Transform the differential equation $$ \begin{cases} u'''(t) = \sin(u''(t)) - u^2(t), & t > 0, \\ u^{(i)} = u_i \quad \text{for } i \in \{0, 1, 2\} \end{cases} $$ into an equivalent system of first order ordinary differential equations.
I know how to carry out this procedure for linear systems but I am stuck because of the $\sin(u''(t))$ term.
For far I tried setting $x :\equiv u''$ und $y :\equiv u'$ and then substituting into the equation to obtain multiple differential equations in $x$ and $y$ but don't know how to simplify them to obtain the desired result: $$ \begin{cases} y'(t) = x(t), \\ x'(t) = \sin(x(t)) - u^2(t), \\ y''(t) = \sin(y'(t)) - u^2(t). \end{cases} $$ For context: This is the first part of a two part question where in the second part one is asked to find the iterative rule for Eulers method.
Givn the equation
$u'''(t) = \sin (u''(t)) - (u(t))^2, \tag 1$
we set
$v(t) = u'(t), \tag 2$
and
$w(t) = v'(t) = u''(t); \tag 3$
so that
$w'(t) = v''(t) = u'''(t); \tag 4$
then the equation (1) may be written
$w'(t) = \sin (w(t)) - (u(t))^2, \tag 5$
which together with
$u'(t) = v(t), \tag 6$
$v'(t) = w(t), \tag 7$
becomes a first-order system in the three variables $u$, $v$, $w$.
We may in fact "unravel" (5)-(7) to recover (1), simply by substituting (3) and (4) back into (5).
Thus, equation (1) is equivalent to the system (5)-(7).