An autonomous system of differential equations
$$
\dot x= f(x),\quad x\in\Omega\subseteq\mathbb R^n,\quad f\in C^n(\Omega)
$$
can be transformed to the form of a n-th order differential equation iff there is a $C^{n}(\Omega)$ function
$\varphi:\; \mathbb R^n\to\mathbb R$ such that the system of functions
$$
\Phi(x)=(\varphi(x),\mathbf F \varphi(x),\ldots,\mathbf F^{n-1} \varphi(x))
$$
is functionally independent on $\Omega$. Here
$$
\mathbf F \varphi(x)=\sum_{i=1}^{n} f_i(x)\frac{\partial \varphi}{\partial x_i},
\;\mathbf F^2 \varphi(x)= \mathbf F( \mathbf F\varphi(x)),\;\ldots,\;
\mathbf F^{n-1} \varphi(x)= \underbrace{\mathbf F(\ldots \mathbf F(\mathbf F}_{n-1}\varphi(x))\ldots).
$$
The mapping $\mathbf F \varphi(x)$ (or $L_f \varphi(x)$) is sometimes called the Lie derivative and sometimes the derivative along the trajectories of the system (depending on the field of science involved).
If such a function $\varphi(x)$ is known, then the equivalent differential equation is
$$\tag{1}
y^{(n)}=\mathbf F^n \varphi(x){\Large|}_{x=\Phi^{-1}(\bar y)},\quad
\bar y= (y,\dot y,\ldots,y^{(n-1)}).
$$
It can be obtained by using the change of variables
$$
y= \varphi(x),\;\dot y= \mathbf F \varphi(x),\;\ddot y= \mathbf F^2 \varphi(x),\;\ldots\; y^{(n-1)}=\mathbf F^{n-1} \varphi(x).
$$
For example, consider the Rossler system
$$
\left\{
\begin{array}{rcl}
\dot x_1&=&-x_2-x_3,\\
\dot x_2&=&x_1+ax_2,\\
\dot x_3&=&x_1x_3-bx_3+c.
\end{array}
\right.
$$
If we guess that we can choose $\varphi(x)=x_2$, then we can get the differential equation. The change of variables is
$$
y=x_2,
$$
$$
\dot y= \mathbf F y= \mathbf F x_2=
(-x_2-x_3)\cdot 0+ (x_1+ax_2)\cdot1+ (x_1x_3-bx_3+c)\cdot0=
x_1+ax_2,
$$
$$
\ddot y= \mathbf F^2 x_2= \mathbf F (x_1+ax_2)= ax_1+(a^2-1)x_2-x_3;
$$
we can also express $x_1,x_2,x_3$ in terms of $y,\dot y,\ddot y$:
$$\tag{2}
x_1=\dot y-ay,\quad x_2=y,\quad x_3=-\ddot y+a\dot y-y.
$$
Finally, we can obtain the right part of the differential equation (1):
$$
\mathbf F^3 x_2= \mathbf F(ax_1+(a^2-1)x_2-x_3)
$$
$$
=a(-x_2-x_3)+(a^2-1)(x_1+ax_2)-(x_1x_3-bx_3+c).
$$
Applying (2) to express $x_1,x_2,x_3$ in terms of $y,\dot y,\ddot y$ , we obtain the equation
$$
\dddot y=a\ddot y-\dot y+(\ddot y-a\dot y+y)(\dot y-ay-b)-c.
$$
Best Answer
You have one too many equations there. Everything is a function if $t$, you don't need to make $y_1 = t$. Call $u = y$, $v = y'$. Then it becomes: $$\begin{cases} u' = v \\ v' = \sin(u) - \cos(t) \\ u(0) = y_0, v(0) = v_0\end{cases}.$$