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From my notes:
Consider the following system of first order ODEs for $n$ unknown functions $y_1(t),y_2(t),...,y_n(t)$
\begin{cases} \dot{y_1}=f_1(t,y_1,...,y_n) \\ \dot{y_2}=f_2(t,y_1,...,y_n) \end{cases}

We can write this in vector notation:
$\dot{\underline{y}}=\underline{f}(t,\underline{y})$ where $\underline{y}=\begin{pmatrix}y_1\\y_2\end{pmatrix}$ and $\underline{f}=\begin{pmatrix}f_1(t,y_1,y_2)\\f_2(t,y_1,y_2)\end{pmatrix}$

Now, I can see how $\dot{\underline{y}}=\begin{pmatrix}\dot{y_1}\\\dot{y_2}\end{pmatrix}$
Please explain how $\underline{f}(t,\underline{y})=\begin{cases} f_1(t,y_1,...,y_n)\\f_2(t,y_1,...,y_n) \end{cases}$

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  • $\begingroup$ Denote $ y = x_{1} $, $ y' = x_{2} $ , $ x = x_{3} $ , $ x' = x_{4} $ and so on. You can then find equalities to simplify e.g $ x_{1} = x_{3} $ etc. You can then represent it as a matrix in the form $ Y' = X_{n} \cdot Y $ $\endgroup$ – Cppg Feb 12 '17 at 21:18
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You "can think of" $f_i(t,y_1,y_2)$ as $f_i(t,y)$, where $y=(y_1,y_2)^T$ with $T$ denoting transposition. This means that you should really use the function $F_i\colon\mathbb R\times \mathbb R^2\to\mathbb R$ defined by $$ F_i(t,y)=f_i(t,y_1,y_2), $$ so that all is completely rigorous (but we never do that, too much complication).

The system $$ \begin{cases} \dot{y_1}=f_1(t,y_1,...,y_n), \\ \dot{y_2}=f_2(t,y_1,...,y_n) \end{cases} $$ is equivalent to $$ \dot y=F(t,y),\quad \text{where} \ F(t,y)=\begin{pmatrix}F_1(t,y)\\F_2(t,y)\end{pmatrix}. $$ Almost the same as in your notes, but not quite.

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You seem to be mixing the case for arbitrary $n$ with $n=2$. For arbitrary $n$, you have equations $$\dot{y}_1 = f_1(t,y_1, \dots, y_n)\\\dot{y}_2 = f_2(t,y_1, \dots, y_n)\\\vdots\\\dot{y}_n = f_n(t,y_1, \dots, y_n)$$ (I dropped the "cases" bracket, because these are not cases. It is just a list of related equations, and the "cases" bracket was only there because someone wanted to group them together visually.)

Also $$\underline y :=\begin{pmatrix}y_1\\y_2\\\vdots\\y_n\end{pmatrix}$$ and we can first use this to simplify the notation for the $f_i$: $$f_i(t, \underline y) := f_i(t, y_1, y_2, \dots, y_n)$$ And $\underline f$ is defined by $$\underline f(t,\underline y) :=\begin{pmatrix}f_1(t, \underline y)\\f_2(t, \underline y)\\\vdots\\f_n(t, \underline y)\end{pmatrix}$$

Now, $$\underline{\dot y} =\begin{pmatrix}\dot y_1\\\dot y_2\\\vdots\\\dot y_n\end{pmatrix} = \begin{pmatrix}f_1(t, \underline y)\\f_2(t, \underline y)\\\vdots\\f_n(t, \underline y)\end{pmatrix} = \underline f(t,\underline y)$$

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