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I am studying Numerical Analysis with the book of Richard L.Burden. A question which I'm struggling with right now is following.

Transform the second-order initial-value problem

$y'' - 2y' + 2y = e^{2t}\sin t$ for $0 \leq t \leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$

into a system of first order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.

Then, $$u_1(t) = y(t), u_2(t) = y'(t)$$ $$u_1'(t) = u_2(t)$$ $$u_2'(t) = e^{2t}\sin t - 2u_1(t) + u_2(t)$$ $$u_1(0) = -0.4, u_2(0) = -0.6$$

This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$

I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$

$f_1 = u_1'= u_2(t)$, So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)

However, I can't understand the following. $$k_{2,1} = hf_1(t_0 + \frac{h}{2}, w_{1,0} + \frac{1}{2}k_{1,1}, w_{2,0} + \frac{1}{2}k_{1,2}) = h\left[w_{2,0} + \frac{1}{2}k_{1,2}\right]$$

Why does $f_1(t_0 + \frac{h}{2}, w_{1,0} + \frac{1}{2}k_{1,1}, w_{2,0} + \frac{1}{2}k_{1,2})$ equal to $w_{2,0} + \frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.

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In this problem, I think what you might be confusing is that we have

$$\begin{align} u_1'(t) &= u_2(t) = f_1(t, u_1, u_2) \\ u_2'(t) &= e^{2t} \sin t - 2 u_1(t) + 2 u_2(t) = f_2(t, u_1, u_2) \end{align}$$

From this, we can see that for all the iterations of $j$ on $f_1$, we have

$$f_1(t, u_1, u_2) = f_1(u_2) = w_{2,j} \tag{1}$$

That is, $f_1$ will only ever have $u_2$ terms, which do not depend on $t$ explicitly or $u_1$, thus the iteration formula reduces to

$$k_{2,1} = hf_1\left(t_0 + \frac{h}{2}, w_{1,0} + \frac{1}{2}k_{1,1}, w_{2,0} + \frac{1}{2}k_{1,2}\right) = hf_1\left(w_{2,0} + \frac{1}{2}k_{1,2}\right)$$

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  • $\begingroup$ Why does f_1 depend only on u_2? Does the formula u_2(t) = f_1(t, u_1, u_2) mean that f_1 depends only on u_2? $\endgroup$ – James Nov 20 '18 at 0:48
  • $\begingroup$ Yes, because $f_1 = u_2$, there is nothing like $f_1 = e^t(u_1 + u_2)$, for example, in which case $f_1$ would depend on all three, namely $t, u_1 ~ \text{and}~ u_2$. In this problem, we have $f_1 = u_2$ only - it is how it was setup in the very beginning. Clear? $\endgroup$ – Moo Nov 20 '18 at 1:12
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    $\begingroup$ I got it. Thanks for answering $\endgroup$ – James Nov 20 '18 at 2:15

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