Rectification of vector field

Question

The following is from page 91 of Arnold's Dynamical System I. Can you please explain it to me?

$$ \frac{\partial h}{\partial x} = E + th_1 + \cdots $$

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I would think that

$$ \frac{\partial h}{\partial x} = E + t \frac{\partial h_1}{\partial x}+ \cdots $$

but it doesn't match what follows:

Suppose given an equation $\dot{x} = v(t,x)$. We shall seek a rectifying diffeomorphism $h=h(t,x)$ for which $y=x$ when $t=0$ (time is not changed). From the condition $\dot{y}=0$ we obtain for $h$ the equation $\partial y/\partial t+(\partial y/\partial x)v \equiv 0$. We expand $v$ and $h$ in series of powers of $t$: $$ h=h_0 + t h_1 + \cdots,\quad v = v_0+t v_1 + \cdots. $$ Then $h_0(x) \equiv x$, and so $\partial h /\partial x = E + t h_1 + \cdots$. We then substitute the series (...)

Answer 1

First: the quote is from Arnold's Ordinary Differential Equations, not from Dynamical Systems I. Regardless, you're right: there is a typo in that line, it should read $\partial h/\partial x = E + t \partial h_1 /\partial x + \cdots$. Therefore, the equations obtained when expanded in orders of $t$ are $h_1 + v_0 = 0$ and $2 h_2 + v_1 + v_0\, \partial h_1/\partial x = 0$. Hence, the sentence below makes more sense (and is generally true for any method involving solving differential equations by substituting series):

The equation for $h_k$ contains, besides $h_k$, only the derivatives of the $h_m$ with smaller indices. Therefore we can sequentially (''recursively'') find first $h_1$, then $h_2$ and hence all the terms of the series being sought.