)$$

$$k_1=hf(t_n,y_n)$$

$$k_2=hf(t_n+{h \over 2},y_n+{k_1 \over 2})$$

$$n=0, t_0=0, y_0=2,$$

$$f(t_0, y_0)=2, k_1=0.2,$$

$$k_2=0.1(2+{0.2 \over 2})\cos(0+{0.1 \over 2})=0.21\cos(0.05)$$

$$y_1=2+0.21\cos(0.05)\approx2.2097$$

$$n=1, t_1=0.1, y_1=2.2097,$$

$$f(t_1, y_1)=2.1987, k_1=0.21987,$$

$$k_2=0.1(2.2097+{0.21987 \over 2})\cos(0.1+{0.1 \over 2})$$

$$=0.22936$$

$$y_2=2.2097+0.22936\approx2.4391$$

$...$

Complete the table

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} n& t_n & y_n & f(t_n,y_n) & y_{n+1} \\ \hline 0 & 0 & 2 & 2 & 2.2097\\ \hdashline 1 & 0.1 & 2.2097 & 2.1987 & 2.4391 \\ \hdashline 2 & 0.2 & 2.4391 & 2.3905 & 2.6870\\ \hdashline 3 & 0.3 & 2.6870 & 2.5670 & 2.9515 \\ \hdashline 4 & 0.4 & 2.9515 & 2.7185 & 3.2295\\ \hdashline 5 & 0.5 & 3.2295 & 2.8341 & 3.5169 \\ \hdashline 6 & 0.6 & 3.5169 & 2.9026 & 3.8084 \\ \hdashline 7 & 0.7 & 3.8084 & 2.9128 & 4.0977 \\ \hdashline 8 & 0.8 & 4.0977 & 2.8549 & 4.3776 \\ \hdashline 9 & 0.9 & 4.3776 & 2.7212 & 4.6401 \\ \hdashline 10 & 4.6401 & \\ \hdashline \end{array}$$