I have equations\begin{align} 7z'(x) &= x^2 + y(x)^2 + z(x)^2 \\ y''(x) &= -(y'(x) + 7y(x))\sin(z(x)) \end{align}where $$(y(0),y'(0),z(0))=(1.5,-2.6,0.5)$$
The task is to solve this problem using a computer. Can someone help with explaining what goes in $u'(x)$? Thanks.
I am guessing $u(x) = (y'',z')$.
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$\begingroup$Set $y'(x) = u(x)$. Therefore $y''(x) = u'(x).$
Accordingly, the equation:
$$y''(x) = -(y'(x) + 7y(x))\sin(z(x)),$$
is transformed to:
$$u'(x) = -(u(x) + 7y(x))\sin(z(x)).$$
Finally, you get a set of first-order ODE:
$$\begin{cases} y'(x) & = u(x) \\ u'(x) & = -(u(x) + 7y(x))\sin(z(x))\\ z'(x) & = \displaystyle\frac{1}{7}\left(x^2 + y(x)^2 + z(x)^2\right) \end{cases}.$$
Initial condition is just $(y(0),u(0),z(0))=(1.5,-2.6,0.5).$
$\endgroup$ $\begingroup$For the example of the book:$$y''=y'\cos x +2y$$
Substitute $u_1=y , u_2=y'$$$\implies u'_1=y'=u_2 $$And,$$ u'_2=y''=2u_1 + u_2\cos x$$With matrix notation we have:$$\pmatrix {u_1 \\ u_2 }'=\pmatrix {0 & 1 \\ 2 & \cos x}\pmatrix {u_1 \\ u_2 }$$$$\pmatrix {u_1 \\ u_2 }'=\pmatrix {u_2 \\ 2u_1 + u_2\cos x}$$Or if you prefer with $u=\pmatrix {u_1 \\ u_2 }$$$u'=\pmatrix {0 & 1 \\ 2 & \cos x}u$$
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